checkpoint 1

concepts/em/u10/e10-1-conductors.tex

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+\subsection{Conductors in Electrostatic Equilibrium}
+
+This subsection summarizes the standard properties of conductors in electrostatics and shows how to apply them in AP-style reasoning.
+
+\dfn{Electrostatic equilibrium in a conductor}{Let a conductor contain mobile charges, and let $\vec{E}$ denote the electric field inside the conducting material. The conductor is in \emph{electrostatic equilibrium} when the free charges have redistributed so that there is no sustained macroscopic charge motion. In an ideal conductor, this requires
+\[
+\vec{E}=\vec{0}
+\]
+everywhere inside the conducting material. Otherwise a nonzero electric field would exert a force on the free charges and they would continue to move.}
+
+\nt{Free charges in a conductor move in response to any interior electric field. Electrons drift until their rearranged surface charges create an induced field that cancels the original field inside the metal. Equilibrium therefore requires no tangential field along the surface and no field anywhere inside the conducting material. If either existed, the charges would keep moving, so the situation would not be electrostatic.}
+
+\ex{Illustrative example}{A neutral metal sphere is placed in an initially uniform external electric field directed to the right. At first the sphere's free electrons move to the left, leaving an induced positive region on the right surface and an induced negative region on the left surface. This redistribution continues until the induced field cancels the external field everywhere inside the metal. In the final equilibrium state,
+\[
+\vec{E}=\vec{0}
+\]
+inside the sphere, and the electric field just outside the surface is perpendicular to the surface.}
+
+\mprop{Standard properties of conductors in electrostatic equilibrium}{Let a conductor be in electrostatic equilibrium. Let $A$ and $B$ be any two points in the conducting material. Let $\hat{n}$ denote the outward unit normal just outside the surface, and let $\sigma$ denote the surface charge density at that point on the surface. Then:
+\begin{enumerate}[label=(\alph*)]
+\item The electric field inside the conducting material is zero:
+\[
+\vec{E}_{\mathrm{inside}}=\vec{0}.
+\]
+\item The conductor is an equipotential, so
+\[
+V_A=V_B.
+\]
+In particular, the entire surface of a connected conductor is at one potential.
+\item Any excess charge placed on an isolated conductor resides on its surface rather than in the interior bulk material.
+\item The electric field just outside the surface is perpendicular to the surface. Its tangential component is zero; otherwise surface charges would move.
+\item For an ideal conductor, the field just outside satisfies
+\[
+\vec{E}_{\mathrm{outside}}=\frac{\sigma}{\varepsilon_0}\hat{n},
+\]
+because the field just inside is zero.
+\end{enumerate}}
+
+\qs{Worked AP-style problem}{An isolated solid conducting sphere has radius
+\[
+R=0.20\,\mathrm{m}
+\]
+and net charge
+\[
+Q=+6.0\times 10^{-9}\,\mathrm{C}.
+\]
+Let point $A$ be at the center of the sphere, let point $B$ be just inside the metal surface, and let point $C$ be just outside the surface. Let $\hat{n}$ denote the outward unit normal at point $C$. Take
+\[
+k=8.99\times 10^9\,\mathrm{N\,m^2/C^2}
+\]
+and proton charge
+\[
+q_p=+1.60\times 10^{-19}\,\mathrm{C}.
+\]
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the electric field at $A$ and at $B$,
+\item the magnitude and direction of the electric field at $C$,
+\item the potential difference $V_A-V_B$, and
+\item the magnitude and direction of the electric force on a proton placed at $C$.
+\end{enumerate}}
+
+\sol Because the sphere is a conductor in electrostatic equilibrium, the electric field everywhere inside the conducting material is zero. Therefore,
+\[
+\vec{E}(A)=\vec{0},
+\qquad
+\vec{E}(B)=\vec{0}.
+\]
+
+For a charged conducting sphere, the external field is the same as that of a point charge $Q$ located at the center. Point $C$ is just outside the surface, so its distance from the center is essentially
+\[
+r=R=0.20\,\mathrm{m}.
+\]
+Thus the field magnitude there is
+\[
+E(C)=k\frac{Q}{R^2}.
+\]
+Substitute the given values:
+\[
+E(C)=\left(8.99\times 10^9\right)\frac{6.0\times 10^{-9}}{(0.20)^2}\,\mathrm{N/C}.
+\]
+Since
+\[
+(0.20)^2=0.040,
+\]
+we get
+\[
+E(C)=\frac{53.94}{0.040}\,\mathrm{N/C}=1.35\times 10^3\,\mathrm{N/C}.
+\]
+Because $Q$ is positive, the field points radially outward, perpendicular to the surface. So
+\[
+\vec{E}(C)=(1.35\times 10^3\,\mathrm{N/C})\hat{n},
+\]
+where $\hat{n}$ is the outward normal.
+
+Next, the entire conductor is an equipotential in electrostatic equilibrium. Since both $A$ and $B$ lie in the conductor,
+\[
+V_A=V_B.
+\]
+Therefore,
+\[
+V_A-V_B=0\,\mathrm{V}.
+\]
+
+Finally, the electric force on a proton at $C$ has magnitude
+\[
+F=q_pE(C).
+\]
+Substitute the values:
+\[
+F=\left(1.60\times 10^{-19}\,\mathrm{C}\right)\left(1.35\times 10^3\,\mathrm{N/C}\right).
+\]
+This gives
+\[
+F=2.16\times 10^{-16}\,\mathrm{N}.
+\]
+Because the proton has positive charge, its force is in the same direction as $\vec{E}$, so the force is radially outward:
+\[
+\vec{F}=(2.16\times 10^{-16}\,\mathrm{N})\hat{n}.
+\]
+
+Therefore,
+\[
+\vec{E}(A)=\vec{0},
+\qquad
+\vec{E}(B)=\vec{0},
+\]
+\[
+\vec{E}(C)=(1.35\times 10^3\,\mathrm{N/C})\hat{n},
+\]
+\[
+V_A-V_B=0\,\mathrm{V},
+\qquad
+\vec{F}=(2.16\times 10^{-16}\,\mathrm{N})\hat{n}.
+\]

concepts/em/u10/e10-2-charge-redistribution.tex

@@ -0,0 +1,92 @@
+\subsection{Charge Redistribution, Contact, Induction, and Grounding}
+
+This subsection explains how charge moves on conductors, why touching conductors exchange charge until they reach one potential, and how induction plus grounding can leave an object with a net charge without direct contact.
+
+\dfn{Charge redistribution, contact, induction, and grounding}{Let $q_A,q_B,\dots$ denote the net charges on conductors $A,B,\dots$, and let $V_A,V_B,\dots$ denote their electric potentials.
+
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item \emph{Charge redistribution}: mobile charge in a conductor moves through the material until electrostatic equilibrium is reached.
+\item \emph{Charging by contact}: if conductors touch or are connected by a wire, charge can flow between them until the connected conductors reach one common potential.
+\item \emph{Charging by induction}: a nearby charged object causes charges in another object to separate without direct contact.
+\item \emph{Grounding}: connecting an object to Earth allows charge to flow between the object and Earth, which acts as a very large charge reservoir.
+\end{enumerate}}
+
+\nt{Charge redistribution in a conductor continues until electrostatic equilibrium is reached. In electrostatics, that means the electric field inside the conductor is zero and all parts of one connected conductor are at the same potential. Therefore contact problems are usually solved with charge conservation plus an equal-potential idea. For identical small conducting spheres, symmetry makes equal potential equivalent to equal final charge, but for conductors of different size or shape equal potential does not generally mean equal charge.}
+
+\ex{Illustrative example}{Two identical small conducting spheres $A$ and $B$ are far apart initially. Let their initial charges be
+\[
+q_{A,i}=+6.0\,\mathrm{nC},
+\qquad
+q_{B,i}=0.
+\]
+If the spheres are touched together and then separated, charge is conserved and the identical spheres must finish with equal charge. The total charge is
+\[
+q_{\mathrm{tot}}=q_{A,i}+q_{B,i}=+6.0\,\mathrm{nC},
+\]
+so each sphere ends with
+\[
+q_{A,f}=q_{B,f}=\frac{q_{\mathrm{tot}}}{2}=+3.0\,\mathrm{nC}.
+\]
+This is a contact example: the charge does not disappear or appear; it redistributes until the connected conductors reach electrostatic equilibrium.}
+
+\mprop{Practical relations and qualitative rules}{Let $q_{\mathrm{tot}}$ denote the total charge of a chosen isolated system, let $q_{A,i},q_{B,i}$ and $q_{A,f},q_{B,f}$ denote initial and final charges, let $q_{\mathrm{object},f}$ denote the final charge of a grounded object, and let $\Delta q_{\mathrm{Earth}}$ denote the charge change of Earth.
+
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item For any isolated system,
+\[
+q_{\mathrm{tot},f}=q_{\mathrm{tot},i}.
+\]
+
+\item If two identical small conducting spheres touch and then separate,
+\[
+q_{A,f}=q_{B,f}=\frac{q_{A,i}+q_{B,i}}{2}.
+\]
+
+\item Induction without grounding redistributes charge but does not change the net charge of the induced object. For an initially neutral conductor, the near side becomes opposite in sign to the external object, the far side becomes the same sign, and the net charge remains zero.
+
+\item Grounding is charge exchange with Earth. If a positive object is nearby, electrons can flow from Earth onto the grounded conductor. If a negative object is nearby, electrons can flow from the conductor to Earth.
+
+\item In the usual induction-charging sequence, the ground connection is removed first and the external charged object is removed second. Then the conductor is left with a net charge opposite in sign to the inducing object. If the object-Earth system is initially neutral, charge conservation for that system gives
+\[
+q_{\mathrm{object},f}+\Delta q_{\mathrm{Earth}}=0.
+\]
+\end{enumerate}}
+
+\qs{Worked AP-style problem}{A neutral metal sphere $S$ is mounted on an insulating stand. A negatively charged rod is brought near the left side of the sphere but does not touch it. While the rod remains in place, the sphere is briefly connected to Earth by a wire. The grounding wire is then removed, and finally the rod is taken away.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the signs of the induced charges on the left and right sides of the sphere before the grounding wire is attached,
+\item the direction of electron flow while the sphere is grounded,
+\item the net charge left on the sphere after the full sequence, and
+\item whether charge conservation is violated by the sphere ending with a net charge.
+\end{enumerate}}
+
+\sol Before the sphere is grounded, the rod is negative, so it repels mobile electrons in the metal sphere. Those electrons shift toward the right side, farther from the rod.
+
+Therefore, before grounding,
+\begin{itemize}
+\item the left side of the sphere is induced to be positive, and
+\item the right side of the sphere is induced to be negative.
+\end{itemize}
+
+Even though the charges have separated, the sphere is still overall neutral at this stage because no charge has entered or left the sphere.
+
+Now the sphere is connected to Earth while the negative rod remains nearby. The excess electrons on the sphere are repelled by the negative rod, and the grounding wire gives those electrons a path to leave. Thus electrons flow
+\[
+\text{from the sphere to Earth}.
+\]
+
+Next, the grounding wire is removed while the rod is still present. Because the sphere is no longer connected to Earth, the electrons that left cannot return. The sphere has lost some electrons, so it now has a net positive charge.
+
+Finally, the rod is taken away. With the rod gone, the remaining positive charge is no longer pulled to one side, so it redistributes over the outer surface of the sphere. The final result is that the sphere is left
+\[
+\text{positively charged}.
+\]
+
+Charge conservation is not violated. The correct isolated system is the sphere together with Earth. During grounding, electrons moved from the sphere to Earth, so the sphere became positive and Earth gained an equal amount of negative charge. If the final charge on the sphere is $q_{S,f}>0$, then
+\[
+q_{S,f}+\Delta q_{\mathrm{Earth}}=0.
+\]
+
+So the total charge of the combined system is unchanged; the process is charge transfer, not charge creation.

concepts/em/u10/e10-3-capacitance.tex

@@ -0,0 +1,190 @@
+\subsection{Capacitance and Capacitor Geometries}
+
+This subsection defines capacitance, shows how capacitor geometry controls it, and derives the parallel-plate result from Gauss's law and the field-potential relation.
+
+\dfn{Capacitor and capacitance}{A \emph{capacitor} is a system of two conductors that can hold equal and opposite charges. Let the conductors carry charges $+Q$ and $-Q$, and let
+\[
+\Delta V=V_{\text{high}}-V_{\text{low}}
+\]
+denote the magnitude of the potential difference between them. The \emph{capacitance} $C$ of the system is defined by
+\[
+C=\frac{Q}{\Delta V}.
+\]
+Capacitance measures how much charge is stored per unit potential difference. Its SI unit is the farad:
+\[
+1\,\mathrm{F}=1\,\mathrm{C/V}.
+\]}
+
+\thm{Capacitance formula and common geometries}{Let a capacitor carry plate charges $\pm Q$, and let $\Delta V$ be the magnitude of the potential difference between its conductors. Then
+\[
+C=\frac{Q}{\Delta V}.
+\]
+
+For a vacuum parallel-plate capacitor with plate area $A$ and plate separation $d$, define the surface charge density by
+\[
+\sigma=\frac{Q}{A}.
+\]
+Ignoring edge effects, Gauss's law gives the nearly uniform electric field between the plates:
+\[
+\vec{E}=\frac{\sigma}{\varepsilon_0}\,\hat{n},
+\qquad
+E=\frac{\sigma}{\varepsilon_0}.
+\]
+Using the potential-drop relation with a path across the gap,
+\[
+\Delta V=\left| -\int \vec{E}\cdot d\vec{\ell} \right|=Ed,
+\]
+so
+\[
+\Delta V=\frac{\sigma d}{\varepsilon_0}=\frac{Qd}{\varepsilon_0 A}.
+\]
+Substitute into $C=Q/\Delta V$:
+\[
+C=\frac{Q}{Qd/(\varepsilon_0 A)}=\varepsilon_0\frac{A}{d}.
+\]
+Thus, for an ideal vacuum parallel-plate capacitor,
+\[
+C=\varepsilon_0\frac{A}{d}.
+\]
+
+Two other useful vacuum results are
+\[
+C_{\text{spherical}}=4\pi\varepsilon_0\frac{ab}{b-a}
+\]
+for concentric spheres of radii $a$ and $b$ with $b>a$, and
+\[
+C_{\text{cylindrical}}=\frac{2\pi\varepsilon_0 L}{\ln(b/a)}
+\]
+for coaxial cylinders of length $L$ and radii $a$ and $b$ with $b>a$. In every case, capacitance is determined by geometry and the material between the conductors.}
+
+\ex{Illustrative example}{A vacuum parallel-plate capacitor has plate area
+\[
+A=3A_0
+\]
+and separation
+\[
+d=2d_0.
+\]
+If a reference capacitor with area $A_0$ and separation $d_0$ has capacitance $C_0$, then
+\[
+C_0=\varepsilon_0\frac{A_0}{d_0}.
+\]
+For the new capacitor,
+\[
+C=\varepsilon_0\frac{3A_0}{2d_0}=\frac{3}{2}C_0.
+\]
+So increasing plate area increases capacitance, while increasing plate separation decreases it.}
+
+\nt{For an ideal capacitor, $C$ is a property of the physical setup, not of the momentary charge or battery setting. In a vacuum parallel-plate capacitor, changing $A$ or $d$ changes $C$, and inserting a dielectric would also change $C$. But if the same capacitor is connected to a larger battery, then $\Delta V$ increases and $Q$ increases proportionally, so the ratio $Q/\Delta V$ stays the same.}
+
+\qs{Worked AP-style problem}{A vacuum parallel-plate capacitor has plate area
+\[
+A=2.0\times 10^{-2}\,\mathrm{m^2}
+\]
+and plate separation
+\[
+d=1.5\times 10^{-3}\,\mathrm{m}.
+\]
+It is connected to a battery that maintains a potential difference
+\[
+\Delta V=12.0\,\mathrm{V}.
+\]
+Take
+\[
+\varepsilon_0=8.85\times 10^{-12}\,\mathrm{F/m}.
+\]
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the capacitance $C$,
+\item the charge magnitude $Q$ on each plate,
+\item the electric field magnitude $E$ between the plates, and
+\item the surface charge density $\sigma$ on either plate.
+\end{enumerate}}
+
+\sol For part (a), use the parallel-plate formula
+\[
+C=\varepsilon_0\frac{A}{d}.
+\]
+Substitute the given values:
+\[
+C=(8.85\times 10^{-12})\frac{2.0\times 10^{-2}}{1.5\times 10^{-3}}\,\mathrm{F}.
+\]
+First evaluate the geometry factor:
+\[
+\frac{2.0\times 10^{-2}}{1.5\times 10^{-3}}=13.3.
+\]
+So
+\[
+C=(8.85\times 10^{-12})(13.3)\,\mathrm{F}=1.18\times 10^{-10}\,\mathrm{F}.
+\]
+Therefore,
+\[
+C=1.18\times 10^{-10}\,\mathrm{F}.
+\]
+
+For part (b), use the definition of capacitance:
+\[
+Q=C\Delta V.
+\]
+Substitute the capacitance and the battery voltage:
+\[
+Q=(1.18\times 10^{-10}\,\mathrm{F})(12.0\,\mathrm{V}).
+\]
+This gives
+\[
+Q=1.42\times 10^{-9}\,\mathrm{C}.
+\]
+So the plates carry charges $+Q$ and $-Q$ with magnitude
+\[
+Q=1.42\times 10^{-9}\,\mathrm{C}.
+\]
+
+For part (c), the field between ideal parallel plates is approximately uniform, so the potential difference satisfies
+\[
+\Delta V=Ed.
+\]
+Solve for $E$:
+\[
+E=\frac{\Delta V}{d}.
+\]
+Substitute the values:
+\[
+E=\frac{12.0\,\mathrm{V}}{1.5\times 10^{-3}\,\mathrm{m}}=8.0\times 10^3\,\mathrm{V/m}.
+\]
+Since $1\,\mathrm{V/m}=1\,\mathrm{N/C}$,
+\[
+E=8.0\times 10^3\,\mathrm{N/C}.
+\]
+
+For part (d), use
+\[
+\sigma=\frac{Q}{A}.
+\]
+Substitute the charge and area:
+\[
+\sigma=\frac{1.42\times 10^{-9}\,\mathrm{C}}{2.0\times 10^{-2}\,\mathrm{m^2}}.
+\]
+This gives
+\[
+\sigma=7.1\times 10^{-8}\,\mathrm{C/m^2}.
+\]
+
+As a check, the field relation for parallel plates predicts
+\[
+E=\frac{\sigma}{\varepsilon_0}=\frac{7.1\times 10^{-8}}{8.85\times 10^{-12}}\,\mathrm{N/C}
+\approx 8.0\times 10^3\,\mathrm{N/C},
+\]
+which agrees with part (c).
+
+Therefore,
+\[
+C=1.18\times 10^{-10}\,\mathrm{F},
+\qquad
+Q=1.42\times 10^{-9}\,\mathrm{C},
+\]
+\[
+E=8.0\times 10^3\,\mathrm{N/C},
+\qquad
+\sigma=7.1\times 10^{-8}\,\mathrm{C/m^2}.
+\]

concepts/em/u10/e10-4-capacitor-energy.tex

@@ -0,0 +1,172 @@
+\subsection{Energy Stored in Capacitors and Fields}
+
+This subsection develops the standard formulas for capacitor energy and connects them to the electric-field energy density in vacuum.
+
+\dfn{Capacitor energy and the field-energy viewpoint}{Let a capacitor have capacitance $C$, let the magnitude of the charge on either conductor be $Q$, and let
+\[
+\Delta V=V_{\text{high}}-V_{\text{low}}
+\]
+denote the magnitude of the potential difference between the conductors. The \emph{energy stored in the capacitor}, denoted $U_C$, is the electric potential energy gained while the capacitor is charged from $0$ to $Q$.
+
+In the field viewpoint, let $u_E$ denote the \emph{electric-field energy density}, measured in joules per cubic meter. For a vacuum region with electric-field magnitude $E$, the stored energy can be regarded as distributed through the field-filled volume.}
+
+\thm{Equivalent capacitor-energy formulas and vacuum field-energy density}{Let a capacitor have capacitance $C$, plate charges $\pm Q$, and potential-difference magnitude $\Delta V$. Then the stored energy can be written in any of the equivalent forms
+\[
+U_C=\frac12 Q\Delta V=\frac12 C(\Delta V)^2=\frac{Q^2}{2C}.
+\]
+For a vacuum region in which the electric-field magnitude is $E$, the electric-field energy density is
+\[
+u_E=\frac12 \varepsilon_0 E^2.
+\]
+For an ideal vacuum parallel-plate capacitor of plate area $A$ and separation $d$, the field is approximately uniform, so
+\[
+U_C=u_E(Ad).
+\]}
+
+\nt{The energy grows quadratically with charge or voltage because charging is cumulative. For a capacitor with fixed $C$, the potential difference is not constant while it charges: it rises in proportion to the accumulated charge, since $\Delta V=Q/C$. That means later bits of charge are harder to add than earlier ones. The average potential difference during charging is therefore half the final value, which is why $U_C=\tfrac12 Q\Delta V$ and why doubling $Q$ or $\Delta V$ makes the stored energy four times as large for the same capacitor.}
+
+\pf{Short derivation from charging work}{Let $q$ denote the instantaneous charge on the capacitor during a slow charging process from $q=0$ to $q=Q$. At that instant, the potential difference is
+\[
+\Delta V(q)=\frac{q}{C}.
+\]
+To move an additional small charge $dq$ onto the capacitor, the external work required is
+\[
+dU=\Delta V(q)\,dq=\frac{q}{C}\,dq.
+\]
+Integrate from $0$ to $Q$:
+\[
+U_C=\int_0^Q \frac{q}{C}\,dq=\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q=\frac{Q^2}{2C}.
+\]
+Using
+\[
+Q=C\Delta V,
+\]
+this becomes
+\[
+U_C=\frac12 Q\Delta V=\frac12 C(\Delta V)^2.
+\]
+For an ideal vacuum parallel-plate capacitor,
+\[
+C=\varepsilon_0\frac{A}{d}
+\qquad \text{and} \qquad
+E=\frac{\Delta V}{d}.
+\]
+Substituting into $U_C=\tfrac12 C(\Delta V)^2$ gives
+\[
+U_C=\frac12 \left(\varepsilon_0\frac{A}{d}\right)(Ed)^2=\frac12 \varepsilon_0 E^2(Ad).
+\]
+Dividing by the volume $Ad$ yields
+\[
+u_E=\frac{U_C}{Ad}=\frac12 \varepsilon_0 E^2.
+\]}
+
+\qs{Worked AP-style problem}{A vacuum parallel-plate capacitor has plate area
+\[
+A=2.0\times 10^{-2}\,\mathrm{m^2}
+\]
+and plate separation
+\[
+d=1.0\times 10^{-3}\,\mathrm{m}.
+\]
+It is connected to a battery that maintains a potential difference
+\[
+\Delta V=20.0\,\mathrm{V}.
+\]
+Take
+\[
+\varepsilon_0=8.85\times 10^{-12}\,\mathrm{F/m}.
+\]
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the capacitance $C$,
+\item the charge magnitude $Q$ on each plate,
+\item the stored energy $U_C$, and
+\item the electric-field energy density $u_E$ between the plates, then verify that $U_C=u_E(Ad)$.
+\end{enumerate}}
+
+\sol For part (a), use the parallel-plate formula
+\[
+C=\varepsilon_0\frac{A}{d}.
+\]
+Substitute the given values:
+\[
+C=(8.85\times 10^{-12})\frac{2.0\times 10^{-2}}{1.0\times 10^{-3}}\,\mathrm{F}.
+\]
+The geometry factor is
+\[
+\frac{2.0\times 10^{-2}}{1.0\times 10^{-3}}=20.0,
+\]
+so
+\[
+C=(8.85\times 10^{-12})(20.0)\,\mathrm{F}=1.77\times 10^{-10}\,\mathrm{F}.
+\]
+
+For part (b), use
+\[
+Q=C\Delta V.
+\]
+Then
+\[
+Q=(1.77\times 10^{-10}\,\mathrm{F})(20.0\,\mathrm{V})=3.54\times 10^{-9}\,\mathrm{C}.
+\]
+
+For part (c), use any equivalent energy formula. Using $U_C=\tfrac12 C(\Delta V)^2$,
+\[
+U_C=\frac12 (1.77\times 10^{-10})(20.0)^2\,\mathrm{J}.
+\]
+Since
+\[
+(20.0)^2=400,
+\]
+we obtain
+\[
+U_C=\frac12 (1.77\times 10^{-10})(400)\,\mathrm{J}=3.54\times 10^{-8}\,\mathrm{J}.
+\]
+As a check,
+\[
+U_C=\frac12 Q\Delta V=\frac12 (3.54\times 10^{-9})(20.0)\,\mathrm{J}=3.54\times 10^{-8}\,\mathrm{J},
+\]
+which agrees.
+
+For part (d), first find the field magnitude between the plates:
+\[
+E=\frac{\Delta V}{d}=\frac{20.0\,\mathrm{V}}{1.0\times 10^{-3}\,\mathrm{m}}=2.0\times 10^4\,\mathrm{V/m}.
+\]
+Now use the vacuum field-energy density formula:
+\[
+u_E=\frac12 \varepsilon_0 E^2.
+\]
+Substitute the values:
+\[
+u_E=\frac12 (8.85\times 10^{-12})(2.0\times 10^4)^2\,\mathrm{J/m^3}.
+\]
+Because
+\[
+(2.0\times 10^4)^2=4.0\times 10^8,
+\]
+we get
+\[
+u_E=\frac12 (8.85\times 10^{-12})(4.0\times 10^8)\,\mathrm{J/m^3}=1.77\times 10^{-3}\,\mathrm{J/m^3}.
+\]
+
+To verify the field viewpoint, compute the volume between the plates:
+\[
+Ad=(2.0\times 10^{-2})(1.0\times 10^{-3})\,\mathrm{m^3}=2.0\times 10^{-5}\,\mathrm{m^3}.
+\]
+Then
+\[
+u_E(Ad)=(1.77\times 10^{-3})(2.0\times 10^{-5})\,\mathrm{J}=3.54\times 10^{-8}\,\mathrm{J}.
+\]
+This matches the capacitor-energy result.
+
+Therefore,
+\[
+C=1.77\times 10^{-10}\,\mathrm{F},
+\qquad
+Q=3.54\times 10^{-9}\,\mathrm{C},
+\]
+\[
+U_C=3.54\times 10^{-8}\,\mathrm{J},
+\qquad
+u_E=1.77\times 10^{-3}\,\mathrm{J/m^3}.
+\]

concepts/em/u10/e10-5-dielectrics.tex

@@ -0,0 +1,224 @@
+\subsection{Dielectrics and Polarization}
+
+This subsection explains how dielectric materials polarize in an electric field and how that polarization changes capacitor behavior in the two common AP settings: fixed charge and fixed voltage.
+
+\dfn{Dielectric, polarization, and dielectric constant}{A \emph{dielectric} is an insulating material placed between capacitor plates. Its charges are not free to flow through the material as they do in a conductor, but the positive and negative charges within its atoms or molecules can shift slightly.
+
+This internal charge separation is called \emph{polarization}. In a polarized dielectric, the side nearer the positive plate becomes slightly negative and the side nearer the negative plate becomes slightly positive, so the dielectric produces an induced electric field that opposes the original field between the plates.
+
+Let $C_0$ denote the capacitance of a capacitor when the gap is vacuum or air, and let $C$ denote the capacitance when a dielectric fully fills the gap. The \emph{dielectric constant} $\kappa$ of the material is the factor by which the capacitance increases:
+\[
+C=\kappa C_0,
+\qquad
+\kappa>1.
+\]
+For an ideal parallel-plate capacitor of plate area $A$ and separation $d$,
+\[
+C_0=\varepsilon_0\frac{A}{d}
+\qquad \text{and therefore} \qquad
+C=\kappa\varepsilon_0\frac{A}{d}.
+\]}
+
+\nt{The dielectric does not cancel the capacitor's field completely. Instead, polarization creates \emph{bound} charges on the dielectric surfaces, and those bound charges produce a field opposite the field from the capacitor plates. That opposition reduces the net interior field.
+
+The key AP distinction is what stays fixed during insertion.
+\begin{itemize}
+\item If the capacitor is isolated after charging, then the free charge on the plates cannot change, so this is a fixed-$Q$ situation. The dielectric lowers the field and voltage.
+\item If the capacitor remains connected to a battery, then the battery keeps the potential difference fixed, so this is a fixed-$V$ situation. The dielectric still tends to reduce the field, but the battery pushes additional charge onto the plates until the original voltage is restored.
+\end{itemize}}
+
+\ex{Illustrative example}{An isolated capacitor has initial capacitance
+\[
+C_0=10.0\,\mathrm{pF}
+\]
+and is charged to
+\[
+V_0=30.0\,\mathrm{V}.
+\]
+The battery is disconnected, and a dielectric with
+\[
+\kappa=2.5
+\]
+is inserted so that it fully fills the gap.
+
+The new capacitance is
+\[
+C=\kappa C_0=(2.5)(10.0\,\mathrm{pF})=25.0\,\mathrm{pF}.
+\]
+Because the capacitor is isolated, the charge stays constant:
+\[
+Q=Q_0=C_0V_0=(10.0\,\mathrm{pF})(30.0\,\mathrm{V})=300\,\mathrm{pC}.
+\]
+The new voltage is therefore
+\[
+V=\frac{Q}{C}=\frac{300\,\mathrm{pC}}{25.0\,\mathrm{pF}}=12.0\,\mathrm{V}.
+\]
+So inserting the dielectric increases the capacitance while reducing the voltage for the same stored charge.}
+
+\mprop{Main capacitor relations for a dielectric that completely fills the gap}{Let $C_0$, $Q_0$, $V_0$, $E_0$, and $U_0$ denote the capacitance, plate-charge magnitude, potential difference, electric-field magnitude, and stored energy before insertion. Let $C$, $Q$, $V$, $E$, and $U$ denote the corresponding quantities after a dielectric of constant $\kappa$ fully fills the gap.
+
+For an ideal parallel-plate capacitor,
+\[
+C=\kappa C_0=\kappa\varepsilon_0\frac{A}{d}.
+\]
+
+If the capacitor is \emph{isolated} after charging (fixed $Q$), then
+\[
+Q=Q_0,
+\qquad
+V=\frac{V_0}{\kappa},
+\qquad
+E=\frac{E_0}{\kappa},
+\qquad
+U=\frac{U_0}{\kappa}.
+\]
+
+If the capacitor remains \emph{connected to a battery} (fixed $V$), then
+\[
+V=V_0,
+\qquad
+Q=\kappa Q_0,
+\qquad
+E=E_0,
+\qquad
+U=\kappa U_0.
+\]
+
+In both cases, the dielectric increases capacitance by the factor $\kappa$. What changes is which other quantity the circuit constraint forces to stay constant.}
+
+\qs{Worked AP-style problem}{A vacuum parallel-plate capacitor has plate area
+\[
+A=1.50\times 10^{-2}\,\mathrm{m^2}
+\]
+and plate separation
+\[
+d=1.00\times 10^{-3}\,\mathrm{m}.
+\]
+It remains connected to a battery that maintains a potential difference
+\[
+V_0=12.0\,\mathrm{V}.
+\]
+A dielectric with dielectric constant
+\[
+\kappa=3.00
+\]
+is inserted so that it completely fills the gap. Take
+\[
+\varepsilon_0=8.85\times 10^{-12}\,\mathrm{F/m}.
+\]
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the initial capacitance $C_0$,
+\item the final capacitance $C$,
+\item the initial and final plate-charge magnitudes $Q_0$ and $Q$,
+\item the electric-field magnitude between the plates before and after insertion, and
+\item the initial and final stored energies $U_0$ and $U$.
+\end{enumerate}}
+
+\sol Because the dielectric fully fills the gap, the capacitance increases by the factor $\kappa$:
+\[
+C=\kappa C_0.
+\]
+Because the battery remains connected, the voltage stays fixed at
+\[
+V=V_0=12.0\,\mathrm{V}.
+\]
+
+For part (a), first find the initial vacuum capacitance:
+\[
+C_0=\varepsilon_0\frac{A}{d}.
+\]
+Substitute the given values:
+\[
+C_0=(8.85\times 10^{-12})\frac{1.50\times 10^{-2}}{1.00\times 10^{-3}}\,\mathrm{F}.
+\]
+The geometry factor is
+\[
+\frac{1.50\times 10^{-2}}{1.00\times 10^{-3}}=15.0,
+\]
+so
+\[
+C_0=(8.85\times 10^{-12})(15.0)\,\mathrm{F}=1.33\times 10^{-10}\,\mathrm{F}.
+\]
+
+For part (b), multiply by $\kappa=3.00$:
+\[
+C=\kappa C_0=(3.00)(1.33\times 10^{-10}\,\mathrm{F})=3.98\times 10^{-10}\,\mathrm{F}.
+\]
+
+For part (c), before insertion the plate-charge magnitude is
+\[
+Q_0=C_0V_0.
+\]
+Thus,
+\[
+Q_0=(1.33\times 10^{-10}\,\mathrm{F})(12.0\,\mathrm{V})=1.59\times 10^{-9}\,\mathrm{C}.
+\]
+After insertion, the voltage is unchanged but the capacitance is larger, so
+\[
+Q=CV=(3.98\times 10^{-10}\,\mathrm{F})(12.0\,\mathrm{V})=4.78\times 10^{-9}\,\mathrm{C}.
+\]
+This is also consistent with
+\[
+Q=\kappa Q_0=(3.00)(1.59\times 10^{-9}\,\mathrm{C})=4.78\times 10^{-9}\,\mathrm{C}.
+\]
+
+For part (d), the electric-field magnitude in an ideal parallel-plate capacitor is related to the maintained potential difference by
+\[
+E=\frac{V}{d}.
+\]
+Before insertion,
+\[
+E_0=\frac{V_0}{d}=\frac{12.0\,\mathrm{V}}{1.00\times 10^{-3}\,\mathrm{m}}=1.20\times 10^4\,\mathrm{V/m}.
+\]
+Because the battery keeps the same voltage across the same plate spacing, the final field magnitude is the same:
+\[
+E=\frac{V}{d}=\frac{12.0\,\mathrm{V}}{1.00\times 10^{-3}\,\mathrm{m}}=1.20\times 10^4\,\mathrm{V/m}.
+\]
+
+For part (e), the initial stored energy is
+\[
+U_0=\frac12 C_0V_0^2.
+\]
+Substitute the values:
+\[
+U_0=\frac12 (1.33\times 10^{-10})(12.0)^2\,\mathrm{J}.
+\]
+Since
+\[
+(12.0)^2=144,
+\]
+we get
+\[
+U_0=\frac12 (1.33\times 10^{-10})(144)\,\mathrm{J}=9.56\times 10^{-9}\,\mathrm{J}.
+\]
+After insertion,
+\[
+U=\frac12 CV^2=\frac12 (3.98\times 10^{-10})(12.0)^2\,\mathrm{J}=2.87\times 10^{-8}\,\mathrm{J}.
+\]
+Equivalently,
+\[
+U=\kappa U_0=(3.00)(9.56\times 10^{-9}\,\mathrm{J})=2.87\times 10^{-8}\,\mathrm{J}.
+\]
+
+Therefore,
+\[
+C_0=1.33\times 10^{-10}\,\mathrm{F},
+\qquad
+C=3.98\times 10^{-10}\,\mathrm{F},
+\]
+\[
+Q_0=1.59\times 10^{-9}\,\mathrm{C},
+\qquad
+Q=4.78\times 10^{-9}\,\mathrm{C},
+\]
+\[
+E_0=E=1.20\times 10^4\,\mathrm{V/m},
+\]
+and
+\[
+U_0=9.56\times 10^{-9}\,\mathrm{J},
+\qquad
+U=2.87\times 10^{-8}\,\mathrm{J}.
+\]

concepts/em/u11/e11-1-current-density.tex

@@ -0,0 +1,128 @@
+\subsection{Current, Drift Velocity, and Current Density}
+
+This subsection connects the macroscopic current $I$ in a wire to the microscopic motion of charge carriers through the current density $\vec{J}$ and the drift velocity $\vec{v}_d$.
+
+\dfn{Current, drift velocity, and current density}{Let $\Delta q$ denote the net charge that crosses a chosen surface in a time interval $\Delta t$. The \emph{electric current} through that surface is
+\[
+I=\frac{\Delta q}{\Delta t}
+\]
+in the limit of very small time intervals.
+
+Let $\vec{v}_d$ denote the average drift velocity of the charge carriers, let $n$ denote the number of carriers per unit volume, and let $q$ denote the charge of each carrier. The \emph{current density} $\vec{J}$ is the vector that describes current per unit area, with direction defined by the motion of positive charge. For an oriented surface element $d\vec{A}$,
+\[
+dI=\vec{J}\cdot d\vec{A}.
+\]
+Thus $\vec{J}$ links the local flow of charge to the total current through a cross section.}
+
+\nt{Conventional current is defined to point in the direction that positive charges would move. Therefore $\vec{J}$ points in the conventional-current direction. In a metal wire, the mobile charge carriers are electrons, so $q=-e$ and the electron drift velocity $\vec{v}_d$ points opposite to $\vec{J}$. If the carriers were positive instead, then $\vec{v}_d$ and $\vec{J}$ would point in the same direction.}
+
+\mprop{Microscopic-macroscopic current relations}{Let $n$ denote the carrier number density, let $q$ denote the charge of each carrier, let $\vec{v}_d$ denote the drift velocity, and let $S$ be a surface with oriented area element $d\vec{A}$. Then the current density is
+\[
+\vec{J}=nq\vec{v}_d.
+\]
+The total current through $S$ is
+\[
+I=\iint_S \vec{J}\cdot d\vec{A}.
+\]
+If $\vec{J}$ is uniform across a flat cross section of area $A$ and parallel to the area normal, then
+\[
+I=JA
+\qquad \text{and} \qquad
+J=\frac{I}{A}.
+\]
+For a straight wire with uniform carrier density, the corresponding magnitude relation is
+\[
+I=n|q|Av_d,
+\]
+where $v_d=|\vec{v}_d|$.}
+
+\qs{Worked AP-style problem}{A long copper wire carries a steady current
+\[
+I=3.0\,\mathrm{A}
+\]
+to the right. The wire has radius
+\[
+r=0.80\,\mathrm{mm}=8.0\times 10^{-4}\,\mathrm{m},
+\]
+so its cross-sectional area is $A=\pi r^2$. Assume the conduction-electron number density is
+\[
+n=8.5\times 10^{28}\,\mathrm{m^{-3}},
+\]
+and the charge of each electron is
+\[
+q_e=-1.60\times 10^{-19}\,\mathrm{C}.
+\]
+Let $+\hat{\imath}$ point to the right.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the current density magnitude $J$,
+\item the current density vector $\vec{J}$, and
+\item the electron drift velocity vector $\vec{v}_d$.
+\end{enumerate}}
+
+\sol First compute the wire's cross-sectional area:
+\[
+A=\pi r^2=\pi(8.0\times 10^{-4}\,\mathrm{m})^2.
+\]
+Since
+\[
+(8.0\times 10^{-4})^2=6.4\times 10^{-7},
+\]
+we get
+\[
+A=\pi(6.4\times 10^{-7})\,\mathrm{m^2}=2.01\times 10^{-6}\,\mathrm{m^2}.
+\]
+
+For part (a), use
+\[
+J=\frac{I}{A}.
+\]
+Then
+\[
+J=\frac{3.0\,\mathrm{A}}{2.01\times 10^{-6}\,\mathrm{m^2}}=1.49\times 10^6\,\mathrm{A/m^2}.
+\]
+
+For part (b), the current is to the right, so conventional current and $\vec{J}$ point to the right:
+\[
+\vec{J}=(1.49\times 10^6\,\mathrm{A/m^2})\hat{\imath}.
+\]
+
+For part (c), use the microscopic relation
+\[
+\vec{J}=nq_e\vec{v}_d.
+\]
+Solve for the drift velocity:
+\[
+\vec{v}_d=\frac{\vec{J}}{nq_e}.
+\]
+Because $q_e$ is negative, $\vec{v}_d$ must point opposite to $\vec{J}$, so it points left. Its magnitude is
+\[
+v_d=\frac{J}{n|q_e|}.
+\]
+Substitute the values:
+\[
+v_d=\frac{1.49\times 10^6}{(8.5\times 10^{28})(1.60\times 10^{-19})}\,\mathrm{m/s}.
+\]
+The denominator is
+\[
+(8.5\times 10^{28})(1.60\times 10^{-19})=1.36\times 10^{10},
+\]
+so
+\[
+v_d=\frac{1.49\times 10^6}{1.36\times 10^{10}}\,\mathrm{m/s}=1.10\times 10^{-4}\,\mathrm{m/s}.
+\]
+Therefore the drift velocity vector is
+\[
+\vec{v}_d=-(1.10\times 10^{-4}\,\mathrm{m/s})\hat{\imath}.
+\]
+
+Therefore,
+\[
+J=1.49\times 10^6\,\mathrm{A/m^2},
+\qquad
+\vec{J}=(1.49\times 10^6\,\mathrm{A/m^2})\hat{\imath},
+\]
+\[
+\vec{v}_d=-(1.10\times 10^{-4}\,\mathrm{m/s})\hat{\imath}.
+\]

concepts/em/u11/e11-2-resistance-ohm.tex

@@ -0,0 +1,161 @@
+\subsection{Resistance, Resistivity, and Ohm's Law}
+
+This subsection introduces resistance and resistivity, connects the macroscopic Ohm's law $V = IR$ to the microscopic relation $\vec{J} = \sigma \vec{E}$, and derives the geometric expression $R = \rho L / A$ for a uniform conductor.
+
+\dfn{Resistance}{Let a conducting element have a potential difference $V$ across its ends and carry a steady current $I$. The \emph{resistance} $R$ of the element is
+\[
+R = \frac{V}{I}.
+\]
+The SI unit of resistance is the \emph{ohm}, denoted $\Omega$, where $1\,\Omega = 1\,\mathrm{V/A}$.}
+
+\dfn{Resistivity, conductivity}{The \emph{resistivity} $\rho$ of a material is an intrinsic property that quantifies how strongly that material opposes the flow of electric current. The \emph{conductivity} $\sigma$ is the reciprocal of resistivity:
+\[
+\sigma = \frac{1}{\rho}.
+\]
+The SI unit of resistivity is the ohm-meter ($\Omega\!\cdot\!\mathrm{m}$). The SI unit of conductivity is $(\Omega\!\cdot\!\mathrm{m})^{-1}$, also called siemens per meter ($\mathrm{S/m}$).}
+
+\nt{A perfect conductor has $\rho = 0$ and $\sigma \to \infty$. A perfect insulator has $\rho \to \infty$ and $\sigma \to 0$. Metals have very low resistivity (typically $10^{-8}\,\Omega\!\cdot\!\mathrm{m}$); good insulators like glass have resistivity on the order of $10^{10}\,\Omega\!\cdot\!\mathrm{m}$ or higher.}
+
+\thm{Microscopic Ohm's law}{Let $\vec{E}$ denote the electric field inside a conducting material and let $\vec{J}$ denote the resulting current density at the same point. If the material is an \emph{ohmic conductor} -- one for which $\rho$ is independent of the magnitude of $\vec{E}$ -- then at every point inside the material,
+\[
+\vec{J} = \sigma \vec{E}
+\qquad \text{or equivalently} \qquad
+\vec{E} = \rho \vec{J},
+\]
+where $\sigma = 1/\rho$ is the conductivity.}
+
+\pf{Microscopic Ohm's law from the macroscopic form}{Consider a straight wire of uniform cross-sectional area $A$ and length $L$, made of a material with resistivity $\rho$. Suppose a potential difference $V$ is applied across the ends, producing a uniform field $\vec{E}$ along the wire and a uniform current density $\vec{J}$.
+
+From the macroscopic definition of resistance, $R = V/I$. For a uniform wire, the field and current density are related to the macroscopic quantities by $E = V/L$ and $J = I/A$. Substituting these into the resistance formula gives
+\[
+R = \frac{EL}{JA}.
+\]
+The resistance of a uniform wire is also known from experiment and geometry to be $R = \rho L/A$. Equating the two expressions for $R$ yields
+\[
+\frac{\rho L}{A} = \frac{EL}{JA},
+\]
+which simplifies to $\rho J = E$, or $\vec{E} = \rho \vec{J}$. This is the microscopic form of Ohm's law. The same relation holds pointwise even if the field and current density vary spatially, because resistivity is a local material property.}
+
+\cor{Ohm's law is empirical}{Not all materials obey Ohm's law. Semiconductors, diodes, and superconductors are non-ohmic: their $I$-$V$ characteristic is not linear. The microscopic relation $\vec{J} = \sigma \vec{E}$ holds only for ohmic conductors where $\sigma$ is independent of $\vec{E}$.}
+
+\mprop{Resistance of a uniform conductor}{A straight wire of length $L$, uniform cross-sectional area $A$, and resistivity $\rho$ has resistance
+\[
+R = \rho \,\frac{L}{A}.
+\]
+This relation follows from combining $V = IR$ with the microscopic Ohm's law $\vec{E} = \rho \vec{J}$ applied to a geometry where $\vec{E}$ and $\vec{J}$ are uniform and parallel to the wire axis.}
+
+\nt{The resistance of a uniform wire increases linearly with length and decreases inversely with cross-sectional area. This is the electrical analogue of fluid flow through a pipe: a longer pipe gives more resistance, and a wider pipe gives less.}
+
+\thm{Temperature dependence of resistivity}{For many materials, over a limited temperature range, the resistivity varies approximately as
+\[
+\rho = \rho_0 \,\bigl[1 + \alpha(T - T_0)\bigr],
+\]
+where $\rho_0$ is the resistivity at a reference temperature $T_0$, and $\alpha$ is the \emph{temperature coefficient of resistivity} for that material. The SI unit of $\alpha$ is $\mathrm{K}^{-1}$ (or equivalently $^\circ\mathrm{C}^{-1}$). For most metals, $\alpha > 0$, so resistivity increases with temperature.}
+
+\ex{Illustrative example}{A copper wire and an aluminum wire of the same length and cross-sectional area are connected to the same potential difference. Since copper has a lower resistivity than aluminum, the copper wire will carry more current and dissipate less power. The ratio of their resistivities at room temperature is approximately $\rho_{\mathrm{Cu}}/\rho_{\mathrm{Al}} \approx 1.7\times 10^{-8} / 2.8\times 10^{-8} \approx 0.61$.}
+
+\qs{Worked example}{A cylindrical copper wire of length
+\[
+L = 50\,\mathrm{m}
+\]
+and radius
+\[
+r = 1.0\,\mathrm{mm} = 1.0 \times 10^{-3}\,\mathrm{m}
+\]
+has a potential difference
+\[
+V = 10\,\mathrm{V}
+\]
+applied across its ends. The resistivity of copper at room temperature is
+\[
+\rho = 1.7 \times 10^{-8}\,\Omega\!\cdot\!\mathrm{m}.
+\]
+Assume the conduction-electron number density of copper is
+\[
+n = 8.5 \times 10^{28}\,\mathrm{m^{-3}},
+\]
+the elementary charge is $e = 1.60 \times 10^{-19}\,\mathrm{C}$, and the wire is uniform. Let the current flow from the high-potential end toward the low-potential end, and let $+\hat{\imath}$ point in the direction of the current.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the resistance $R$ of the wire,
+\item the current $I$ through the wire,
+\item the current density magnitude $J$ and vector $\vec{J}$,
+\item the drift velocity magnitude $v_d$ of the electrons, and
+\item the power dissipated in the wire.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The cross-sectional area of the cylindrical wire is
+\[
+A = \pi r^2 = \pi (1.0 \times 10^{-3}\,\mathrm{m})^2 = \pi \times 1.0 \times 10^{-6}\,\mathrm{m^2} = 3.14 \times 10^{-6}\,\mathrm{m^2}.
+\]
+The resistance is
+\[
+R = \rho \,\frac{L}{A}.
+\]
+Substitute the values:
+\[
+R = \left(1.7 \times 10^{-8}\,\Omega\!\cdot\!\mathrm{m}\right)\,\frac{50\,\mathrm{m}}{3.14 \times 10^{-6}\,\mathrm{m^2}}.
+\]
+Compute the numerator:
+\[
+\left(1.7 \times 10^{-8}\right)(50) = 8.5 \times 10^{-7}\,\Omega\!\cdot\!\mathrm{m^2}.
+\]
+Then
+\[
+R = \frac{8.5 \times 10^{-7}}{3.14 \times 10^{-6}}\,\Omega = 0.271\,\Omega.
+\]
+
+\textbf{Part (b).} Ohm's law gives the current:
+\[
+I = \frac{V}{R} = \frac{10\,\mathrm{V}}{0.271\,\Omega} = 36.9\,\mathrm{A}.
+\]
+
+\textbf{Part (c).} The current density magnitude is
+\[
+J = \frac{I}{A} = \frac{36.9\,\mathrm{A}}{3.14 \times 10^{-6}\,\mathrm{m^2}} = 1.18 \times 10^7\,\mathrm{A/m^2}.
+\]
+The current flows in the $+\hat{\imath}$ direction (from high to low potential), so the current density vector is
+\[
+\vec{J} = \left(1.18 \times 10^7\,\mathrm{A/m^2}\right)\hat{\imath}.
+\]
+
+\textbf{Part (d).} The drift velocity of electrons relates to the current density by
+\[
+\vec{J} = nq_e\vec{v}_d,
+\]
+where $q_e = -e$ is the charge of an electron. Since the electrons are negatively charged, their drift velocity is opposite to the current direction. The magnitude is
+\[
+v_d = \frac{J}{ne} = \frac{1.18 \times 10^7\,\mathrm{A/m^2}}{\left(8.5 \times 10^{28}\,\mathrm{m^{-3}}\right)\left(1.60 \times 10^{-19}\,\mathrm{C}\right)}.
+\]
+The denominator is
+\[
+\left(8.5 \times 10^{28}\right)\left(1.60 \times 10^{-19}\right) = 1.36 \times 10^{10}\,\mathrm{C/m^3},
+\]
+so
+\[
+v_d = \frac{1.18 \times 10^7}{1.36 \times 10^{10}}\,\mathrm{m/s} = 8.69 \times 10^{-4}\,\mathrm{m/s}.
+\]
+Electrons drift opposite to $\vec{J}$, so $\vec{v}_d = -\left(8.69 \times 10^{-4}\,\mathrm{m/s}\right)\hat{\imath}$.
+
+\textbf{Part (e).} The power dissipated in the wire is
+\[
+P = IV = \left(36.9\,\mathrm{A}\right)\left(10\,\mathrm{V}\right) = 369\,\mathrm{W}.
+\]
+Equivalently, $P = I^2R = (36.9\,\mathrm{A})^2(0.271\,\Omega) = 369\,\mathrm{W}$, or $P = V^2/R = (10\,\mathrm{V})^2/(0.271\,\Omega) = 369\,\mathrm{W}$. All three give the same result.
+
+Therefore,
+\[
+R = 0.271\,\Omega,
+\qquad
+I = 36.9\,\mathrm{A},
+\qquad
+J = 1.18 \times 10^7\,\mathrm{A/m^2},
+\]
+\[
+\vec{J} = \left(1.18 \times 10^7\,\mathrm{A/m^2}\right)\hat{\imath},
+\qquad
+v_d = 8.69 \times 10^{-4}\,\mathrm{m/s},
+\qquad
+P = 369\,\mathrm{W}.
+\]

concepts/em/u11/e11-3-power.tex

@@ -0,0 +1,99 @@
+\subsection{Electric Power and Dissipation}
+
+Electric power quantifies the rate at which electrical energy is transferred or dissipated in a circuit element. When a charge $q$ moves through a potential difference $\Delta V$, its electric potential energy changes by $\Delta U = q\,\Delta V$. The power delivered to (or dissipated by) a circuit element is the rate of this energy transfer.
+
+\dfn{Electric power}{Let a circuit element have a potential difference $\Delta V$ across it and carry a current $I$ through it. The \emph{electric power} delivered to the element is
+\[
+P = I\,\Delta V.
+\]
+The SI unit of power is the watt ($1\,\mathrm{W}=1\,\mathrm{J/s}=1\,\mathrm{A\!\cdot\!V}$). When the element is a resistor, electrical energy is dissipated as thermal energy (Joule heating).}
+
+\nt{Power is the rate of energy transfer. A battery \emph{supplies} power to the circuit, while resistors \emph{dissipate} it. In a steady circuit, the total power supplied equals the total power dissipated.}
+
+\ex{Illustrative example}{A $60\,\mathrm{W}$ incandescent lightbulb is connected to a $120\,\mathrm{V}$ household outlet. Find (a) the current through the bulb and (b) the resistance of its filament.
+
+\sol (a) From $P = I\,\Delta V$,
+\[
+I = \frac{P}{\Delta V} = \frac{60\,\mathrm{W}}{120\,\mathrm{V}} = 0.50\,\mathrm{A}.
+\]
+
+(b) From Ohm's law,
+\[
+R = \frac{\Delta V}{I} = \frac{120\,\mathrm{V}}{0.50\,\mathrm{A}} = 240\,\Omega. \qedhere
+\]}
+
+\nt{The $I^{2}R$ form shows why transmission lines use very high voltages: for a fixed power $P=IV$, raising the voltage lowers the current, and since resistive losses scale as $I^{2}R$, the dissipation drops dramatically.}
+
+\thm{Microscopic power density}{In a continuous medium, the local rate of energy dissipation per unit volume is the dot product of the current density $\vec{J}$ and the electric field $\vec{E}$ at that point:
+\[
+p = \vec{J}\cdot\vec{E}.
+\]
+The total power delivered to a volume $\mathcal{V}$ is the integral
+\[
+P = \iiint_{\mathcal{V}} \vec{J}\cdot\vec{E}\; d\tau.
+\]}
+
+\pf{Microscopic to macroscopic}{Consider a straight wire of length $L$ and cross-sectional area $A$, with a uniform electric field $\vec{E}$ along its axis and a uniform current $I$. The field is related to the potential difference by $\Delta V = E\,L$. The current density is $J = I/A$, and by Ohm's law $E = \rho\,J$ (where $\rho$ is the resistivity). The total power dissipated is
+\[
+P = \iiint_{\mathcal{V}} \vec{J}\cdot\vec{E}\; d\tau = J\,E\,(A\,L)
+\]
+since $\vec{J}$ and $\vec{E}$ are parallel and uniform. Substituting $J = I/A$ and $E = \Delta V/L$,
+\[
+P = \left(\frac{I}{A}\right)\!\left(\frac{\Delta V}{L}\right)(A\,L) = I\,\Delta V,
+\]
+recovering the macroscopic expression. Alternatively, using $E = \rho\,J = \rho\,(I/A)$ and $R = \rho\,L/A$,
+\[
+P = J\,E\,(A\,L) = \frac{I^{2}}{A^{2}}\cdot\rho\frac{I}{A}\cdot A\,L
+= I^{2}\,\frac{\rho\,L}{A} = I^{2}R. \qedhere
+\]
+
+\qs{Worked example}{A resistor is connected across a $12.0\,\mathrm{V}$ battery. The current through the resistor is measured to be $2.00\,\mathrm{A}$.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the power dissipated by the resistor,
+\item the resistance of the resistor, and
+\item the total energy dissipated as heat over a time interval of $5.00\,\mathrm{min}$.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The power dissipated by the resistor is given by
+\[
+P = I\,\Delta V.
+\]
+Substitute the given values:
+\[
+P = (2.00\,\mathrm{A})(12.0\,\mathrm{V}) = 24.0\,\mathrm{W}.
+\]
+
+\textbf{Part (b).} By Ohm's law,
+\[
+R = \frac{\Delta V}{I}.
+\]
+Substitute the values:
+\[
+R = \frac{12.0\,\mathrm{V}}{2.00\,\mathrm{A}} = 6.00\,\Omega.
+\]
+As a check, verify using $P = I^{2}R$:
+\[
+P = (2.00\,\mathrm{A})^{2}(6.00\,\Omega) = 4.00 \times 6.00 = 24.0\,\mathrm{W},
+\]
+which agrees with part (a).}
+
+\textbf{Part (c).} The total energy dissipated is power multiplied by time:
+\[
+E = P\,t.
+\]
+Convert the time to seconds:
+\[
+t = 5.00\,\mathrm{min}\times 60.0\,\mathrm{s/min} = 300\,\mathrm{s}.
+\]
+Then
+\[
+E = (24.0\,\mathrm{W})(300\,\mathrm{s}) = 7200\,\mathrm{J}.
+\]
+Alternatively, using $E = I\,\Delta V\,t$ directly:
+\[
+E = (2.00\,\mathrm{A})(12.0\,\mathrm{V})(300\,\mathrm{s}) = 7200\,\mathrm{J}.
+\]
+
+\nt{Energy can also be written as $E = I^{2}Rt$ or $E = (\Delta V)^{2}t/R$. All three are equivalent and follow from $E = P\,t$ together with Ohm's law.}

concepts/em/u11/e11-4-equivalent-resistance.tex

@@ -0,0 +1,164 @@
+\subsection{Equivalent Resistance of Series and Parallel Circuits}
+
+This subsection defines equivalent resistance for resistive networks, derives the series and parallel combination rules from Kirchhoff's laws and Ohm's law, and shows how to reduce compound circuits step by step to a single equivalent resistor.
+
+\dfn{Equivalent resistance}{Consider a network of resistors connected to an ideal battery of emf $\mathcal{E}$. The \emph{equivalent resistance} $R_{\mathrm{eq}}$ of the network is defined by
+\[
+R_{\mathrm{eq}} = \frac{\mathcal{E}}{I_{\mathrm{total}}},
+\]
+where $I_{\mathrm{total}}$ is the total current delivered by the battery. The equivalent resistance is the resistance of a single resistor that would draw the same current from the same battery as the entire network.}
+
+\nt{Equivalent resistance is a bookkeeping device: it replaces an entire resistive sub-network by a single resistor whose effect on the rest of the circuit is identical. The replacement is always done between two terminals, and it preserves the $I$--$V$ relationship at those terminals.}
+
+\dfn{Series combination}{Two (or more) resistors are in \emph{series} when they share exactly one common node and no other element is connected to that node. Equivalently, the same current $I$ flows through each resistor in a series chain, and the total voltage is the sum of individual voltage drops:
+\[
+V = V_1 + V_2 + \cdots + V_n.
+\]
+The equivalent resistance of $n$ resistors in series is
+\[
+R_{\mathrm{eq}} = R_1 + R_2 + \cdots + R_n = \sum_{i=1}^{n} R_i.
+\]}
+
+\thm{Voltage divider rule}{Consider two resistors $R_1$ and $R_2$ in series connected to a potential difference $V_{\mathrm{in}}$. The voltage across $R_2$ alone is
+\[
+V_{\mathrm{out}} = V_{R_2} = V_{\mathrm{in}}\,\frac{R_2}{R_1 + R_2}.
+\]
+This relation follows from Ohm's law and the fact that the same current $I = V_{\mathrm{in}}/(R_1+R_2)$ flows through both resistors.}
+
+\nt{The voltage divider distributes the input voltage proportionally to each resistance. A larger resistance drops more voltage. This rule is the electrical analogue of the weighted average: $V_{R_i}$ is the fraction of $V_{\mathrm{in}}$ carried by $R_i$.}
+
+\dfn{Parallel combination}{Two (or more) resistors are in \emph{parallel} when they are connected between the same two nodes, so the potential difference $V$ across each is identical. The total current splits among the branches:
+\[
+I_{\mathrm{total}} = I_1 + I_2 + \cdots + I_n.
+\]
+The equivalent resistance of $n$ resistors in parallel is
+\[
+\frac{1}{R_{\mathrm{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} = \sum_{i=1}^{n} \frac{1}{R_i}.
+\]
+For two resistors in parallel, this simplifies to
+\[
+R_{\mathrm{eq}} = \frac{R_1 R_2}{R_1 + R_2}.
+\]}
+
+\thm{Current divider rule}{Consider two resistors $R_1$ and $R_2$ in parallel connected to a total current $I_{\mathrm{total}}$. The current through $R_1$ is
+\[
+I_1 = I_{\mathrm{total}}\,\frac{R_2}{R_1 + R_2}.
+\]
+This follows from KCL and Ohm's law: the common voltage is $V = I_{\mathrm{total}}/G_{\mathrm{eq}}$ where $G_{\mathrm{eq}} = 1/R_1 + 1/R_2$, and $I_1 = V/R_1$.}
+
+\nt{Current divides inversely to resistance: the smaller resistor carries more current. In the limit $R_1 \ll R_2$, essentially all current flows through $R_1$; we say $R_1$ \emph{shunts} $R_2$.}
+
+\mprop{Series and parallel equivalent resistance}{For $n$ resistors, the equivalent resistance satisfies
+\[
+\text{Series:}\quad R_{\mathrm{eq}} = \sum_{i=1}^{n} R_i
+\qquad\text{and}\qquad
+\text{Parallel:}\quad \frac{1}{R_{\mathrm{eq}}} = \sum_{i=1}^{n} \frac{1}{R_i}.
+\]
+The current through and voltage across each individual resistor are obtained by applying Ohm's law $V_i = I_i R_i$ once $I_i$ or $V_i$ is known from the reduction analysis.}
+
+\pf{Series combination}{Kirchhoff's voltage law (the loop rule) states that the sum of potential differences around any closed loop is zero. For two resistors in series with a battery of emf $\mathcal{E}$,
+\[
+\mathcal{E} - V_1 - V_2 = 0,
+\]
+so $V_1 + V_2 = \mathcal{E}$. Since the same current $I$ flows through both resistors, $V_1 = IR_1$ and $V_2 = IR_2$. Substituting gives $I(R_1 + R_2) = \mathcal{E}$, or $R_{\mathrm{eq}} = \mathcal{E}/I = R_1 + R_2$. The result extends immediately to $n$ resistors by induction.}
+
+\pf{Parallel combination}{Kirchhoff's current law (the node rule) states that the sum of currents entering a node equals the sum of currents leaving. At the top node of the parallel connection, $I_{\mathrm{total}} = I_1 + I_2 + \cdots + I_n$. Since each branch has the same voltage $V$, we have $I_i = V/R_i$. Thus
+\[
+I_{\mathrm{total}} = V\sum_{i=1}^{n} \frac{1}{R_i}.
+\]
+But $I_{\mathrm{total}} = V/R_{\mathrm{eq}}$, so $1/R_{\mathrm{eq}} = \sum 1/R_i$.}
+
+\cor{Bounds on equivalent resistance}{For any combination of series and parallel resistors, the equivalent resistance of a parallel block is always less than the smallest resistance in that block:
+\[
+R_{\mathrm{parallel}} < \min_i(R_i).
+\]
+Conversely, the equivalent resistance of a series chain is always greater than the largest individual resistance. These bounds follow from the positivity of all resistances and provide a useful sanity check on computed results.}
+
+\ex{Illustrative example}{When resistors are identical, the formulas simplify. $n$ identical resistors of resistance $R$ in series give $R_{\mathrm{eq}} = nR$. In parallel, $R_{\mathrm{eq}} = R/n$. Thus three $12\,\Omega$ resistors in parallel give $R_{\mathrm{eq}} = 4\,\Omega$, while in series they give $36\,\Omega$.}
+
+\nt{The strategy for analysing a mixed (compound) circuit is: (1) identify the innermost series or parallel groups. (2) Replace each group by its equivalent resistance. (3) Repeat until the entire network is reduced to a single resistor $R_{\mathrm{eq}}$. (4) Use $I_{\mathrm{total}} = \mathcal{E}/R_{\mathrm{eq}}$ to find the battery current, then work backwards through the reduction steps to find individual branch currents and voltages.}
+
+\nt{On the AP Physics C E\&M exam, equivalent resistance problems test your ability to correctly identify series versus parallel connections and to apply Kirchhoff's laws systematically. The most common errors are misidentifying which elements share the same current (series) and which share the same voltage (parallel). Always trace the current paths and label the nodes to verify your classification.}
+
+\qs{Worked example}{An ideal battery of emf
+\[
+\mathcal{E} = 24\,\mathrm{V}
+\]
+is connected to a resistive network consisting of four resistors arranged as follows. Resistors $R_1 = 2.0\,\Omega$ and $R_2 = 4.0\,\Omega$ are connected in series. This series combination is connected in parallel with resistor $R_3 = 6.0\,\Omega$. Finally, resistor $R_4 = 3.0\,\Omega$ is connected in series with this entire parallel block, and the combination is connected to the battery.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the total equivalent resistance $R_{\mathrm{eq}}$ of the network,
+\item the total current $I_{\mathrm{total}}$ delivered by the battery,
+\item the voltage drop across each resistor, and
+\item the current through each resistor.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} Begin by reducing the network from the inside out. Resistors $R_1$ and $R_2$ are in series, so their combined resistance is
+\[
+R_{12} = R_1 + R_2 = 2.0\,\Omega + 4.0\,\Omega = 6.0\,\Omega.
+\]
+This series combination is in parallel with $R_3 = 6.0\,\Omega$. The equivalent resistance of the parallel block is
+\[
+\frac{1}{R_p} = \frac{1}{R_{12}} + \frac{1}{R_3} = \frac{1}{6.0\,\Omega} + \frac{1}{6.0\,\Omega} = \frac{2}{6.0\,\Omega} = \frac{1}{3.0\,\Omega}.
+\]
+Hence $R_p = 3.0\,\Omega$.
+
+The parallel block is in series with $R_4$, so the total equivalent resistance is
+\[
+R_{\mathrm{eq}} = R_p + R_4 = 3.0\,\Omega + 3.0\,\Omega = 6.0\,\Omega.
+\]
+
+\textbf{Part (b).} Ohm's law gives the total current from the battery:
+\[
+I_{\mathrm{total}} = \frac{\mathcal{E}}{R_{\mathrm{eq}}} = \frac{24\,\mathrm{V}}{6.0\,\Omega} = 4.0\,\mathrm{A}.
+\]
+
+\textbf{Parts (c) and (d).} Now work backwards through the reduction. The current through $R_4$ equals the total current, since $R_4$ is in series with the battery:
+\[
+I_4 = I_{\mathrm{total}} = 4.0\,\mathrm{A}.
+\]
+The voltage drop across $R_4$ is
+\[
+V_4 = I_4 R_4 = (4.0\,\mathrm{A})(3.0\,\Omega) = 12\,\mathrm{V}.
+\]
+The remaining voltage appears across the parallel block:
+\[
+V_p = \mathcal{E} - V_4 = 24\,\mathrm{V} - 12\,\mathrm{V} = 12\,\mathrm{V}.
+\]
+Since the resistors $R_1$, $R_2$, and $R_3$ are all connected to the parallel block, the voltage across each is $V_p = 12\,\mathrm{V}$ (for $R_3$ directly) and $V_p = 12\,\mathrm{V}$ (for the series pair $R_1$--$R_2$).
+
+The current through $R_3$ is
+\[
+I_3 = \frac{V_p}{R_3} = \frac{12\,\mathrm{V}}{6.0\,\Omega} = 2.0\,\mathrm{A}.
+\]
+The current through the $R_1$--$R_2$ branch is
+\[
+I_{12} = \frac{V_p}{R_{12}} = \frac{12\,\mathrm{V}}{6.0\,\Omega} = 2.0\,\mathrm{A}.
+\]
+Since $R_1$ and $R_2$ are in series, the same current flows through both:
+\[
+I_1 = I_2 = I_{12} = 2.0\,\mathrm{A}.
+\]
+The individual voltage drops are
+\[
+V_1 = I_1 R_1 = (2.0\,\mathrm{A})(2.0\,\Omega) = 4.0\,\mathrm{V},
+\]
+\[
+V_2 = I_2 R_2 = (2.0\,\mathrm{A})(4.0\,\Omega) = 8.0\,\mathrm{V}.
+\]
+
+\noindent\textbf{Checks:}
+\begin{itemize}
+\item KVL on the left loop: $V_1 + V_2 = 4.0\,\mathrm{V} + 8.0\,\mathrm{V} = 12\,\mathrm{V} = V_p$.
+\item KCL at the junction: $I_{12} + I_3 = 2.0\,\mathrm{A} + 2.0\,\mathrm{A} = 4.0\,\mathrm{A} = I_{\mathrm{total}}$.
+\item KVL on the full loop: $V_p + V_4 = 12\,\mathrm{V} + 12\,\mathrm{V} = 24\,\mathrm{V} = \mathcal{E}$.
+\end{itemize}
+
+Therefore,
+\begin{align*}
+R_{\mathrm{eq}} &= 6.0\,\Omega,\\
+I_{\mathrm{total}} &= 4.0\,\mathrm{A},\\
+(V_1, V_2, V_3, V_4) &= (4.0\,\mathrm{V}, 8.0\,\mathrm{V}, 12\,\mathrm{V}, 12\,\mathrm{V}),\\
+(I_1, I_2, I_3, I_4) &= (2.0\,\mathrm{A}, 2.0\,\mathrm{A}, 2.0\,\mathrm{A}, 4.0\,\mathrm{A}).
+\end{align*}

concepts/em/u11/e11-5-kirchhoff.tex

@@ -0,0 +1,213 @@
+\subsection{Kirchhoff's Junction and Loop Rules}
+
+This subsection introduces Kirchhoff's two rules for analyzing electric circuits. The junction rule (Kirchhoff's current law) expresses charge conservation at circuit nodes. The loop rule (Kirchhoff's voltage law) expresses energy conservation around any closed loop. Together they provide a systematic method for finding unknown currents in multi-loop circuits.
+
+\dfn{Electric circuit junction (node)}{A \emph{junction} (or \emph{node}) is a point in a circuit where three or more conductors meet. The current in each conductor leading to the junction is called a \emph{branch current}.}
+
+\nt{In a steady-state dc circuit the charge at any junction is constant: no charge accumulates at a node. This is the physical reason the junction rule holds.}
+
+\thm{Kirchhoff's junction rule (KCL)}{At any junction in a steady-state dc circuit, the sum of currents entering the junction equals the sum of currents leaving the junction:
+\begin{equation}
+\sum I_{\text{in}} = \sum I_{\text{out}}.
+\end{equation}
+Equivalently, the algebraic sum of all currents at a junction is zero:
+\begin{equation}
+\sum I = 0,
+\end{equation}
+where currents entering the junction are taken as positive and currents leaving are taken as negative.
+
+\textbf{Calc 3 connection:} The junction rule is the circuit analogue of the steady-state continuity equation $\nabla \cdot \vec{J} = 0$ (charge conservation with no time-varying charge density). Integrating $\nabla \cdot \vec{J} = 0$ over a small volume enclosing the junction gives $\oiint \vec{J} \cdot d\vec{A} = 0$, which is exactly $\sum I = 0$.}
+
+\dfn{Electric circuit loop}{A \emph{loop} is any closed path through a circuit, traversing a sequence of circuit elements and returning to the starting point.}
+
+\dfn{Sign convention for traversing circuit elements}{When applying the loop rule, traverse the loop in a chosen direction and assign potential changes as follows:
+\begin{itemize}
+\item \emph{Resistor:} Traversing in the \emph{same} direction as the assumed current gives a potential \emph{drop} of $IR$ (contribute $-IR$). Traversing \emph{opposite} to the assumed current gives a potential \emph{rise} of $IR$ (contribute $+IR$).
+\item \emph{Battery (emf):} Traversing from the \emph{negative} terminal to the \emph{positive} terminal gives a potential \emph{rise} of $\mathcal{E}$ (contribute $+\mathcal{E}$). Traversing from \emph{positive} to \emph{negative} gives a potential \emph{drop} of $\mathcal{E}$ (contribute $-\mathcal{E}$).
+\end{itemize}
+The assumed direction of each branch current must be declared before writing equations. If the solved value of a current is negative, the actual current flows opposite to the assumed direction.}
+
+\thm{Kirchhoff's loop rule (KVL)}{For any closed loop in a circuit, the algebraic sum of the potential differences across all elements in the loop is zero:
+\begin{equation}
+\sum \Delta V = 0.
+\end{equation}
+
+\textbf{Calc 3 connection:} The loop rule follows from energy conservation: a test charge $q$ that moves around a closed path and returns to its starting point must have zero net change in potential energy, so $\oint \vec{E} \cdot d\vec{\ell} = 0$. In electrostatics this is a consequence of $\vec{E}$ being a conservative field. More generally, in quasi-static circuits with no changing magnetic flux through the loop, Faraday's law gives $\nabla \times \vec{E} = -\partial \vec{B}/\partial t = 0$, so the integral over any closed loop vanishes.}
+
+\nt{The loop rule holds for \emph{any} closed loop, not just the obvious ``mesh'' loops of a circuit diagram. Any closed path through the elements counts. In practice one typically chooses the minimal loops (meshes) because they lead to the most economical system of equations.}
+
+\nt{The sign of a solved current encodes direction. A negative result does \emph{not} mean the current is unphysical --- it simply means the actual current is opposite to the assumed direction. Always state the assumed direction and report the final direction clearly.}
+
+\mprop{Algorithm for solving multi-loop circuits}{Given a multi-loop circuit with unknown branch currents:
+\begin{enumerate}
+\item Label every branch with a current variable and an assumed direction.
+\item Identify all junctions. For $N$ junctions, write $N-1$ independent junction equations.
+\item Choose enough independent loops (typically meshes) so that the total number of equations (junction + loop) equals the number of unknown currents.
+\item Apply the loop rule to each chosen loop, using the sign conventions above.
+\item Solve the resulting system of linear equations.
+\item Check: any current with a negative value flows opposite to the assumed direction. Verify that all junction equations are satisfied.
+\end{enumerate}
+The number of independent equations needed equals the number of unknown branch currents.}
+
+\ex{Illustrative example}{Consider a junction with three wires meeting. Current $I_1 = 3.0\,\mathrm{A}$ enters the junction and current $I_2 = 1.5\,\mathrm{A}$ leaves. The third branch carries current $I_3$. By the junction rule, $I_1 = I_2 + I_3$, so $I_3 = 1.5\,\mathrm{A}$ leaves the junction.}
+
+\nt{For circuits with a single battery and resistors in simple series or parallel, the equivalent-resistance method is faster. Kirchhoff's rules are needed whenever the circuit cannot be reduced to simple series/parallel combinations --- for instance, when there are two or more batteries arranged in different branches, forming multiple loops.}
+
+\qs{Worked example}{Consider the two-loop circuit shown in the diagram below. The circuit consists of a left loop and a right loop sharing a common middle branch.
+
+\begin{itemize}
+\item The \emph{left branch} contains a battery of emf $\mathcal{E}_1 = 12\,\mathrm{V}$ (positive terminal up) in series with a resistor $R_1 = 4.0\,\Omega$.
+\item The \emph{middle branch} contains a resistor $R_3 = 3.0\,\Omega$.
+\item The \emph{right branch} contains a battery of emf $\mathcal{E}_2 = 6.0\,\mathrm{V}$ (positive terminal up) in series with a resistor $R_2 = 6.0\,\Omega$.
+\end{itemize}
+
+\begin{center}
+\textbf{Diagram description:} Two rectangular loops share a vertical middle branch. The left vertical branch has the $12\,\mathrm{V}$ battery (positive up) and $4.0\,\Omega$ resistor. The middle vertical branch has the $3.0\,\Omega$ resistor. The right vertical branch has the $6.0\,\mathrm{V}$ battery (positive up) and $6.0\,\Omega$ resistor. All three vertical branches connect at top and bottom horizontal wires (ideal conductors with zero resistance).
+\end{center}
+
+Let the top junction be $A$ and the bottom junction be $B$. Define three branch currents:
+\begin{itemize}
+\item $I_1$ flows \emph{upward} through the left branch (from $B$ to $A$).
+\item $I_2$ flows \emph{upward} through the right branch (from $B$ to $A$).
+\item $I_3$ flows \emph{downward} through the middle branch (from $A$ to $B$).
+\end{itemize}
+Assume these directions when applying Kirchhoff's rules.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the junction equation at node $A$,
+\item the two independent loop equations (left loop and right loop),
+\item the values of all three currents $I_1$, $I_2$, and $I_3$, and
+\item the direction of each current (consistent with the assumed direction).
+\end{enumerate}}
+
+\sol \textbf{Part (a). Junction equation.} At junction $A$, the currents $I_1$ and $I_2$ both enter (they flow upward from $B$ to $A$ in their respective branches). The current $I_3$ leaves $A$ (it flows downward from $A$ to $B$). By the junction rule:
+\[
+I_1 + I_2 = I_3.
+\]
+This is our first equation.
+
+\textbf{Part (b). Loop equations.}
+
+\emph{Left loop} (traverse clockwise starting from junction $B$):
+\begin{itemize}
+\item Go up through the left branch, in the same direction as $I_1$: the battery $\mathcal{E}_1$ is traversed from $-$ to $+$, contributing $+\mathcal{E}_1 = +12\,\mathrm{V}$. The resistor $R_1$ is traversed in the same direction as $I_1$, contributing $-I_1 R_1 = -4.0\,I_1$.
+\item Go across the top wire from the left branch to the middle branch (ideal wire, $\Delta V = 0$).
+\item Go down through the middle branch, in the same direction as $I_3$: resistor $R_3$ is traversed in the direction of $I_3$, contributing $-I_3 R_3 = -3.0\,I_3$.
+\item Go across the bottom wire back to $B$ (ideal wire, $\Delta V = 0$).
+\end{itemize}
+Summing around the loop:
+\[
++12 - 4.0\,I_1 - 3.0\,I_3 = 0,
+\]
+\[
+4.0\,I_1 + 3.0\,I_3 = 12.
+\]
+\label{eq:loop1}
+
+\emph{Right loop} (traverse clockwise starting from junction $A$):
+\begin{itemize}
+\item Go down through the right branch, in the \emph{opposite} direction to $I_2$: the resistor $R_2$ is traversed opposite to $I_2$, contributing $+I_2 R_2 = +6.0\,I_2$. The battery $\mathcal{E}_2$ is traversed from $+$ to $-$, contributing $-\mathcal{E}_2 = -6\,\mathrm{V}$.
+\item Go across the bottom wire from right to middle (ideal wire, $\Delta V = 0$).
+\item Go up through the middle branch, in the \emph{opposite} direction to $I_3$: resistor $R_3$ is traversed opposite to $I_3$, contributing $+I_3 R_3 = +3.0\,I_3$.
+\item Go across the top wire back to $A$ (ideal wire, $\Delta V = 0$).
+\end{itemize}
+Summing around the loop:
+\[
++6.0\,I_2 - 6 + 3.0\,I_3 = 0,
+\]
+\[
+6.0\,I_2 + 3.0\,I_3 = 6.
+\]
+\label{eq:loop2}
+
+\textbf{Part (c). Solving for the currents.} We have three equations:
+\begin{align}
+I_3 &= I_1 + I_2, &\text{(junction)} \\
+4.0\,I_1 + 3.0\,I_3 &= 12, &\text{(left loop)} \\
+6.0\,I_2 + 3.0\,I_3 &= 6. &\text{(right loop)}
+\end{align}
+
+Substitute equation (1) into equation (2):
+\[
+4.0\,I_1 + 3.0\,(I_1 + I_2) = 12,
+\]
+\[
+4.0\,I_1 + 3.0\,I_1 + 3.0\,I_2 = 12,
+\]
+\[
+7.0\,I_1 + 3.0\,I_2 = 12.
+\]
+\label{eq:a}
+
+Substitute equation (1) into equation (3):
+\[
+6.0\,I_2 + 3.0\,(I_1 + I_2) = 6,
+\]
+\[
+6.0\,I_2 + 3.0\,I_1 + 3.0\,I_2 = 6,
+\]
+\[
+3.0\,I_1 + 9.0\,I_2 = 6.
+\]
+Dividing by $3.0$:
+\[
+I_1 + 3.0\,I_2 = 2.0,
+\]
+so
+\[
+I_1 = 2.0 - 3.0\,I_2.
+\]
+\label{eq:b}
+
+Substitute equation (\ref{eq:b}) into equation (\ref{eq:a}):
+\[
+7.0\,(2.0 - 3.0\,I_2) + 3.0\,I_2 = 12,
+\]
+\[
+14.0 - 21.0\,I_2 + 3.0\,I_2 = 12,
+\]
+\[
+14.0 - 18.0\,I_2 = 12,
+\]
+\[
+-18.0\,I_2 = -2.0,
+\]
+\[
+I_2 = \frac{2.0}{18.0} = \frac{1}{9}\,\mathrm{A}.
+\]
+
+From equation (\ref{eq:b}):
+\[
+I_1 = 2.0 - 3.0\left(\frac{1}{9}\right) = 2.0 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}\,\mathrm{A}.
+\]
+
+From the junction equation (1):
+\[
+I_3 = I_1 + I_2 = \frac{5}{3} + \frac{1}{9} = \frac{15}{9} + \frac{1}{9} = \frac{16}{9}\,\mathrm{A}.
+\]
+
+\textbf{Verification.} Check against the left-loop equation:
+\[
+4.0\left(\frac{5}{3}\right) + 3.0\left(\frac{16}{9}\right) = \frac{20}{3} + \frac{16}{3} = \frac{36}{3} = 12.
+\]
+Check against the right-loop equation:
+\[
+6.0\left(\frac{1}{9}\right) + 3.0\left(\frac{16}{9}\right) = \frac{6}{9} + \frac{48}{9} = \frac{54}{9} = 6.
+\]
+Both are satisfied.
+
+\textbf{Part (d). Directions.} All three currents are positive, confirming they flow in the assumed directions:
+\begin{itemize}
+\item $I_1 = 5/3\,\mathrm{A}$ upward in the left branch,
+\item $I_2 = 1/9\,\mathrm{A}$ upward in the right branch,
+\item $I_3 = 16/9\,\mathrm{A}$ downward in the middle branch.
+\end{itemize}
+
+Therefore, the branch currents are:
+\[
+I_1 = \frac{5}{3}\,\mathrm{A},\qquad
+I_2 = \frac{1}{9}\,\mathrm{A},\qquad
+I_3 = \frac{16}{9}\,\mathrm{A},
+\]
+all flowing in the assumed directions.

concepts/em/u11/e11-6-rc-circuits.tex

@@ -0,0 +1,278 @@
+\subsection{RC Transients and the Time Constant}
+
+This subsection introduces RC circuits -- circuits containing resistors and capacitors -- derives the first-order differential equation governing charge evolution during charging and discharging, solves it by separation of variables, and defines the time constant $\tau = RC$ as the characteristic timescale of the transient response.
+
+\dfn{RC circuit and transient response}{A series \emph{RC circuit} consists of a resistor $R$, a capacitor $C$, and (optionally) a battery of emf $\mathcal{E}$, all connected in a single closed loop. When the circuit is first connected (or when the battery is disconnected), the capacitor neither holds its initial charge nor its final charge instantaneously; instead, its charge evolves over time. This time-dependent behavior is called a \emph{transient response}.}
+
+\nt{At the instant a capacitor begins charging, it behaves like a short circuit: its voltage is zero and all of the battery voltage appears across the resistor. As the capacitor charges, its voltage increases and the current decreases. In the limit $t \to \infty$, the capacitor is fully charged to voltage $\mathcal{E}$ and the current drops to zero, so the capacitor acts like an open circuit.}
+
+\thm{Charging a capacitor}{A capacitor of capacitance $C$ that is initially uncharged is connected in series at $t = 0$ with a resistor of resistance $R$ and a battery of emf $\mathcal{E}$, forming a single-loop circuit. The charge on the capacitor at time $t$ is
+\[
+q(t) = C\mathcal{E}\,\bigl(1 - e^{-t/RC}\bigr),
+\]
+and the current flowing through the resistor is
+\[
+I(t) = \frac{\mathcal{E}}{R}\,e^{-t/RC}.
+\]
+Here $q(0) = 0$ and $I(0) = \mathcal{E}/R$. As $t \to \infty$, $q \to C\mathcal{E}$ and $I \to 0$.}
+
+\pf{Derivation of charging equations}{Apply Kirchhoff's voltage law around the loop. The potential drops across the resistor and capacitor sum to the battery emf:
+\[
+\mathcal{E} - IR - \frac{q}{C} = 0,
+\]
+where $I = \dfrac{dq}{dt}$ is the current (the rate at which charge accumulates on the capacitor). Substituting gives the first-order linear ODE
+\[
+R\,\frac{dq}{dt} + \frac{q}{C} = \mathcal{E}.
+\]
+Divide by $R$:
+\[
+\frac{dq}{dt} + \frac{1}{RC}\,q = \frac{\mathcal{E}}{R}.
+\]
+This is a first-order linear ODE. Use the integrating factor $\mu(t) = e^{t/RC}$:
+\[
+\frac{d}{dt}\!\left(q\,e^{t/RC}\right) = \frac{\mathcal{E}}{R}\,e^{t/RC}.
+\]
+Integrate both sides from $0$ to $t$, with $q(0) = 0$:
+\[
+q(t)\,e^{t/RC} - q(0) = \frac{\mathcal{E}}{R}\int_0^t e^{t'/RC}\,dt' = \mathcal{E}C\left(e^{t/RC} - 1\right).
+\]
+Solving for $q(t)$:
+\[
+q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right).
+\]
+The current is obtained from $I = dq/dt$:
+\[
+I(t) = \frac{d}{dt}\!\left[C\mathcal{E}\left(1 - e^{-t/RC}\right)\right]
+     = \frac{\mathcal{E}}{R}\,e^{-t/RC}.
+\]
+The initial current is $I(0) = \mathcal{E}/R$, and as $t \to \infty$ the exponential vanishes, so $q \to C\mathcal{E}$ and $I \to 0$.}
+
+\thm{Discharging a capacitor}{A capacitor of capacitance $C$ that is initially charged to charge $q_0$ is connected at $t = 0$ with a resistor of resistance $R$, forming a single-loop circuit with no battery. The charge on the capacitor at time $t$ is
+\[
+q(t) = q_0\,e^{-t/RC},
+\]
+and the rate of change of charge is
+\[
+\frac{dq}{dt} = -\,\frac{q_0}{RC}\,e^{-t/RC}.
+\]
+The negative sign indicates that charge is \emph{decreasing}: the capacitor is discharging. The magnitude of the current through the resistor is
+\[
+|I(t)| = \left|\frac{dq}{dt}\right| = \frac{q_0}{RC}\,e^{-t/RC}.
+\]
+As $t \to \infty$, both $q \to 0$ and $I \to 0$.}
+
+\pf{Derivation of discharging equations}{With no battery in the loop, Kirchhoff's voltage law gives
+\[
+-\,\frac{q}{C} - IR = 0,
+\]
+where $I$ is the current through the resistor and $q/C$ is the voltage across the capacitor. During discharge, charge flows off the capacitor, so $I = -\,dq/dt$ (current is positive while charge is decreasing). Substituting:
+\[
+-\,\frac{q}{C} - R\!\left(-\frac{dq}{dt}\right) = 0,
+\]
+which simplifies to
+\[
+\frac{dq}{dt} = -\,\frac{q}{RC}.
+\]
+Separate variables:
+\[
+\frac{dq}{q} = -\frac{dt}{RC}.
+\]
+Integrate from $0$ to $t$, with $q(0) = q_0$:
+\[
+\int_{q_0}^{q(t)}\frac{dq'}{q'} = -\frac{1}{RC}\int_0^t dt',
+\qquad
+\ln\!\left(\frac{q(t)}{q_0}\right) = -\frac{t}{RC}.
+\]
+Exponentiate:
+\[
+q(t) = q_0\,e^{-t/RC}.
+\]
+Differentiating gives the rate of change of charge:
+\[
+\frac{dq}{dt} = -\,\frac{q_0}{RC}\,e^{-t/RC}.
+\]
+The current flowing through the resistor (in the direction that discharges the capacitor) is $I = -\,dq/dt$, so
+\[
+I(t) = \frac{q_0}{RC}\,e^{-t/RC}.
+\]
+The voltage across the capacitor is $V_C = q/C = (q_0/C)e^{-t/RC}$, and the voltage across the resistor is $V_R = IR = (q_0/C)e^{-t/RC}$, so $V_R = V_C$ at every instant, consistent with the loop equation.}
+
+\nt{During charging, the current is positive and flows onto the positively charged plate of the capacitor. During discharging, the current flows off the capacitor, and the instantaneous current through the resistor has the same magnitude as the rate at which charge leaves the capacitor: $|dq/dt| = I$.}
+
+\dfn{Time constant of an RC circuit}{The \emph{time constant} of an RC circuit is
+\[
+\tau = RC,
+\]
+where $R$ is the resistance and $C$ is the capacitance. The SI unit of $\tau$ is the second (s). The time constant sets the characteristic timescale of the transient response: at $t = \tau$, the capacitor charge during charging reaches $1 - e^{-1} \approx 63.2\%$ of its final value $C\mathcal{E}$, and during discharging the charge falls to $e^{-1} \approx 36.8\%$ of its initial value $q_0$. After $5\tau$, the transient is essentially over: $1 - e^{-5} \approx 0.993$ of the final charge has been reached during charging, and $e^{-5} \approx 0.0067$ of the initial charge remains during discharging.}
+
+\nt{The time constant is the product of two quantities with SI units $\Omega$ (ohms) and F (farads). Since $\Omega = \mathrm{V/A}$ and $\mathrm{F} = \mathrm{C/V}$, the product is $(\mathrm{V/A})(\mathrm{C/V}) = \mathrm{C/A} = \mathrm{C/(C/s)} = \mathrm{s}$, confirming that $\tau$ has units of time. A larger resistance slows the charge flow, giving a longer time constant. A larger capacitance stores more charge per volt, also requiring more time to charge or discharge, giving a longer time constant.}
+
+\mprop{Charging and discharging at one time constant}{For a charging capacitor, at $t = \tau = RC$:
+\[
+q(\tau) = C\mathcal{E}\left(1 - \frac{1}{e}\right) \approx 0.632\,C\mathcal{E},
+\qquad
+I(\tau) = \frac{\mathcal{E}}{R}\cdot\frac{1}{e} \approx 0.368\,\frac{\mathcal{E}}{R}.
+\]
+For a discharging capacitor, at $t = \tau = RC$:
+\[
+q(\tau) = \frac{q_0}{e} \approx 0.368\,q_0,
+\qquad
+|I(\tau)| = \frac{q_0}{RC}\cdot\frac{1}{e} \approx 0.368\,\frac{q_0}{RC}.
+\]
+After $n$ time constants ($t = n\tau$):
+\[
+q_{\text{charge}} = C\mathcal{E}\left(1 - e^{-n}\right)
+\qquad \text{and} \qquad
+q_{\text{discharge}} = q_0\,e^{-n}.
+\]
+Thus, after $n = 5$, charging reaches $1 - e^{-5} \approx 99.3\%$ of full charge and discharging leaves only $e^{-5} \approx 0.67\%$ of the initial charge.}
+
+\nt{The time constant is independent of the initial conditions and of the battery emf. It depends only on the circuit geometry (through $R$ and $C$). This is a hallmark of first-order linear systems: the timescale of the exponential decay is set by the coefficients of the differential equation, not by the particular solution's initial values.}
+
+\thm{Energy during charging}{When an initially uncharged capacitor $C$ is charged through a resistor $R$ by a battery of emf $\mathcal{E}$, the total energy supplied by the battery is
+\[
+U_{\text{battery}} = C\mathcal{E}^2.
+\]
+The energy finally stored in the capacitor is
+\[
+U_C = \frac{1}{2}\,C\mathcal{E}^2.
+\]
+The remaining half,
+\[
+U_R = \frac{1}{2}\,C\mathcal{E}^2,
+\]
+is dissipated as Joule heat in the resistor during the charging process. The fraction dissipated in the resistor is exactly $50\%$, independent of the value of $R$.}
+
+\pf{Energy during charging}{The battery supplies energy at rate $dU_{\text{battery}}/dt = \mathcal{E}I(t)$, so the total energy delivered during charging is
+\[
+U_{\text{battery}} = \int_0^\infty \mathcal{E}\,I(t)\,dt = \mathcal{E}\int_0^\infty \frac{\mathcal{E}}{R}\,e^{-t/RC}\,dt.
+\]
+The integral is
+\[
+\int_0^\infty e^{-t/RC}\,dt = RC,
+\]
+so
+\[
+U_{\text{battery}} = \frac{\mathcal{E}^2}{R}\cdot RC = C\mathcal{E}^2.
+\]
+The energy stored in the capacitor at full charge ($q = C\mathcal{E}$) is
+\[
+U_C = \frac{q^2}{2C} = \frac{(C\mathcal{E})^2}{2C} = \frac{1}{2}\,C\mathcal{E}^2.
+\]
+By energy conservation, the energy dissipated in the resistor is
+\[
+U_R = U_{\text{battery}} - U_C = C\mathcal{E}^2 - \frac{1}{2}\,C\mathcal{E}^2 = \frac{1}{2}\,C\mathcal{E}^2.
+\]
+One can verify this directly:
+\[
+U_R = \int_0^\infty I(t)^2R\,dt = \int_0^\infty \frac{\mathcal{E}^2}{R}\,e^{-2t/RC}\,dt
+     = \frac{\mathcal{E}^2}{R}\cdot\frac{RC}{2} = \frac{1}{2}\,C\mathcal{E}^2.
+\]
+The result is independent of $R$, because a larger $R$ gives less current but a proportionally longer charging time.}
+
+\cor{Energy during discharging}{During complete discharge, the energy initially stored in the capacitor,
+\[
+U_{\text{initial}} = \frac{q_0^2}{2C},
+\]
+is entirely dissipated as Joule heat in the resistor:
+\[
+U_R = \frac{q_0^2}{2C}.
+\]
+This follows from
+\[
+U_R = \int_0^\infty I(t)^2R\,dt = \int_0^\infty \left(\frac{q_0}{RC}\,e^{-t/RC}\right)^{\!2}\!R\,dt
+    = \frac{q_0^2}{RC^2}\int_0^\infty e^{-2t/RC}\,dt = \frac{q_0^2}{RC^2}\cdot\frac{RC}{2} = \frac{q_0^2}{2C}.
+\]}
+
+\ex{Illustrative example}{If a capacitor of capacitance $C$ is charged through a resistor $R$ by a battery of emf $\mathcal{E}$, the time at which the current has dropped to half its initial value is found from $I(t) = (\mathcal{E}/R)e^{-t/RC} = (\mathcal{E}/2R)$, giving $e^{-t/RC} = 1/2$ and $t = RC\ln(2) \approx 0.693\,\tau$. At this instant, the charge on the capacitor is $q = C\mathcal{E}(1 - 1/2) = C\mathcal{E}/2$, exactly half of its final value. The energy stored in the capacitor is $\frac{1}{2}C(\mathcal{E}/2)^2 = \frac{1}{8}C\mathcal{E}^2 = 25\%$ of the energy that will ultimately be stored, while the battery has supplied $C\mathcal{E}\cdot(\mathcal{E}/2) = \frac{1}{2}C\mathcal{E}^2$, exactly half of the total energy it will supply.}
+
+\qs{Worked example}{A series RC circuit consists of a battery of emf
+\[
+\mathcal{E} = 12.0\,\mathrm{V},
+\]
+a resistor of resistance
+\[
+R = 2.00\,\mathrm{k\Omega} = 2.00 \times 10^3\,\Omega,
+\]
+and an initially uncharged capacitor of capacitance
+\[
+C = 3.00\,\mathrm{\mu F} = 3.00 \times 10^{-6}\,\mathrm{F}.
+\]
+At $t = 0$, a switch is closed connecting all three elements in series. Assume ideal wires and components.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the time constant $\tau$ of the circuit,
+\item the time $t_{1/2}$ at which the capacitor reaches $50.0\%$ of its final (maximum) charge, and
+\item the current $I(t_{1/2})$ at that instant.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The time constant is
+\[
+\tau = RC = \left(2.00 \times 10^3\,\Omega\right)\!\left(3.00 \times 10^{-6}\,\mathrm{F}\right) = 6.00 \times 10^{-3}\,\mathrm{s}.
+\]
+In more convenient units,
+\[
+\tau = 6.00\,\mathrm{ms}.
+\]
+
+\textbf{Part (b).} The charge on the capacitor during charging is
+\[
+q(t) = C\mathcal{E}\,\bigl(1 - e^{-t/\tau}\bigr).
+\]
+The final (maximum) charge is
+\[
+q_{\text{max}} = C\mathcal{E} = \left(3.00 \times 10^{-6}\,\mathrm{F}\right)\!\left(12.0\,\mathrm{V}\right) = 36.0 \times 10^{-6}\,\mathrm{C} = 36.0\,\mathrm{\mu C}.
+\]
+We want the time $t_{1/2}$ at which $q(t_{1/2}) = q_{\text{max}}/2$. Setting the charge equation equal to $q_{\text{max}}/2$:
+\[
+C\mathcal{E}\,\bigl(1 - e^{-t_{1/2}/\tau}\bigr) = \frac{C\mathcal{E}}{2}.
+\]
+Cancel $C\mathcal{E}$ and solve:
+\[
+1 - e^{-t_{1/2}/\tau} = \frac{1}{2},
+\qquad
+e^{-t_{1/2}/\tau} = \frac{1}{2}.
+\]
+Taking the natural logarithm:
+\[
+-\frac{t_{1/2}}{\tau} = \ln\!\left(\frac{1}{2}\right) = -\ln(2),
+\qquad
+t_{1/2} = \tau\ln(2).
+\]
+Substitute $\tau = 6.00\,\mathrm{ms}$:
+\[
+t_{1/2} = (6.00\,\mathrm{ms})\ln(2) = (6.00 \times 10^{-3}\,\mathrm{s})\,(0.6931) = 4.16 \times 10^{-3}\,\mathrm{s}.
+\]
+Thus,
+\[
+t_{1/2} = 4.16\,\mathrm{ms}.
+\]
+
+\textbf{Part (c).} The current during charging is
+\[
+I(t) = \frac{\mathcal{E}}{R}\,e^{-t/\tau}.
+\]
+At $t = t_{1/2}$, we found $e^{-t_{1/2}/\tau} = 1/2$, so
+\[
+I(t_{1/2}) = \frac{\mathcal{E}}{R}\cdot\frac{1}{2} = \frac{1}{2}\cdot\frac{12.0\,\mathrm{V}}{2.00 \times 10^3\,\Omega}.
+\]
+Compute the initial current:
+\[
+\frac{\mathcal{E}}{R} = \frac{12.0}{2.00 \times 10^3}\,\mathrm{A} = 6.00 \times 10^{-3}\,\mathrm{A} = 6.00\,\mathrm{mA}.
+\]
+Therefore,
+\[
+I(t_{1/2}) = \frac{6.00\,\mathrm{mA}}{2} = 3.00\,\mathrm{mA}.
+\]
+
+\textbf{Check.} At $t_{1/2} = \tau\ln(2)$, the charge is $q = q_{\text{max}}(1 - e^{-\ln 2}) = q_{\text{max}}(1 - 1/2) = q_{\text{max}}/2 = 18.0\,\mathrm{\mu C}$, and the current is $I = (\mathcal{E}/R)e^{-\ln 2} = (\mathcal{E}/R)(1/2) = 3.00\,\mathrm{mA}$. Both are consistent with the expected behavior of a charging RC circuit.
+
+Therefore,
+\[
+\tau = 6.00\,\mathrm{ms},
+\qquad
+t_{1/2} = 4.16\,\mathrm{ms},
+\qquad
+I(t_{1/2}) = 3.00\,\mathrm{mA}.
+\]

concepts/em/u11/e11-7-internal-resistance.tex

@@ -0,0 +1,189 @@
+\subsection{Internal Resistance and Measurement Devices}
+
+This subsection introduces the real-battery model (ideal emf in series with an internal resistance), derives the terminal-voltage relation under load, and discusses how practical measurement devices (ammeters and voltmeters) affect the circuits they measure.
+
+\dfn{Real battery (emf and internal resistance)}{A \emph{real battery} is modelled as an ideal electromotive-force source $\mathcal{E}$ in series with an \emph{internal resistance} $r$. The emf $\mathcal{E}$ represents the work per unit charge the battery can deliver when no current flows. The internal resistance $r$ accounts for the finite conductivity of the electrolyte, electrode materials, and other dissipative processes inside the battery. The SI unit of emf is the volt (V), which is dimensionally equivalent to $\mathrm{J/C}$.}
+
+\thm{Terminal voltage of a real battery}{Let a real battery with emf $\mathcal{E}$ and internal resistance $r$ deliver a current $I$ to an external load. The \emph{terminal voltage} $V$ across the battery's terminals is
+\[
+V = \mathcal{E} - Ir.
+\]
+When the battery is delivering current (discharging), the terminal voltage is \emph{less} than the emf by the voltage drop $Ir$ across the internal resistance. When $I = 0$ (open circuit), the terminal voltage equals the emf: $V = \mathcal{E}$. If the current is driven backwards through the battery (charging), then $I$ is negative and $V > \mathcal{E}$.}
+
+\pf{Derivation of the terminal-voltage relation}{Consider a real battery connected to an external load resistor $R$. The equivalent circuit consists of the emf $\mathcal{E}$, the internal resistance $r$, and the load $R$ all in series. By Kirchhoff's loop rule, traversing the loop in the direction of current $I$:
+\[
+\mathcal{E} - Ir - IR = 0.
+\]
+Solving for the current gives
+\[
+I = \frac{\mathcal{E}}{r + R}.
+\]
+The terminal voltage $V$ is the potential drop across the load, which also equals the potential drop from the battery's positive to negative terminal:
+\[
+V = IR = \mathcal{E} - Ir.
+\]
+This proves the relation $V = \mathcal{E} - Ir$.}
+
+\thm{Power delivered by a real battery}{The power $P$ delivered to an external load $R$ by a real battery of emf $\mathcal{E}$ and internal resistance $r$ is
+\[
+P = I^2 R = \mathcal{E}\,I - I^2 r,
+\]
+where $I = \mathcal{E}/(r + R)$. The first term $\mathcal{E}I$ is the total rate at which the battery converts chemical energy to electrical energy; the second term $I^2 r$ is the rate of internal dissipation as heat within the battery. The difference is the power delivered to the external circuit.}
+
+\thm{Power transfer theorem (maximum power transfer)}{For a real battery with fixed $\mathcal{E}$ and $r$ connected to a variable load $R$, the power delivered to the load is
+\[
+P(R) = \frac{\mathcal{E}^2\,R}{(r + R)^2}.
+\]
+This power is maximized when $R = r$, giving $P_{\max} = \mathcal{E}^2 / (4r)$.}
+
+\pf{Maximum power transfer}{From the preceding theorem, $P(R) = I^2 R$ with $I = \mathcal{E}/(r + R)$, so
+\[
+P(R) = \frac{\mathcal{E}^2 R}{(r + R)^2}.
+\]
+To find the maximum, differentiate with respect to $R$ and set the derivative to zero:
+\[
+\frac{dP}{dR} = \mathcal{E}^2 \,\frac{(r + R)^2 - R \cdot 2(r + R)}{(r + R)^4}
+= \mathcal{E}^2 \,\frac{(r + R) - 2R}{(r + R)^3}
+= \mathcal{E}^2 \,\frac{r - R}{(r + R)^3}.
+\]
+Setting $dP/dR = 0$ gives $R = r$. For $R < r$ the derivative is positive (power increases); for $R > r$ it is negative (power decreases). Hence $R = r$ is a maximum. Substituting $R = r$ into the power expression gives $P_{\max} = \mathcal{E}^2/(4r)$.}
+
+\cor{Short circuit and open circuit limits}{When the load is zero ($R = 0$, \emph{short circuit}), the current is $I_{\text{sc}} = \mathcal{E}/r$ and the terminal voltage is $V = 0$. All power is dissipated internally: $P_{\text{load}} = 0$, and the battery heats up. When the load is infinite ($R \to \infty$, \emph{open circuit}), the current is $I = 0$ and $V = \mathcal{E}$. No power is delivered.}
+
+\dfn{Ammeter}{An \emph{ammeter} measures the current through a branch of a circuit. It is inserted in \emph{series} with the branch. An ideal ammeter has zero resistance so it does not affect the circuit. A real ammeter has a small but finite resistance $R_A$ (the \emph{ammeter resistance}).}
+
+\dfn{Voltmeter}{A \emph{voltmeter} measures the potential difference between two points in a circuit. It is connected in \emph{parallel} between those points. An ideal voltmeter has infinite resistance so no current flows through it. A real voltmeter has a large but finite resistance $R_V$ (the \emph{voltmeter resistance}).}
+
+\thm{Loading effects of measurement devices}{When an ammeter of resistance $R_A$ is inserted in series with a circuit of total resistance $R_{\text{eq}}$ (not including $R_A$), the current measured is
+\[
+I_{\text{measured}} = \frac{I_{\text{true}}\,R_{\text{eq}}}{R_{\text{eq}} + R_A},
+\]
+where $I_{\text{true}}$ is the current that would flow without the ammeter. The reading is smaller than the true value by a factor of $R_{\text{eq}}/(R_{\text{eq}} + R_A)$.
+
+When a voltmeter of resistance $R_V$ is connected across a component of resistance $R$ that has voltage $V$ across it (without the voltmeter), the measured voltage is
+\[
+V_{\text{measured}} = V \,\frac{R\,R_V/(R + R_V)}{R\,R_V/(R + R_V) + R_{\text{series}}}
+= V \,\frac{R_V}{R_V + R_{\text{series}}\,(1 + R_V/R)}^{-1},
+\]
+where $R_{\text{series}}$ is the resistance in series with the component. In the common case where the component of interest is connected to a source with small internal resistance $r \ll R_V$, the correction is small: $V_{\text{measured}} \approx V\,(1 - r/R_V)$. The reading is smaller than the true voltage.}
+
+\thm{Efficiency of power transfer}{The \emph{efficiency} $\eta$ of a real battery delivering power to a load $R$ is the ratio of power delivered to the load to the total power generated by the emf:
+\[
+\eta = \frac{P_{\text{load}}}{P_{\text{total}}}
+= \frac{I^2 R}{\mathcal{E}I}
+= \frac{IR}{\mathcal{E}}
+= \frac{R}{R + r}.
+\]
+Efficiency increases as $R$ becomes large compared to $r$. When $R = r$, the efficiency is $50\%$: half the power is dissipated in the internal resistance.}
+
+\nt{An ideal ammeter ($R_A = 0$) does not change the current in the branch it measures. A real ammeter always slightly \emph{reduces} the current. An ideal voltmeter ($R_V = \infty$) draws no current and does not disturb the circuit. A real voltmeter always slightly \emph{lowers} the voltage it is measuring because it provides an additional parallel current path. In well-designed circuits, $R_A \ll R_{\text{branch}}$ and $R_V \gg R_{\text{parallel}}$, so these perturbations are negligible.}
+
+\ex{Illustrative example}{A $9\,\mathrm{V}$ battery with internal resistance $r = 1\,\Omega$ is connected to a load $R = 8\,\Omega$. The current is $I = 9\,\mathrm{V}/(1\,\Omega + 8\,\Omega) = 1.0\,\mathrm{A}$, and the terminal voltage is $V = 9\,\mathrm{V} - (1.0\,\mathrm{A})(1\,\Omega) = 8\,\mathrm{V}$.}
+
+\qs{Worked example}{A battery has an emf
+\[
+\mathcal{E} = 12.0\,\mathrm{V}
+\]
+and an internal resistance
+\[
+r = 2.0\,\Omega.
+\]
+The battery is connected to an external load resistor $R = 10.0\,\Omega$, as shown in the circuit diagram below.
+
+\begin{center}
+\emph{(Diagram description: A single loop consisting of an ideal emf source $\mathcal{E}$, an internal resistor $r$ in series, and an external load resistor $R$ in series, forming a closed circuit.)}
+\end{center}
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the current $I$ in the circuit,
+\item the terminal voltage $V$ across the battery,
+\item the power $P_{\text{load}}$ delivered to the load resistor,
+\item the power $P_{\text{int}}$ dissipated in the internal resistance,
+\item the total power $P_{\text{total}}$ generated by the emf,
+\item the efficiency $\eta$ of the battery, and
+\item the value of $R$ that maximizes the power delivered to the load.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The circuit consists of $\mathcal{E}$, $r$, and $R$ in series. By Kirchhoff's loop rule,
+\[
+\mathcal{E} - Ir - IR = 0,
+\]
+so
+\[
+I = \frac{\mathcal{E}}{r + R}
+= \frac{12.0\,\mathrm{V}}{2.0\,\Omega + 10.0\,\Omega}
+= \frac{12.0\,\mathrm{V}}{12.0\,\Omega}
+= 1.0\,\mathrm{A}.
+\]
+
+\textbf{Part (b).} The terminal voltage is
+\[
+V = \mathcal{E} - Ir
+= 12.0\,\mathrm{V} - (1.0\,\mathrm{A})(2.0\,\Omega)
+= 12.0\,\mathrm{V} - 2.0\,\mathrm{V}
+= 10.0\,\mathrm{V}.
+\]
+Equivalently, $V = IR = (1.0\,\mathrm{A})(10.0\,\Omega) = 10.0\,\mathrm{V}$, which confirms the result.
+
+\textbf{Part (c).} The power delivered to the load resistor is
+\[
+P_{\text{load}} = I^2 R
+= (1.0\,\mathrm{A})^2(10.0\,\Omega)
+= 10.0\,\mathrm{W}.
+\]
+Alternatively, $P_{\text{load}} = VI = (10.0\,\mathrm{V})(1.0\,\mathrm{A}) = 10.0\,\mathrm{W}$, giving the same result.
+
+\textbf{Part (d).} The power dissipated in the internal resistance is
+\[
+P_{\text{int}} = I^2 r
+= (1.0\,\mathrm{A})^2(2.0\,\Omega)
+= 2.0\,\mathrm{W}.
+\]
+
+\textbf{Part (e).} The total power generated by the emf is
+\[
+P_{\text{total}} = \mathcal{E} I
+= (12.0\,\mathrm{V})(1.0\,\mathrm{A})
+= 12.0\,\mathrm{W}.
+\]
+Check: $P_{\text{total}} = P_{\text{load}} + P_{\text{int}} = 10.0\,\mathrm{W} + 2.0\,\mathrm{W} = 12.0\,\mathrm{W}$, so energy is conserved.
+
+\textbf{Part (f).} The efficiency of the battery is
+\[
+\eta = \frac{P_{\text{load}}}{P_{\text{total}}}
+= \frac{10.0\,\mathrm{W}}{12.0\,\mathrm{W}}
+= 0.833
+= 83.3\%.
+\]
+Equivalently, $\eta = R/(R + r) = 10.0\,\Omega/(10.0\,\Omega + 2.0\,\Omega) = 10/12 = 0.833 = 83.3\%$.
+
+\textbf{Part (g).} By the maximum power transfer theorem, the power delivered to the load is maximized when the load resistance equals the internal resistance:
+\[
+R = r = 2.0\,\Omega.
+\]
+At this value, the maximum power delivered to the load is
+\[
+P_{\max} = \frac{\mathcal{E}^2}{4r}
+= \frac{(12.0\,\mathrm{V})^2}{4(2.0\,\Omega)}
+= \frac{144\,\mathrm{V}^2}{8.0\,\Omega}
+= 18.0\,\mathrm{W}.
+\]
+
+Therefore,
+\[
+I = 1.0\,\mathrm{A},
+\qquad
+V = 10.0\,\mathrm{V},
+\qquad
+P_{\text{load}} = 10.0\,\mathrm{W},
+\]
+\[
+P_{\text{int}} = 2.0\,\mathrm{W},
+\qquad
+P_{\text{total}} = 12.0\,\mathrm{W},
+\qquad
+\eta = 83.3\%,
+\qquad
+R_{\text{max power}} = 2.0\,\Omega.
+\]

concepts/em/u12/e12-1-magnetic-force-charge.tex

@@ -0,0 +1,128 @@
+\subsection{Magnetic Force on a Moving Charge}
+
+A charge that moves through a magnetic field experiences a force perpendicular to both its velocity and the field. This is the magnetic part of the Lorentz force. Unlike the electric force, the magnetic force acts only on moving charges and is always perpendicular to the direction of motion.
+
+\dfn{Magnetic force on a point charge}{A particle of charge $q$ moving with velocity $\vec{v}$ through a magnetic field $\vec{B}$ experiences a magnetic force
+\[
+\vec{F}_B = q\,\vec{v}\times\vec{B}.
+\]
+The magnitude of this force is
+\[
+F_B = |q|\,v\,B\,\sin\theta,
+\]
+where $\theta$ is the angle between the vectors $\vec{v}$ and $\vec{B}$ measured in the plane they span. The direction of $\vec{F}_B$ is perpendicular to that plane and is determined by the right-hand rule, reversed for negative charge.}
+
+\nt{The magnetic force vanishes when the charge is at rest ($\vec{v}=\vec{0}$), when $\vec{v}$ is parallel or antiparallel to $\vec{B}$ ($\theta=0^\circ$ or $180^\circ$), or when $B=0$. The force is maximal when $\vec{v}\perp\vec{B}$ ($\theta=90^\circ$).}
+
+\thm{Lorentz magnetic force law}{Let $q$ be the charge of a particle, $\vec{v}$ its velocity vector, and $\vec{B}$ the magnetic field at the particle's position. Then the magnetic force on the particle is
+\[
+\vec{F}_B = q\,\vec{v}\times\vec{B}.
+\]
+\begin{itemize}
+\item \textbf{Magnitude:} $F_B = |q|\,v\,B\,\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.
+\item \textbf{Direction:} Point the fingers of your right hand along $\vec{v}$, then curl them toward $\vec{B}$. Your thumb points in the direction of $\vec{F}_B$ if $q>0$. If $q<0$, the force is opposite to your thumb.
+\item \textbf{SI unit of $B$:} The tesla, $\mathrm{T} = \dfrac{\mathrm{N}}{\mathrm{C}\cdot\mathrm{m/s}} = \dfrac{\mathrm{N}}{\mathrm{A}\cdot\mathrm{m}} = \dfrac{\mathrm{kg}}{\mathrm{C}\cdot\mathrm{s}}$.
+\end{itemize}}
+
+\pf{Lorentz magnetic force law from cross-product geometry}{The vector cross product $\vec{v}\times\vec{B}$ is defined to have magnitude $vB\sin\theta$ and direction given by the right-hand rule. Multiplying by $q$ scales the magnitude by $|q|$ and reverses direction if $q<0$. Thus
+\[
+\vec{F}_B = q\,(\vec{v}\times\vec{B})
+\]
+has magnitude $|q|vB\sin\theta$ and the correct directional behaviour. This is the experimentally determined magnetic force law for a point charge.}
+
+\cor{Charge at rest or parallel to field}{When $\vec{v}=\vec{0}$ or when $\vec{v}\parallel\vec{B}$, we have $\sin\theta=0$ and therefore $F_B=0$. The magnetic field exerts no force on a stationary charge or on a charge moving exactly along the field lines.}
+
+\mprop{Magnetic force vs.\ electric force}{For the same charge $q$ placed in both an electric field $\vec{E}$ and a magnetic field $\vec{B}$, the total Lorentz force is
+\[
+\vec{F} = q\,\vec{E} + q\,\vec{v}\times\vec{B}.
+\]
+Key differences:
+\begin{itemize}
+\item $\vec{F}_E$ points parallel (or antiparallel) to $\vec{E}$, regardless of motion.
+\item $\vec{F}_B$ is always perpendicular to $\vec{v}$, so it does zero work on the charge.
+\item $\vec{F}_B$ vanishes when $\vec{v}=\vec{0}$; $\vec{F}_E$ does not.
+\end{itemize}}
+
+\thm{Magnetic force does no work}{Since $\vec{F}_B\perp\vec{v}$ at every instant, the instantaneous power delivered by the magnetic force is
+\[
+P = \vec{F}_B\cdot\vec{v} = q\,(\vec{v}\times\vec{B})\cdot\vec{v} = 0.
+\]
+The magnetic force can change the direction of a particle's velocity but never its kinetic energy. This is the mathematical expression of the scalar triple-product identity $(\vec{a}\times\vec{b})\cdot\vec{a}=0$.}
+
+\ex{Illustrative example}{When a charged particle enters a uniform magnetic field perpendicularly, it follows a circular path. The magnetic force provides the centripetal force:
+\[
+|q|\,v\,B = \frac{m\,v^2}{R} \quad\Rightarrow\quad R = \frac{m\,v}{|q|\,B},
+\]
+where $R$ is the radius of the circular path. The period of revolution is
+\[
+T = \frac{2\pi R}{v} = \frac{2\pi m}{|q|\,B},
+\]
+which is independent of the particle's speed. This circular-motion analysis is developed fully in section~E12.2.}
+
+\qs{Worked example}{A proton (charge $q=+1.60\times 10^{-19}\,\mathrm{C}$, mass $m=1.67\times 10^{-27}\,\mathrm{kg}$) moves with speed $v=3.0\times 10^6\,\mathrm{m/s}$ in the $+\hat{\imath}$ direction. It enters a region with a uniform magnetic field $\vec{B}=(0.50\,\mathrm{T})\,\hat{\jmath}$. The proton's velocity is perpendicular to the field.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the magnitude of the magnetic force on the proton,
+\item the direction of the magnetic force as a unit vector, and
+\item the initial acceleration vector of the proton.
+\end{enumerate}}
+
+\sol \textbf{(a) Force magnitude.} The magnetic force is $\vec{F}_B=q\,\vec{v}\times\vec{B}$. With $\vec{v}=v\,\hat{\imath}$ and $\vec{B}=B\,\hat{\jmath}$, the angle between them is $\theta=90^\circ$ and $\sin\theta=1$. The magnitude is
+\[
+F_B = |q|\,v\,B\,\sin 90^\circ = |q|\,v\,B.
+\]
+Substitute the given values:
+\[
+F_B = (1.60\times 10^{-19}\,\mathrm{C})(3.0\times 10^6\,\mathrm{m/s})(0.50\,\mathrm{T}).
+\]
+Compute step by step:
+\[
+(1.60\times 10^{-19})(3.0\times 10^6) = 4.8\times 10^{-13},
+\]
+\[
+(4.8\times 10^{-13})(0.50) = 2.4\times 10^{-13}\,\mathrm{N}.
+\]
+Thus
+\[
+F_B = 2.4\times 10^{-13}\,\mathrm{N}.
+\]
+
+\textbf{(b) Force direction.} Use the cross product directly:
+\[
+\vec{F}_B = q\,(\vec{v}\times\vec{B}) = q\,(v\,\hat{\imath})\times(B\,\hat{\jmath}) = q\,v\,B\,(\hat{\imath}\times\hat{\jmath}).
+\]
+Since $\hat{\imath}\times\hat{\jmath}=\hat{k}$,
+\[
+\vec{F}_B = q\,v\,B\,\hat{k}.
+\]
+Because $q>0$ and $v,B>0$, the force points in the $+\hat{k}$ direction. In unit-vector form:
+\[
+\vec{F}_B = (2.4\times 10^{-13}\,\mathrm{N})\,\hat{k}.
+\]
+The right-hand rule confirms this: fingers along $+\hat{\imath}$, curl toward $+\hat{\jmath}$, thumb points along $+\hat{k}$.
+
+\textbf{(c) Acceleration vector.} By Newton's second law,
+\[
+\vec{a} = \frac{\vec{F}_B}{m}.
+\]
+Substitute:
+\[
+\vec{a} = \frac{(2.4\times 10^{-13}\,\mathrm{N})\,\hat{k}}{1.67\times 10^{-27}\,\mathrm{kg}}.
+\]
+Compute the magnitude:
+\[
+\frac{2.4\times 10^{-13}}{1.67\times 10^{-27}} = \frac{2.4}{1.67}\times 10^{14} \approx 1.44\times 10^{14}\,\mathrm{m/s^2}.
+\]
+Thus
+\[
+\vec{a} = (1.44\times 10^{14}\,\mathrm{m/s^2})\,\hat{k}.
+\]
+
+\bigskip
+\textbf{Final answers:}
+\begin{enumerate}[label=(\alph*)]
+\item $F_B = 2.4\times 10^{-13}\,\mathrm{N}$
+\item $\vec{F}_B = (2.4\times 10^{-13}\,\mathrm{N})\,\hat{k}$
+\item $\vec{a} = (1.44\times 10^{14}\,\mathrm{m/s^2})\,\hat{k}$
+\end{enumerate}

concepts/em/u12/e12-2-particle-motion-in-b.tex

@@ -0,0 +1,190 @@
+\subsection{Circular and Helical Motion in a Uniform Magnetic Field}
+
+When a charged particle moves through a uniform magnetic field, the magnetic force acts as a centripetal force, bending the particle's path. The resulting motion depends on the angle between the velocity $\vec{v}$ and the field $\vec{B}$: perpendicular entry produces circular motion, while oblique entry produces helical motion.
+
+\dfn{Circular motion of a charge in a uniform B-field}{Let a particle of mass $m$ and charge $q$ move with speed $v$ through a uniform magnetic field $\vec{B}$, with $\vec{v}\perp\vec{B}$. The magnetic force has constant magnitude $|q|vB$ and is always perpendicular to $\vec{v}$, so it provides the centripetal force for uniform circular motion:
+\[
+|q|\,v\,B=\frac{m\,v^{2}}{R}.
+\]
+Solving for the radius,
+\[
+R=\frac{m\,v}{|q|\,B}=\frac{m\,v_{\perp}}{|q|\,B},
+\]
+where $v_{\perp}=v$ is the perpendicular speed. The period of revolution is
+\[
+T=\frac{2\pi R}{v}=\frac{2\pi m}{|q|\,B}.
+\]
+The frequency of revolution (cyclotron frequency) is
+\[
+f=\frac{1}{T}=\frac{|q|\,B}{2\pi m},
+\]
+and the angular frequency is
+\[
+\omega=2\pi f=\frac{|q|\,B}{m}.
+\]
+Note that $T$, $f$, and $\omega$ are independent of the particle's speed.}
+
+\nt{Because the magnetic force does no work (it is always perpendicular to $\vec{v}$), the particle's speed and kinetic energy remain constant throughout the circular motion. The magnetic field only changes the direction of the velocity, not its magnitude. This is why the radius depends on $v$ but the period does not.}
+
+\thm{Cyclotron motion}{Let $m$ be the mass and $q$ the charge of a particle entering a uniform magnetic field $\vec{B}$ with velocity component $v_{\perp}$ perpendicular to $\vec{B}$. The particle undergoes uniform circular motion in the plane perpendicular to $\vec{B}$ with:
+\begin{itemize}
+\item \textbf{Radius:} $R=\dfrac{m\,v_{\perp}}{|q|\,B}$
+\item \textbf{Period:} $T=\dfrac{2\pi m}{|q|\,B}$
+\item \textbf{Cyclotron frequency:} $f=\dfrac{|q|\,B}{2\pi m}$
+\item \textbf{Angular frequency:} $\omega=\dfrac{|q|\,B}{m}$
+\end{itemize}
+The sense of rotation is counter-clockwise for $q>0$ and clockwise for $q<0$ when viewing along the direction of $\vec{B}$.}
+
+\pf{Cyclotron motion from Newton's second law}{The magnetic force on the particle is $\vec{F}_B=q\,\vec{v}\times\vec{B}$. When $\vec{v}\perp\vec{B}$, the force magnitude is $F_B=|q|vB$ and its direction is always perpendicular to $\vec{v}$ (toward the center of curvature). For uniform circular motion, Newton's second law requires
+\[
+F_{\text{net}}=\frac{m\,v^{2}}{R}.
+\]
+Equating the magnetic force to the required centripetal force:
+\[
+|q|\,v\,B=\frac{m\,v^{2}}{R}.
+\]
+Solving for $R$:
+\[
+R=\frac{m\,v^{2}}{|q|\,v\,B}=\frac{m\,v}{|q|\,B}.
+\]
+The period is the circumference divided by the speed:
+\[
+T=\frac{2\pi R}{v}=\frac{2\pi}{v}\cdot\frac{m\,v}{|q|\,B}=\frac{2\pi m}{|q|\,B}.
+\]
+Thus $T$ is independent of $v$ and $R$. The angular frequency is $\omega=2\pi/T=|q|B/m$, known as the cyclotron angular frequency.}
+
+\dfn{Helical motion of a charge in a uniform B-field}{Let a particle of mass $m$ and charge $q$ move with speed $v$ through a uniform magnetic field $\vec{B}$, with the velocity making an angle $\alpha$ with $\vec{B}$ (where $0^\circ<\alpha<90^\circ$). Decompose the velocity into components parallel and perpendicular to $\vec{B}$:
+\[
+v_{\parallel}=v\cos\alpha,\qquad v_{\perp}=v\sin\alpha.
+\]
+The perpendicular component produces circular motion with radius
+\[
+R=\frac{m\,v_{\perp}}{|q|\,B}=\frac{m\,v\,\sin\alpha}{|q|\,B}.
+\]
+The parallel component is unaffected by the magnetic force and produces uniform linear motion along $\vec{B}$. The combination is helical motion.
+
+The \emph{pitch} $p$ of the helix is the distance advanced along $\vec{B}$ during one full revolution:
+\[
+p=v_{\parallel}\,T=(v\cos\alpha)\cdot\frac{2\pi m}{|q|\,B}=\frac{2\pi m\,v\,\cos\alpha}{|q|\,B}.
+\]
+
+The sense of the circular rotation follows the same rule as cyclotron motion: counter-clockwise for $q>0$ and clockwise for $q<0$, when viewing along $\vec{B}$.}
+
+\nt{If $\alpha=0^\circ$, then $v_{\perp}=0$ and the particle travels in a straight line along $\vec{B}$ (no magnetic force). If $\alpha=90^\circ$, then $v_{\parallel}=0$ and the particle undergoes pure circular motion (no drift along $\vec{B}$). Helical motion interpolates between these two extremes. The pitch increases as $\alpha\to 0^\circ$ and approaches zero as $\alpha\to 90^\circ$.}
+
+\mprop{Cyclotron and helical motion parameters}{For a particle of mass $m$ and charge $q$ in a uniform magnetic field $\vec{B}$, with velocity $\vec{v}$ at angle $\alpha$ to $\vec{B}$:
+\begin{align}
+R&=\frac{m\,v\,\sin\alpha}{|q|\,B} && \text{(helix radius)} \\
+T&=\frac{2\pi m}{|q|\,B} && \text{(period of revolution, independent of }v\text{ and }\alpha)\\
+f&=\frac{|q|\,B}{2\pi m} && \text{(cyclotron frequency)} \\
+p&=\frac{2\pi m\,v\,\cos\alpha}{|q|\,B} && \text{(helix pitch)}
+\end{align}}
+
+\mprop{Magnetic force does no work}{The magnetic force $\vec{F}_B=q\,\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$, so
+\[
+P=\vec{F}_B\cdot\vec{v}=0\qquad\text{and}\qquad\Delta K=0.
+\]
+The kinetic energy and speed of the particle remain constant. This holds for both pure circular motion and helical motion.}
+
+\ex{Illustrative example}{An electron and a proton, each with the same speed $v$, enter perpendicular to the same uniform magnetic field. The proton's radius is larger by the mass ratio $m_p/m_e\approx 1836$, but both complete one revolution in the same time $T=2\pi m/|q|B$, because the proton's period is 1836 times longer due to its mass but it travels a proportionally longer path (radius 1836 times larger), so the time cancels out.}
+
+\qs{Worked example}{An electron (mass $m_e=9.11\times 10^{-31}\,\mathrm{kg}$, charge $q=-1.60\times 10^{-19}\,\mathrm{C}$) enters a region with a uniform magnetic field $\vec{B}=(0.040\,\mathrm{T})\,\hat{\jmath}$. At the moment it enters, its velocity is
+\[
+\vec{v}=(4.0\times 10^{5}\,\mathrm{m/s})\,\hat{\imath}+(2.0\times 10^{5}\,\mathrm{m/s})\,\hat{k}.
+\]
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the radius of the helical path,
+\item the period of revolution,
+\item the pitch of the helix, and
+\item the sense of rotation (clockwise or counter-clockwise when viewing along $+\vec{B}$).
+\end{enumerate}}
+
+\sol \textbf{(a) Radius of the helix.} Decompose the velocity into components parallel and perpendicular to $\vec{B}$ (which points along $\hat{\jmath}$):
+\[
+v_{\parallel}=v_{k}=2.0\times 10^{5}\,\mathrm{m/s},\qquad v_{\perp}=\sqrt{v_{i}^{2}+v_{j}^{2}}=\sqrt{(4.0\times 10^{5})^{2}+0^{2}}=4.0\times 10^{5}\,\mathrm{m/s}.
+\]
+Note that $v_j=0$, so the entire $x$-component is perpendicular to $\vec{B}$.
+
+The radius of the helical path is
+\[
+R=\frac{m\,v_{\perp}}{|q|\,B}.
+\]
+Substitute the values:
+\[
+R=\frac{(9.11\times 10^{-31}\,\mathrm{kg})(4.0\times 10^{5}\,\mathrm{m/s})}{(1.60\times 10^{-19}\,\mathrm{C})(0.040\,\mathrm{T})}.
+\]
+Compute the numerator:
+\[
+(9.11\times 10^{-31})(4.0\times 10^{5})=3.644\times 10^{-25}\,\mathrm{kg{\cdot}m/s}.
+\]
+Compute the denominator:
+\[
+(1.60\times 10^{-19})(0.040)=6.4\times 10^{-21}\,\mathrm{C{\cdot}T}.
+\]
+Thus
+\[
+R=\frac{3.644\times 10^{-25}}{6.4\times 10^{-21}}\,\mathrm{m}=5.69\times 10^{-5}\,\mathrm{m}.
+\]
+So
+\[
+R=5.69\times 10^{-5}\,\mathrm{m}=56.9\,\mathrm{\mu m}.
+\]
+
+\textbf{(b) Period of revolution.} The period depends only on the particle's properties and the field strength:
+\[
+T=\frac{2\pi m}{|q|\,B}.
+\]
+Substitute:
+\[
+T=\frac{2\pi(9.11\times 10^{-31}\,\mathrm{kg})}{(1.60\times 10^{-19}\,\mathrm{C})(0.040\,\mathrm{T})}.
+\]
+The denominator is $6.4\times 10^{-21}\,\mathrm{C{\cdot}T}$ as computed above. The numerator is
+\[
+2\pi(9.11\times 10^{-31})=5.724\times 10^{-30}\,\mathrm{kg}.
+\]
+Thus
+\[
+T=\frac{5.724\times 10^{-30}}{6.4\times 10^{-21}}\,\mathrm{s}=8.94\times 10^{-10}\,\mathrm{s}.
+\]
+So
+\[
+T=8.94\times 10^{-10}\,\mathrm{s}=0.894\,\mathrm{ns}.
+\]
+
+\textbf{(c) Pitch of the helix.} The pitch is the distance advanced along $\vec{B}$ in one period:
+\[
+p=v_{\parallel}\,T.
+\]
+Substitute:
+\[
+p=(2.0\times 10^{5}\,\mathrm{m/s})(8.94\times 10^{-10}\,\mathrm{s}).
+\]
+Compute:
+\[
+p=1.788\times 10^{-4}\,\mathrm{m}.
+\]
+So
+\[
+p=1.79\times 10^{-4}\,\mathrm{m}=179\,\mathrm{\mu m}.
+\]
+
+\textbf{(d) Sense of rotation.} The magnetic field points in the $+\hat{\jmath}$ direction. To determine the sense of rotation, note the initial force on the electron at the entry point. The velocity has a $+x$ component, so at $t=0$ the perpendicular velocity is $\vec{v}_{\perp}=(4.0\times 10^{5}\,\mathrm{m/s})\,\hat{\imath}$. The magnetic force is
+\[
+\vec{F}=q\,(\vec{v}\times\vec{B})=(-1.60\times 10^{-19})\,[(4.0\times 10^{5}\,\hat{\imath})\times(0.040\,\hat{\jmath})].
+\]
+Since $\hat{\imath}\times\hat{\jmath}=\hat{k}$,
+\[
+\vec{F}=(-1.60\times 10^{-19})(4.0\times 10^{5})(0.040)\,\hat{k}=-(2.56\times 10^{-15}\,\mathrm{N})\,\hat{k}.
+\]
+The initial force points in the $-z$ direction, meaning the electron curves toward $-z$ from its initial $+x$ direction. Viewing along $+\hat{\jmath}$ (the direction of $\vec{B}$), the $+x$ axis is to the right and the $+z$ axis points toward you. Starting at $+x$ and curving toward $-z$, the electron moves clockwise. This is consistent with the general rule: for negative charge, the rotation is clockwise when viewing along $\vec{B}$.
+
+\bigskip
+\textbf{Final answers:}
+\begin{enumerate}[label=(\alph*)]
+\item $R=5.69\times 10^{-5}\,\mathrm{m}=56.9\,\mathrm{\mu m}$
+\item $T=8.94\times 10^{-10}\,\mathrm{s}=0.894\,\mathrm{ns}$
+\item $p=1.79\times 10^{-4}\,\mathrm{m}=179\,\mathrm{\mu m}$
+\item Clockwise (viewing along $+\hat{\jmath}$)
+\end{enumerate}

concepts/em/u12/e12-3-force-on-current.tex

@@ -0,0 +1,217 @@
+\subsection{Force on Current-Carrying Conductors and Loops}
+
+A current-carrying wire in a magnetic field experiences a force because the moving charge carriers inside the wire each feel a magnetic force. When the current flows through a closed loop, the forces on individual segments can produce a net torque, causing the loop to rotate. This is the operating principle of electric motors.
+
+\dfn{Force on a current-carrying wire}{A wire carrying current $I$ and placed in a magnetic field $\vec{B}$ experiences a magnetic force. For a straight wire segment of length $\ell$ carrying current $I$, with the vector $\vec{\ell}$ pointing in the direction of the current, the force is
+\[
+\vec{F} = I\,\vec{\ell}\times\vec{B}.
+\]
+The magnitude of this force is
+\[
+F = I\,\ell\,B\,\sin\theta,
+\]
+where $\theta$ is the angle between the current direction (the direction of $\vec{\ell}$) and the magnetic field $\vec{B}$. The direction is given by the right-hand rule for cross products, reversed for negative current carriers.}
+
+\nt{The vector $\vec{\ell}$ has magnitude equal to the length of the wire segment and points along the wire in the direction of conventional current. In a curved wire, the total force is obtained by integrating the differential-force expression over the entire path: $\vec{F} = \displaystyle\oint I\,d\vec{\ell}\times\vec{B}$.}
+
+\thm{Magnetic force on a current-carrying conductor}{Let a wire carry steady current $I$ through a magnetic field $\vec{B}$.
+\begin{itemize}
+\item \textbf{Straight wire:} For a straight wire segment of length $\ell$, with $\vec{\ell}$ pointing along the wire in the current direction,
+\[
+\vec{F} = I\,\vec{\ell}\times\vec{B}.
+\]
+The magnitude is $F = I\,\ell\,B\,\sin\theta$, where $\theta$ is the angle between $\vec{\ell}$ and $\vec{B}$.
+\item \textbf{Differential element:} For an arbitrary wire path, the force on a differential element $d\vec{\ell}$ is
+\[
+d\vec{F} = I\,d\vec{\ell}\times\vec{B},
+\]
+and the total force is
+\[
+\vec{F} = \int_{\text{wire}} I\,d\vec{\ell}\times\vec{B}.
+\]
+\item \textbf{Right-hand rule:} Point your right hand's fingers along $\vec{\ell}$ (current direction), then curl them toward $\vec{B}$. Your thumb gives the direction of $\vec{F}$.
+\end{itemize}}
+
+\pf{Force on a current-carrying wire from the force on moving charges}{The force on a single charge $q$ moving with drift velocity $\vec{v}_d$ is $\vec{F}_q = q\,\vec{v}_d\times\vec{B}$. In a wire segment of length $\ell$ and cross-sectional area $A$, the number of charge carriers is $N = n\,A\,\ell$, where $n$ is the carrier number density. The total force is
+\[
+\vec{F} = N\,q\,\vec{v}_d\times\vec{B} = n\,A\,\ell\,q\,\vec{v}_d\times\vec{B}.
+\]
+The current is $I = n\,q\,v_d\,A$, and the direction of $\vec{v}_d$ for positive carriers is the current direction. Letting $\vec{\ell}$ point in that direction with magnitude $\ell$, we have $n\,q\,\vec{v}_d = (I/A)\,\hat{\ell}$, and so
+\[
+\vec{F} = I\,\vec{\ell}\times\vec{B},
+\]
+where we used $\vec{\ell} = \ell\,\hat{\ell}$. This derivation confirms that the macroscopic force on a current-carrying wire follows directly from the Lorentz force on individual charge carriers.}
+
+\cor{Straight wire parallel or perpendicular to field}{When the wire is parallel or antiparallel to $\vec{B}$ ($\theta = 0^\circ$ or $180^\circ$), then $\sin\theta = 0$ and $F = 0$. When the wire is perpendicular to $\vec{B}$ ($\theta = 90^\circ$), the force is maximal: $F = I\,\ell\,B$.}
+
+\cor{Closed loop in a uniform field}{When a closed current loop sits entirely in a uniform magnetic field, the net force is zero:
+\[
+\vec{F}_{\text{net}} = I\oint d\vec{\ell}\times\vec{B} = I\left(\oint d\vec{\ell}\right)\times\vec{B} = \vec{0}.
+\]
+The integral $\oint d\vec{\ell}$ around any closed loop is the zero vector. Thus, no net force acts on a closed loop in a uniform field, though individual segments still feel forces.}
+
+\dfn{Magnetic dipole moment of a current loop}{A planar loop carrying current $I$ with enclosed area $A$ has a \emph{magnetic dipole moment}
+\[
+\vec{\mu} = N\,I\,A\,\hat{n},
+\]
+where $N$ is the number of turns in the loop, $A$ is the area enclosed by one turn, and $\hat{n}$ is a unit vector perpendicular to the plane of the loop. The direction of $\hat{n}$ is given by the right-hand rule: curl the fingers of your right hand around the loop in the direction of the current, and your thumb points along $\hat{n}$.}
+
+\mprop{Torque and potential energy of a current loop in a uniform field}{Let a planar current loop with magnetic dipole moment $\vec{\mu} = N I A\,\hat{n}$ be placed in a uniform magnetic field $\vec{B}$. Then:
+\begin{enumerate}
+\item The torque on the loop is
+\[
+\vec{\tau} = \vec{\mu}\times\vec{B}.
+\]
+The magnitude is
+\[
+\tau = \mu\,B\,\sin\phi = N\,I\,A\,B\,\sin\phi,
+\]
+where $\phi$ is the angle between $\hat{n}$ and $\vec{B}$.
+\item The potential energy of the loop is
+\[
+U = -\vec{\mu}\cdot\vec{B} = -\mu\,B\,\cos\phi = -N\,I\,A\,B\,\cos\phi.
+\]
+\end{enumerate}
+Equilibrium occurs when $\phi = 0^\circ$ (stable, $\hat{n}\parallel\vec{B}$, torque zero, energy minimum) or $\phi = 180^\circ$ (unstable, $\hat{n}$ antiparallel to $\vec{B}$, torque zero, energy maximum).}
+
+\pf{Torque on a current loop in a uniform field}{Consider a rectangular loop of width $a$ (in the $x$-direction) and height $b$ (in the $y$-direction), carrying current $I$ in a uniform field $\vec{B}=B\,\hat{k}$ (out of the plane). Let the normal $\hat{n}$ to the loop make angle $\phi$ with $\hat{k}$. The loop rotates about an axis through its center, perpendicular to $\vec{B}$.
+
+The forces on the top and bottom segments (length $a$) are equal and opposite and collinear, so they cancel. The forces on the two side segments (length $b$) are
+\[
+\vec{F}_1 = I\,\vec{b}_1\times\vec{B}
+\quad\text{and}\quad
+\vec{F}_2 = I\,\vec{b}_2\times\vec{B},
+\]
+with $\vec{b}_1 = -\vec{b}_2$. These forces have magnitude $F = I\,b\,B$ and act at perpendicular distances $(a/2)\,\sin\phi$ from the axis. Each produces a torque of magnitude
+\[
+\tau_{\text{side}} = F\cdot\frac{a}{2}\,\sin\phi = I\,b\,B\cdot\frac{a}{2}\,\sin\phi.
+\]
+Both sides contribute in the same rotational sense, so
+\[
+\tau = 2\cdot I\,b\,B\cdot\frac{a}{2}\,\sin\phi = I\,a\,b\,B\,\sin\phi.
+\]
+Since $A = a\,b$ and $\vec{\mu}$ has magnitude $\mu = I A$ (for $N=1$),
+\[
+\tau = \mu\,B\,\sin\phi.
+\]
+The vector form is $\vec{\tau} = \vec{\mu}\times\vec{B}$. For $N$ turns, multiply by $N$.
+
+For the potential energy, the torque tends to align $\hat{n}$ with $\vec{B}$. The work done by an external agent rotating the loop from angle $\phi_0$ to $\phi$ equals the change in potential energy:
+\[
+\Delta U = -\int_{\phi_0}^{\phi} \tau_{\text{ext}}\,d\phi'
+= -\int_{\phi_0}^{\phi} \mu B\,\sin\phi'\,d\phi'
+= -\mu B\left(\cos\phi_0 - \cos\phi\right).
+\]
+Choosing the reference $U(\phi_0 = 90^\circ) = 0$, we find
+\[
+U = -\mu B\,\cos\phi = -\vec{\mu}\cdot\vec{B}.
+\]
+
+\nt{The torque tries to align $\vec{\mu}$ with $\vec{B}$. The stable equilibrium orientation has $\hat{n}\parallel\vec{B}$, i.e., the plane of the loop is perpendicular to the field. The unstable equilibrium has $\hat{n}$ antiparallel to $\vec{B}$. A DC motor exploits this by periodically reversing the current so the loop continues rotating.}}
+
+\nt{The SI unit of magnetic dipole moment $\vec{\mu}$ is ampere-square meter, $\mathrm{A\cdot m^2}$. This is equivalent to joules per tesla, $\mathrm{J/T}$.}
+
+\ex{Illustrative example}{A coil with $N=50$ turns, each of area $A = 8.0\,\mathrm{cm^2} = 8.0\times 10^{-4}\,\mathrm{m^2}$, carries current $I = 2.0\,\mathrm{A}$ in a field $B = 0.40\,\mathrm{T}$. The maximum torque occurs at $\phi = 90^\circ$:
+\[
+\tau_{\text{max}} = N\,I\,A\,B = (50)(2.0\,\mathrm{A})(8.0\times 10^{-4}\,\mathrm{m^2})(0.40\,\mathrm{T}) = 3.2\times 10^{-2}\,\mathrm{N\cdot m}.}
+
+\qs{Worked example}{A rectangular wire loop has width $a=0.10\,\mathrm{m}$ (horizontal side) and height $b=0.050\,\mathrm{m}$ (vertical side). The loop has $N=100$ turns and carries current $I=2.0\,\mathrm{A}$ in the direction shown: clockwise when viewed from the front (from the $+x$-axis toward the $yz$-plane). The loop sits in a uniform magnetic field $\vec{B} = (0.60\,\mathrm{T})\,\hat{\imath}$ pointing to the right. The normal vector $\hat{n}$ points along the $-\hat{k}$ direction (into the page) by the right-hand rule for clockwise current.
+
+The loop lies in the $xy$-plane, centered at the origin, with its sides parallel to the $x$- and $y$-axes.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the net magnetic force on the loop,
+\item the net magnetic torque on the loop (magnitude and direction), and
+\item the magnetic potential energy of the loop in this orientation, and determine whether the loop will rotate clockwise or counterclockwise when released.
+\end{enumerate}}
+
+\textbf{Given quantities:}
+\begin{itemize}
+\item Width (horizontal): $a = 0.10\,\mathrm{m}$
+\item Height (vertical): $b = 0.050\,\mathrm{m}$
+\item Number of turns: $N = 100$
+\item Current: $I = 2.0\,\mathrm{A}$ (clockwise in the $xy$-plane)
+\item Magnetic field: $\vec{B} = (0.60\,\mathrm{T})\,\hat{\imath}$
+\item Normal vector: $\hat{n} = -\hat{k}$
+\end{itemize}
+
+\sol \textbf{(a) Net force.} The magnetic field is uniform, and the loop is a closed current loop. From the corollary, the net force on a closed loop in a uniform magnetic field is zero. We can verify this segment by segment.
+
+The loop has four segments:
+\begin{itemize}
+\item \textit{Bottom segment} (length $a$, current to the right): $\vec{\ell}_1 = a\,\hat{\imath}$, so $\vec{F}_1 = I\,(a\,\hat{\imath})\times(B\,\hat{\imath}) = \vec{0}$ (parallel to $\vec{B}$).
+\item \textit{Top segment} (length $a$, current to the left): $\vec{\ell}_3 = -a\,\hat{\imath}$, so $\vec{F}_3 = I\,(-a\,\hat{\imath})\times(B\,\hat{\imath}) = \vec{0}$.
+\item \textit{Right segment} (length $b$, current downward): $\vec{\ell}_2 = -b\,\hat{\jmath}$, so
+\[
+\vec{F}_2 = I\,(-b\,\hat{\jmath})\times(B\,\hat{\imath}) = I\,b\,B\,(-\hat{\jmath}\times\hat{\imath}) = I\,b\,B\,\hat{k}.
+\]
+\item \textit{Left segment} (length $b$, current upward): $\vec{\ell}_4 = b\,\hat{\jmath}$, so
+\[
+\vec{F}_4 = I\,(b\,\hat{\jmath})\times(B\,\hat{\imath}) = I\,b\,B\,(\hat{\jmath}\times\hat{\imath}) = -I\,b\,B\,\hat{k}.
+\]
+\end{itemize}
+
+Summing:
+\[
+\vec{F}_{\text{net}} = \vec{F}_1+\vec{F}_2+\vec{F}_3+\vec{F}_4 = \vec{0} + I\,b\,B\,\hat{k} + \vec{0} - I\,b\,B\,\hat{k} = \vec{0}.
+\]
+For $N$ turns, each segment's force is multiplied by $N$, but they still cancel:
+\[
+\boxed{\vec{F}_{\text{net}} = \vec{0}}.
+\]
+
+\textbf{(b) Net torque.} The magnetic dipole moment has magnitude
+\[
+\mu = N\,I\,A = N\,I\,(a\,b).
+\]
+Substitute the values:
+\[
+\mu = (100)(2.0\,\mathrm{A})(0.10\,\mathrm{m})(0.050\,\mathrm{m}) = (100)(2.0)(0.0050)\,\mathrm{A\cdot m^2} = 1.0\,\mathrm{A\cdot m^2}.
+\]
+The dipole moment vector is $\vec{\mu} = \mu\,\hat{n} = -(1.0\,\mathrm{A\cdot m^2})\,\hat{k}$.
+
+The torque is
+\[
+\vec{\tau} = \vec{\mu}\times\vec{B} = [-(1.0)\,\hat{k}]\times[(0.60)\,\hat{\imath}] = -(1.0)(0.60)\,(\hat{k}\times\hat{\imath}).
+\]
+Since $\hat{k}\times\hat{\imath}=\hat{\jmath}$,
+\[
+\vec{\tau} = -(0.60)\,\hat{\jmath}\,\mathrm{N\cdot m}.
+\]
+The magnitude is
+\[
+\tau = 0.60\,\mathrm{N\cdot m}.
+\]
+To interpret the direction, $-\hat{\jmath}$ points downward. Using the right-hand rule for torque, the loop tends to rotate such that $\vec{\mu}$ aligns with $\vec{B}$ — that is, $\hat{n}$ rotates from $-\hat{k}$ toward $+\hat{\imath}$. This corresponds to a rotation about the $y$-axis.
+
+More physically: the right side of the loop (at $+x/2$) feels force $\vec{F}_2 = N\,I\,b\,B\,\hat{k} = 100(2.0)(0.050)(0.60)\,\hat{k} = 6.0\,\mathrm{N}$ upward, while the left side (at $-x/2$) feels $6.0\,\mathrm{N}$ downward. This pair of forces creates a torque that rotates the right side up and the left side down — a rotation about the $y$-axis.
+
+\[
+\boxed{\tau = 0.60\,\mathrm{N\cdot m},\quad \text{rotation about the }-\hat{\jmath}\text{ axis (right side up, left side down)}}.
+\]
+
+\textbf{(c) Potential energy and rotational tendency.} The potential energy is
+\[
+U = -\vec{\mu}\cdot\vec{B} = -[-(1.0)\,\hat{k}]\cdot[(0.60)\,\hat{\imath}].
+\]
+Since $\hat{k}\perp\hat{\imath}$, their dot product is $0$, so
+\[
+\boxed{U = 0}.
+\]
+
+This corresponds to $\phi = 90^\circ$ between $\vec{\mu}$ (pointing in $-\hat{k}$) and $\vec{B}$ (pointing in $+\hat{\imath}$), since $\cos 90^\circ = 0$.
+
+The loop will rotate toward the stable equilibrium orientation where $\vec{\mu}\parallel\vec{B}$. Currently $\vec{\mu}$ points into the page ($-\hat{k}$) while $\vec{B}$ points right ($+\hat{\imath}$). The stable orientation has $\hat{n}$ aligned with $+\hat{\imath}$, meaning the loop's plane is perpendicular to the field (normal pointing along $\vec{B}$). The torque computed in part~(b) drives this rotation: the right side is pushed upward and the left side downward, rotating the loop about the $y$-axis. Viewed from the $+y$ direction (looking down from above), this rotation appears counterclockwise.
+
+\[
+\boxed{\text{Rotates about the }y\text{-axis toward stable equilibrium (right side up, left side down).}}
+\]
+
+\bigskip
+\textbf{Summary of results:}
+\begin{enumerate}[label=(\alph*)]
+\item $\vec{F}_{\text{net}} = \vec{0}$
+\item $\tau = 0.60\,\mathrm{N\cdot m}$, rotation about the $-\hat{\jmath}$ axis
+\item $U = 0$, loop rotates toward stable equilibrium where $\hat{n}\parallel\vec{B}$
+\end{enumerate}

concepts/em/u12/e12-4-biot-savart.tex

@@ -0,0 +1,204 @@
+\subsection{The Biot--Savart Law}
+
+Currents produce magnetic fields. The Biot--Savart law gives the magnetic field at a point in space due to a steady current distribution. It is the magnetic analogue of Coulomb's law in electrostatics: just as Coulomb's law tells you the electric field of a charge distribution by integrating over point charges, the Biot--Savart law tells you the magnetic field of a current distribution by integrating over current elements.
+
+\dfn{Biot--Savart law (differential form)}{Let a steady current $I$ flow through a thin wire. Consider a differential element of the wire of length $d\ell$ carrying current $I$, represented by the vector $d\vec{\ell}$ pointing in the direction of the current. Let $P$ be an observation point, and let $\vec{r}$ be the displacement vector from the current element to $P$, with magnitude $r = |\vec{r}|$ and unit vector $\hat{r} = \vec{r}/r$. The differential magnetic field at $P$ due to this current element is
+\[
+d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{\ell}\times\hat{r}}{r^2}
+         = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{\ell}\times\vec{r}}{r^3}.
+\]
+Here $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\cdot m/A}$ is the permeability of free space (the magnetic constant).}
+
+\nt{The Biot--Savart law is fundamentally a \emph{superposition principle}: the total field is the vector sum (integral) of all the differential contributions from every current element. Because each contribution involves a cross product, $d\vec{B}$ is always perpendicular to both the current element $d\vec{\ell}$ and the displacement $\vec{r}$. The $1/r^2$ dependence mirrors Coulomb's law, making the Biot--Savart law a Green's-function solution to the static Maxwell equations for $\vec{B}$.}
+
+\thm{Biot--Savart law}{For a steady current $I$ flowing along a wire path $C$, the total magnetic field at a point $P$ is
+\[
+\vec{B} = \int_C \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{\ell}\times\hat{r}}{r^2}.
+\]
+\begin{itemize}
+\item \textbf{Constants:} $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\cdot m/A}$ (permeability of free space). In SI units, $d\vec{B}$ has units of tesla ($\mathrm{T}$).
+\item \textbf{Geometry:} $d\vec{\ell}$ points along the wire in the direction of conventional current. The vector $\vec{r}$ points \emph{from} the current element \emph{to} the observation point. The unit vector $\hat{r} = \vec{r}/r$ is the normalized version of this displacement.
+\item \textbf{Direction:} The direction of $d\vec{B}$ is given by the right-hand rule for the cross product $d\vec{\ell}\times\hat{r}$. $d\vec{B}$ is perpendicular to the plane containing $d\vec{\ell}$ and $\vec{r}$.
+\item \textbf{Magnitude:} The magnitude of the differential field is
+\[
+dB = \frac{\mu_0}{4\pi}\,\frac{I\,d\ell\,\sin\theta}{r^2},
+\]
+where $\theta$ is the angle between $d\vec{\ell}$ and $\vec{r}$.
+\item \textbf{Symmetry considerations:} In many symmetric geometries (long straight wires, circular loops, solenoids), symmetry allows you to argue that certain components of $\vec{B}$ cancel upon integration, dramatically simplifying the calculation.
+\end{itemize}}
+
+\pf{Biot--Savart law from the macroscopic field of a long wire and superposition}{The field of a long straight wire carrying current $I$ is known experimentally (from Amp\`ere's law or direct measurement) to be
+\[
+B = \frac{\mu_0 I}{2\pi s},
+\]
+where $s$ is the perpendicular distance from the wire. The Biot--Savart law is the differential statement that, when integrated for an infinite straight wire, reproduces this result.
+
+Consider an infinite straight wire along the $z$-axis carrying current $I$ in the $+\hat{k}$ direction. The observation point is in the $xy$-plane at distance $s$ from the wire. A current element at position $z$ has
+\[
+d\vec{\ell} = dz\,\hat{k}, \qquad
+\vec{r} = s\,\hat{s} - z\,\hat{k}, \qquad
+r = \sqrt{s^2+z^2}.
+\]
+The cross product is
+\[
+d\vec{\ell}\times\vec{r} = dz\,\hat{k}\times(s\,\hat{s}-z\,\hat{k}) = s\,dz\,(\hat{k}\times\hat{s}) = s\,dz\,\hat{\phi},
+\]
+which points in the azimuthal direction. The magnitude of the field contribution is
+\[
+dB = \frac{\mu_0}{4\pi}\,\frac{I\,s\,dz}{(s^2+z^2)^{3/2}}.
+\]
+Integrating from $z=-\infty$ to $z=+\infty$ using $z = s\tan\theta$, $dz = s\,\sec^2\theta\,d\theta$:
+\[
+B = \frac{\mu_0 I}{4\pi}\int_{-\infty}^{\infty}\frac{s\,dz}{(s^2+z^2)^{3/2}}
+  = \frac{\mu_0 I}{4\pi}\int_{-\pi/2}^{\pi/2}\frac{s^2\,\sec^2\theta}{s^3\,\sec^3\theta}\,d\theta
+  = \frac{\mu_0 I}{4\pi s}\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta
+  = \frac{\mu_0 I}{4\pi s}\Bigl[\sin\theta\Bigr]_{-\pi/2}^{\pi/2}
+  = \frac{\mu_0 I}{2\pi s}.
+\]
+This matches the known result for the field of an infinite wire. By linearity, the Biot--Savart law applied to any current distribution gives the correct total field via superposition.
+\qed}
+
+\mprop{Field of a long straight wire}{An infinitely long straight wire carrying steady current $I$ produces a magnetic field at perpendicular distance $s$ given by
+\[
+B = \frac{\mu_0 I}{2\pi s}.
+\]
+The field lines are concentric circles around the wire, with direction given by the right-hand rule: point your right thumb along the current, and your fingers curl in the direction of $\vec{B}$. In cylindrical coordinates $(s,\phi,z)$ with the wire along the $z$-axis,
+\[
+\vec{B} = \frac{\mu_0 I}{2\pi s}\,\hat{\phi}.\]
+}
+
+\mprop{Field at the center of a circular current loop}{A circular loop of radius $R$ carrying steady current $I$ produces a magnetic field at its center given by
+\[
+B = \frac{\mu_0 I}{2R}.
+\]
+The direction is perpendicular to the plane of the loop, given by the right-hand rule: curl your fingers in the direction of the current, and your thumb points along $\vec{B}$. For $N$ tightly wound turns, multiply by $N$:
+\[
+B = \frac{\mu_0 N I}{2R}.\]
+}
+
+\mprop{Field on the axis of a circular current loop}{A circular loop of radius $R$ carrying steady current $I$ produces a magnetic field on its symmetry axis at distance $z$ from the center given by
+\[
+B(z) = \frac{\mu_0 I\,R^2}{2\,(R^2+z^2)^{3/2}}.
+\]
+The field points along the axis in the direction given by the right-hand rule. At the center ($z=0$), this reduces to $B = \mu_0 I/(2R)$. Far from the loop ($z\gg R$), the field falls off as
+\[
+B \approx \frac{\mu_0 I\,R^2}{2\,z^3} = \frac{\mu_0}{4\pi}\,\frac{2\pi R^2 I}{z^3} = \frac{\mu_0}{4\pi}\,\frac{2\mu}{z^3},
+\]
+which is the dipole-field form, where $\mu = \pi R^2 I$ is the magnetic dipole moment of the loop.
+}
+
+\cor{Far-field (dipole) limit of a current loop}{Far from any compact current loop, the magnetic field has the universal dipole form
+\[
+\vec{B}_{\text{dipole}}(\vec{r}) = \frac{\mu_0}{4\pi}\left[\frac{3(\vec{\mu}\cdot\hat{r})\hat{r}-\vec{\mu}}{r^3}\right],
+\]
+where $\vec{\mu} = I\,A\,\hat{n}$ is the magnetic dipole moment ($A$ is the loop area, $\hat{n}$ its normal). This is the analogue of the electric dipole field in electrostatics.}
+
+\cor{Straight wire of finite length}{For a finite straight wire segment of length $L$ carrying current $I$, the field at a point perpendicular to the midpoint of the wire at distance $s$ is
+\[
+B = \frac{\mu_0 I}{2\pi s}\,\frac{L/2}{\sqrt{s^2+(L/2)^2}}.
+\]
+In the limit $L\to\infty$, the fraction approaches $1$, recovering the infinite-wire result $B=\mu_0 I/(2\pi s)$.}
+
+\ex{Illustrative example}{A square loop of side $a$ carries current $I$. Each of the four sides contributes equally. From the finite-wire formula with $L=a$ and the perpendicular distance from the midpoint to the center being $s=a/2$:
+\[
+B_{\text{one side}} = \frac{\mu_0 I}{2\pi(a/2)}\cdot\frac{a/2}{\sqrt{(a/2)^2+(a/2)^2}}
+= \frac{\mu_0 I}{\pi a}\cdot\frac{a/2}{a/\sqrt{2}}
+= \frac{\mu_0 I}{\pi a}\cdot\frac{\sqrt{2}}{2}.
+\]
+Four sides contribute in the same direction (perpendicular to the square's plane), so
+\[
+B_{\text{center}} = 4\cdot\frac{\mu_0 I\sqrt{2}}{2\pi a}
+= \frac{2\sqrt{2}\,\mu_0 I}{\pi a}.\]
+\qed}
+
+\nt{Key symmetry principles for Biot--Savart calculations:}
+\begin{itemize}
+\item \textbf{Straight wire segments aimed directly at (or away from) the observation point} contribute \emph{zero} field: when $d\vec{\ell}\parallel\vec{r}$, the cross product $d\vec{\ell}\times\hat{r}=\vec{0}$. This is a very useful shortcut.
+\item \textbf{Circular arcs} centered on the observation point contribute field proportional to the arc angle. For an arc of angle $\theta$ (in radians) and radius $R$, $B = (\mu_0 I/4\pi R)\cdot\theta$.
+\item \textbf{Perpendicular geometry} maximises the contribution: when $d\vec{\ell}\perp\vec{r}$ at every point (as on a circular arc centered at $P$), the magnitude is $dB = (\mu_0/4\pi)\,I\,d\ell/R^2$ with no angle factor.
+\end{itemize}
+
+\qs{Worked example}{A wire bent into the shape shown carries a steady current $I$ in the direction indicated. The wire consists of three segments:
+\begin{enumerate}[label=(\roman*)]
+\item A straight horizontal segment running from $x=-\infty$ to $x=-R$ along the line $y=0$, approaching the origin.
+\item A circular arc of radius $R$ centered at the origin, extending from the point $(R,0)$ counterclockwise through the upper half-plane to the point $(-R,0)$ (a semicircle).
+\item A straight horizontal segment running from $x=-R$ along the line $y=0$ to $x=+\infty$, extending to the right.
+\end{enumerate}
+Find the magnetic field $\vec{B}$ at the origin $O$ (the center of the circular arc).
+
+Assume the wire lies entirely in the $xy$-plane and the current $I$ flows to the right on the incoming straight segment, then counterclockwise along the arc, then to the right on the outgoing straight segment. Use $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\cdot m/A}$.}
+
+\textbf{Given quantities:}
+\begin{itemize}
+\item Current: $I$
+\item Arc radius: $R$
+\item Arc: semicircle in the upper half-plane ($y\geq 0$), counterclockwise
+\item Observation point: origin $O$
+\end{itemize}
+
+\sol \textbf{Strategy.} By the superposition principle, $\vec{B}_{\text{total}} = \vec{B}_{\text{straight, incoming}} + \vec{B}_{\text{arc}} + \vec{B}_{\text{straight, outgoing}}$. We evaluate each contribution separately.
+
+\medskip
+\noindent\textbf{(i) Incoming straight wire (segment from $x=-\infty$ to $x=-R$, along $y=0$).}
+
+The current flows along the $x$-axis (the line $y=0$) toward the origin. The observation point (the origin) lies \emph{on the line} of the wire. For every current element $d\vec{\ell}$ on this segment, the displacement vector $\vec{r}$ from the element to the origin points along the $+x$ direction, which is \emph{parallel} to $d\vec{\ell}$ (current flows in the $+x$ direction). Therefore
+\[
+d\vec{\ell}\times\hat{r} = 0
+\]
+for all elements of this segment, and
+\[
+\vec{B}_{\text{incoming}} = \vec{0}.
+\]
+The same reasoning applies to the outgoing wire.
+
+\medskip
+\noindent\textbf{(ii) Outgoing straight wire (segment from $x=-R$ to $x=+\infty$, along $y=0$).}
+
+Again, the observation point lies on the line of the wire. The displacement $\vec{r}$ from every current element to the origin is collinear with $d\vec{\ell}$, so $d\vec{\ell}\times\hat{r}=\vec{0}$. Thus
+\[
+\vec{B}_{\text{outgoing}} = \vec{0}.
+\]
+
+\medskip
+\noindent\textbf{(iii) Semicircular arc (radius $R$, upper half-plane, counterclockwise).}
+
+For the circular arc, every current element is at distance $R$ from the origin, and every $d\vec{\ell}$ is tangent to the circle. The displacement vector from each element to the origin points radially inward (toward the center). The angle between $d\vec{\ell}$ (tangential) and $\vec{r}$ (radial) is $90^\circ$, so $\sin 90^\circ = 1$ everywhere.
+
+The magnitude of each differential contribution is
+\[
+dB = \frac{\mu_0}{4\pi}\,\frac{I\,d\ell}{R^2}.
+\]
+The direction: by the right-hand rule, $d\vec{\ell}\times\hat{r}$ for counterclockwise current on the upper semicircle points in the $+\hat{k}$ direction (out of the page) everywhere along the arc.
+
+The total magnitude is
+\[
+B_{\text{arc}} = \int_{\text{arc}} \frac{\mu_0 I}{4\pi R^2}\,d\ell
+= \frac{\mu_0 I}{4\pi R^2}\int_{\text{arc}} d\ell
+= \frac{\mu_0 I}{4\pi R^2}\cdot(\pi R)
+= \frac{\mu_0 I}{4 R}.
+\]
+Here we used that the arc length is $\pi R$ (a semicircle).
+
+In vector form, with $\hat{k}$ pointing out of the $xy$-plane:
+\[
+\vec{B}_{\text{arc}} = \frac{\mu_0 I}{4 R}\,\hat{k}.
+\]
+
+\medskip
+\noindent\textbf{(iv) Total field.}
+
+Summing the three contributions:
+\[
+\vec{B}_{\text{total}} = \vec{0} + \frac{\mu_0 I}{4 R}\,\hat{k} + \vec{0}
+= \frac{\mu_0 I}{4 R}\,\hat{k}.
+\]
+
+\bigskip
+\textbf{Final answer:}
+\[
+\boxed{\vec{B} = \frac{\mu_0 I}{4 R}\,\hat{k}}
+\]
+The field points out of the page (perpendicular to the wire plane, in the $+\hat{k}$ direction by the right-hand rule for the counterclockwise arc current).
+
+\bigskip
+\textbf{Check.} If the arc were a full loop, we would recover $B = \mu_0 I/(2R)$ (the result from the centre-of-loop formula). Since we have a semicircle (half a loop), the field should be half of that: $B = \mu_0 I/(4R)$. This matches our result, confirming consistency.

concepts/em/u12/e12-5-ampere.tex

@@ -0,0 +1,138 @@
+\subsection{Amp\`ere's Law and Symmetry Reduction}
+
+This subsection states Amp\`ere's law and shows how symmetry can reduce a difficult line integral to simple algebra when the current distribution is highly symmetric. It is the magnetic analogue of Gauss's law.
+
+\dfn{Amperian loop and enclosed current}{Let $C$ be any closed curve in space (called an \emph{Amperian loop}). The \emph{enclosed current} $I_{\mathrm{enc}}$ is the algebraic sum of all steady currents passing through any open surface $S$ bounded by $C$. The sign of each current is determined by the right-hand rule: curl the fingers of your right hand along the direction of integration around $C$; if your thumb points in the direction of the current, that current counts as positive. Currents in the opposite direction count as negative.}
+
+\nt{Just as with Gauss's law, Amp\`ere's law is always true, but it is not always useful for finding $\vec{B}$. In a general asymmetric current distribution, knowing only the total enclosed current does not tell you the field at each point on the loop. The main strategy is therefore: first identify strong symmetry, then choose an Amperian loop matched to that symmetry so that $B=|\vec{B}|$ is constant on the field-contributing parts of the loop and the angle between $\vec{B}$ and $d\vec{\ell}$ is everywhere $0^\circ$, $180^\circ$, or $90^\circ$.}
+
+\thm{Amp\`ere's law and when symmetry makes it useful}{Let $C$ be any closed curve and $I_{\mathrm{enc}}$ the net steady current passing through any surface bounded by $C$. Then Amp\`ere's law states
+\[
+\oint_C \vec{B}\cdot d\vec{\ell}=\mu_0 I_{\mathrm{enc}}.
+\]
+This law is always true. It becomes a practical method for solving for the magnetic field when the current distribution has enough symmetry that one can choose an Amperian loop for which the magnitude $B=|\vec{B}|$ is constant on each field-contributing part of the loop and the angle between $\vec{B}$ and $d\vec{\ell}$ is everywhere $0^\circ$, $180^\circ$, or $90^\circ$. Then the line integral reduces to algebraic terms such as $B\ell$, $-B\ell$, or $0$. Common useful cases are cylindrical symmetry (long straight wires), planar symmetry (infinite current sheets), and solenoidal symmetry (ideal solenoids). The direction of $\vec{B}$ follows the right-hand rule relative to the enclosed current: if the thumb of your right hand points in the direction of the current, your fingers curl in the direction of the magnetic field circulation.}
+
+\nt{Amp\`ere's law is the magnetic analogue of Gauss's law. Gauss's law relates the electric field flux through a closed surface to the enclosed charge, $\oint\vec{E}\cdot d\vec{A}=q_{\mathrm{enc}}/\varepsilon_0$. Amp\`ere's law relates the magnetic field circulation around a closed loop to the enclosed current, $\oint\vec{B}\cdot d\vec{\ell}=\mu_0 I_{\mathrm{enc}}$. Both are universally valid but are practically useful for finding fields only when the source distribution has high symmetry. The matching of symmetry to geometry is parallel: spherical symmetry $\to$ spherical Gaussian surface, cylindrical symmetry $\to$ circular Amperian loop, planar symmetry $\to$ rectangular Amperian loop.}
+
+\pf{How symmetry reduces the line integral}{Let a long straight wire carry current $I$ along the $+z$ axis. By cylindrical symmetry, the magnetic field circulates around the wire in concentric circles in planes perpendicular to the wire, and its magnitude $B(r)$ depends only on the radial distance $r$ from the wire axis. Choose a circular Amperian loop of radius $r$ centred on the wire. Along this loop, $\vec{B}$ is everywhere tangent to $d\vec{\ell}$, so $\vec{B}\cdot d\vec{\ell}=B(r)\,d\ell$, and $B(r)$ is constant everywhere on the loop. Therefore,
+\[
+\oint_C \vec{B}\cdot d\vec{\ell}=B(r)\oint_C d\ell=B(r)(2\pi r).
+\]
+If the enclosed current is $I_{\mathrm{enc}}$, Amp\`ere's law gives
+\[
+B(r)(2\pi r)=\mu_0 I_{\mathrm{enc}}.
+\]
+The law itself is general, but the symmetry is what allowed $B(r)$ to be pulled outside the integral.}
+
+\mprop{Magnetic field of a long straight wire}{Let $\mu_0=4\pi\times 10^{-7}\,\mathrm{T\cdot m/A}$ be the permeability of free space. Consider a long straight wire of radius $R$ carrying total steady current $I$ with uniform current density $J$ across its cross section. The magnetic field magnitude is
+\[
+B(r)=\begin{cases}
+\dfrac{\mu_0 I\,r}{2\pi R^2} & \text{for }r<R\;\text{(inside the wire)},\\[1.2ex]
+\dfrac{\mu_0 I}{2\pi r} & \text{for }r>R\;\text{(outside the wire)}.
+\end{cases}
+\]
+Inside the wire the field grows linearly with $r$; outside it falls as $1/r$. In both regions, $\vec{B}$ is tangent to circles centred on the wire axis, in the direction given by the right-hand rule.}
+
+\cor{Continuity of $B$ at the wire surface}{At the surface of the wire ($r=R$), both the inside and outside formulas give the same result:
+\[
+B(R)=\frac{\mu_0 I\,R}{2\pi R^2}=\frac{\mu_0 I}{2\pi R}.
+\]
+The magnetic field is continuous at the boundary even though the functional form changes. The maximum field occurs at the surface.}
+
+\mprop{Magnetic field of an infinite current sheet}{An infinite planar sheet carrying uniform surface current density $K$ (current per unit width, in units of $\mathrm{A/m}$). Choose a rectangular Amperian loop of length $L$ straddling the sheet, with two long sides parallel to the field and two short sides perpendicular. By symmetry, the field has constant magnitude on each side of the sheet and is parallel to the sheet but perpendicular to the current direction. Amp\`ere's law gives
+\[
+2BL=\mu_0(KL)\quad\Rightarrow\quad B=\frac{\mu_0 K}{2}.
+\]
+The field reverses direction on opposite sides of the sheet. The field outside is independent of distance from the sheet.}
+
+\ex{Illustrative example}{A square loop of side $0.1\,\mathrm{m}$ carries a current $I=2\,\mathrm{A}$ in a uniform magnetic field $B=0.5\,\mathrm{T}$ perpendicular to the plane of the loop. The magnetic flux through the loop is
+\[
+\Phi_B = BA = (0.5\,\mathrm{T})(0.1\,\mathrm{m})^2 = 5.0\times 10^{-3}\,\mathrm{T\cdot m^2}.
+\]}
+
+\qs{Worked example}{A long cylindrical wire of radius $R=2.0\times 10^{-3}\,\mathrm{m}=2.0\,\mathrm{mm}$ carries a steady current $I=10\,\mathrm{A}$ uniformly distributed across its cross section. The current flows in the $+\hat{k}$ direction.
+
+Find the magnetic field magnitude and direction at:
+\begin{enumerate}[label=(\alph*)]
+\item $r_1=1.0\times 10^{-3}\,\mathrm{m}=1.0\,\mathrm{mm}$ (inside the wire), and
+\item $r_2=5.0\times 10^{-3}\,\mathrm{m}=5.0\,\mathrm{mm}$ (outside the wire).
+\end{enumerate}}
+
+\sol Let the wire lie along the $z$ axis with current flowing in the $+\hat{k}$ direction. By cylindrical symmetry, the magnetic field circulates around the wire in concentric circles in planes perpendicular to the wire. The field magnitude depends only on the radial distance $r$ from the wire axis.
+
+Choose a circular Amperian loop of radius $r$ centred on the wire axis. Along this loop, $\vec{B}$ is everywhere tangent to $d\vec{\ell}$, so $\vec{B}\cdot d\vec{\ell}=B(r)\,d\ell$. The line integral becomes
+\[
+\oint_C \vec{B}\cdot d\vec{\ell}=B(r)\oint_C d\ell=B(r)(2\pi r).
+\]
+
+The enclosed current depends on whether the loop is inside or outside the wire. The uniform current density is
+\[
+J=\frac{I}{\pi R^2}.
+\]
+
+\textbf{(a) Inside the wire ($r_1<R$).} The enclosed current is the fraction of the total current passing through the area inside the Amperian loop:
+\[
+I_{\mathrm{enc}}=J(\pi r_1^2)=\frac{I}{\pi R^2}(\pi r_1^2)=I\,\frac{r_1^2}{R^2}.
+\]
+Substitute the values:
+\[
+I_{\mathrm{enc}}=(10\,\mathrm{A})\,\frac{(1.0\times 10^{-3}\,\mathrm{m})^2}{(2.0\times 10^{-3}\,\mathrm{m})^2}=(10\,\mathrm{A})\,\frac{1.0\times 10^{-6}}{4.0\times 10^{-6}}.
+\]
+Thus
+\[
+I_{\mathrm{enc}}=(10\,\mathrm{A})\times\frac{1}{4}=2.5\,\mathrm{A}.
+\]
+
+Apply Amp\`ere's law:
+\[
+B(r_1)(2\pi r_1)=\mu_0 I_{\mathrm{enc}}.
+\]
+Solve for $B(r_1)$:
+\[
+B(r_1)=\frac{\mu_0 I_{\mathrm{enc}}}{2\pi r_1}=\frac{(4\pi\times 10^{-7}\,\mathrm{T\cdot m/A})(2.5\,\mathrm{A})}{2\pi(1.0\times 10^{-3}\,\mathrm{m})}.
+\]
+Compute step by step:
+\[
+(4\pi\times 10^{-7})(2.5)=10\pi\times 10^{-7},
+\]
+\[
+2\pi(1.0\times 10^{-3})=2\pi\times 10^{-3},
+\]
+\[
+B(r_1)=\frac{10\pi\times 10^{-7}}{2\pi\times 10^{-3}}\,\mathrm{T}=5.0\times 10^{-4}\,\mathrm{T}.
+\]
+
+The direction follows the right-hand rule: thumb along $+\hat{k}$ (current direction), fingers curl counterclockwise when viewed from above.
+
+\textbf{(b) Outside the wire ($r_2>R$).} The Amperian loop encloses the entire wire, so
+\[
+I_{\mathrm{enc}}=I=10\,\mathrm{A}.
+\]
+
+Apply Amp\`ere's law:
+\[
+B(r_2)(2\pi r_2)=\mu_0 I.
+\]
+Solve for $B(r_2)$:
+\[
+B(r_2)=\frac{\mu_0 I}{2\pi r_2}=\frac{(4\pi\times 10^{-7}\,\mathrm{T\cdot m/A})(10\,\mathrm{A})}{2\pi(5.0\times 10^{-3}\,\mathrm{m})}.
+\]
+Compute step by step:
+\[
+(4\pi\times 10^{-7})(10)=40\pi\times 10^{-7},
+\]
+\[
+2\pi(5.0\times 10^{-3})=10\pi\times 10^{-3},
+\]
+\[
+B(r_2)=\frac{40\pi\times 10^{-7}}{10\pi\times 10^{-3}}\,\mathrm{T}=4.0\times 10^{-4}\,\mathrm{T}.
+\]
+
+The direction is again given by the right-hand rule: counterclockwise when viewed from above.
+
+\bigskip
+\textbf{Final answers:}
+\begin{enumerate}[label=(\alph*)]
+\item $B(r_1)=5.0\times 10^{-4}\,\mathrm{T}=0.50\,\mathrm{mT}$, directed counterclockwise around the wire.
+\item $B(r_2)=4.0\times 10^{-4}\,\mathrm{T}=0.40\,\mathrm{mT}$, directed counterclockwise around the wire.
+\end{enumerate}

concepts/em/u12/e12-6-solenoids-dipoles.tex

@@ -0,0 +1,177 @@
+\subsection{Solenoids, Parallel Currents, and Magnetic Dipoles}
+
+This subsection covers three closely related topics: the magnetic field inside a solenoid and a toroid (derived from Amp\`ere's law), the magnetic force per unit length between two parallel current-carrying wires, and the magnetic dipole moment of a current loop. Together these describe how steady currents in geometrically ordered configurations produce well-defined magnetic fields and forces.
+
+\dfn{Ideal solenoid}{An \emph{ideal solenoid} is a long, tightly wound helical coil of wire. When the winding is close and the length $L$ is much greater than the radius $R$, the magnetic field inside is uniform, axial, and of magnitude
+\[
+B = \mu_0\,n\,I,
+\]
+where $I$ is the current, $n = N/L$ is the number of turns per unit length ($N$ is the total number of turns, $L$ is the length of the solenoid), and $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A}$ is the permeability of free space. Outside the ideal solenoid the field is approximately zero. The field direction follows the right-hand rule: curl the fingers of your right hand in the direction of the current around the coil, and your thumb points along $\vec{B}$ inside the solenoid.}
+
+\nt{The field inside a solenoid depends only on the turn density $n$ and the current $I$; it is independent of the radius of the coil. This is analogous to the electric field inside a uniformly charged infinite sheet, which depends only on the surface charge density and not on the lateral dimensions.}
+
+\thm{Magnetic field of a long solenoid}{For an ideal solenoid of length $L \gg R$ with $N$ turns carrying current $I$, the magnetic field inside is uniform and directed along the axis. Its magnitude is
+\[
+B = \mu_0\,n\,I
+\qquad\text{where}\qquad
+n = \frac{N}{L}.
+\]
+Outside the solenoid $B \approx 0$. The field lines are straight, parallel lines inside and loop back around outside (forming closed loops).}
+
+\dfn{Toroid}{A \emph{toroid} is a solenoid bent into a circle (a doughnut shape). It consists of $N$ turns wound around a circular core of mean radius $r$. By applying Amp\`ere's law to a circular Amperian loop of radius $r$ inside the toroid, the magnetic field inside the torus is
+\[
+B = \frac{\mu_0\,N\,I}{2\pi\,r}.
+\]
+Outside the toroid ($r$ is not within the winding region), $B \approx 0$. The field lines are circles concentric with the toroid axis.}
+
+\nt{The toroid field formula $B = \mu_0 N I/(2\pi r)$ shows a $1/r$ dependence, unlike the uniform solenoid. For a toroid with a large mean radius and small cross-section (so $r$ varies little across the windings), the field is nearly uniform and approximately $B \approx \mu_0 n I$ where $n = N/(2\pi r_{\text{avg}})$.}
+
+\mprop{Solenoid and toroid magnetic fields}{The magnetic field produced by steady currents in these common device geometries is:
+\begin{itemize}
+\item \textbf{Long solenoid (inside):} $B = \mu_0\,n\,I$, where $n = N/L$ is the turn density. The field is uniform and axial.
+\item \textbf{Toroid (inside the windings):} $B = \dfrac{\mu_0\,N\,I}{2\pi\,r}$, where $r$ is the radial distance from the toroid centre. The field circulates along circular field lines and decreases as $1/r$.
+\item \textbf{Outside both devices:} $B \approx 0$ (ideal case).
+\end{itemize}}
+
+\pf{Solenoid field from Amp\`ere's law (outline)}{Consider an ideal solenoid with $n$ turns per unit length. Choose a rectangular Amperian loop with one long side (length $\ell$) inside the solenoid parallel to the axis, the opposite side outside, and the two short sides perpendicular to the axis. The line integral of $\vec{B}\cdot d\vec{\ell}$ around the loop receives contributions only from the inside segment, because $\vec{B}\approx 0$ outside and $\vec{B}\perp d\vec{\ell}$ on the perpendicular segments. Thus
+\[
+\oint \vec{B}\cdot d\vec{\ell} = B\,\ell.
+\]
+The enclosed current is $I_{\text{enc}} = n\,\ell\,I$ (each of the $n\ell$ turns inside the loop carries current $I$). By Amp\`ere's law, $B\,\ell = \mu_0\,n\,\ell\,I$, giving $B = \mu_0 n I$.}
+
+\pf{Toroid field from Amp\`ere's law}{Choose a circular Amperian loop of radius $r$ inside the toroid, concentric with the toroid axis. By symmetry, $\vec{B}$ is tangent to the circle and has constant magnitude $B$ at fixed $r$. The line integral is
+\[
+\oint \vec{B}\cdot d\vec{\ell} = B\,(2\pi r).
+\]
+The loop encloses $N$ turns each carrying current $I$, so $I_{\text{enc}} = N I$. Amp\`ere's law gives $B(2\pi r) = \mu_0 N I$, yielding $B = \mu_0 N I/(2\pi r)$.}
+
+\nt{Both derivations rely on Amp\`ere's law $\oint \vec{B}\cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$ and the presence of sufficient symmetry to pull $B$ out of the integral. The solenoid requires the infinite-length approximation; the toroid requires circular symmetry. These are the two magnetostatic configurations in the AP Physics C curriculum where Amp\`ere's law gives a clean result.}
+
+\dfn{Magnetic force between two parallel current-carrying wires}{Two long, straight, parallel wires separated by distance $d$, carrying steady currents $I_1$ and $I_2$, exert magnetic forces on each other. Wire 1 produces a magnetic field $B_1 = \dfrac{\mu_0 I_1}{2\pi d}$ at the location of wire 2. The force per unit length on wire 2 due to wire 1 is
+\[
+\frac{F_{12}}{L} = \frac{\mu_0\,I_1\,I_2}{2\pi\,d}.
+\]
+The force is \emph{attractive} when the currents flow in the \emph{same} direction and \emph{repulsive} when they flow in \emph{opposite} directions. By Newton's third law, the force per unit length on wire 1 due to wire 2 has the same magnitude and opposite direction.}
+
+\nt{The rule for attraction and repulsion is the \emph{opposite} of what happens with electric charges. Parallel currents (same direction) \emph{attract}, while like electric charges \emph{repel}. A useful mnemonic: ``like currents attract, unlike repel'' -- but remember this refers to current \emph{directions}, not charge types.}
+
+\thm{Force per unit length between two parallel wires}{Two long straight parallel wires separated by distance $d$ carry currents $I_1$ and $I_2$. The magnitude of the magnetic force per unit length is
+\[
+\frac{F}{L} = \frac{\mu_0\,I_1\,I_2}{2\pi\,d}.
+\]
+\begin{itemize}
+\item Currents in the \textbf{same direction} $\;\rightarrow\;$ \textbf{attractive} force.
+\item Currents in \textbf{opposite directions} $\;\rightarrow\;$ \textbf{repulsive} force.
+\end{itemize}
+The direction of the force on either wire is perpendicular to the wire and toward (or away from) the other wire, along the line connecting the wires.}
+
+\pf{Force between parallel wires from the Lorentz force}{Wire 1 (carrying $I_1$) produces a magnetic field at the position of wire 2. By the right-hand rule for a long straight wire, $B_1$ circles wire 1 and has magnitude $B_1 = \mu_0 I_1/(2\pi d)$. The field at wire 2 is perpendicular to wire 2. The Lorentz force on a length $L$ of wire 2 is $\vec{F}_{12} = I_2\,\vec{L}\times\vec{B}_1$. Since $\vec{L}\perp\vec{B}_1$, the magnitude is
+\[
+F_{12} = I_2\,L\,B_1 = I_2\,L\,\frac{\mu_0 I_1}{2\pi d}.
+\]
+Dividing by $L$ gives $F_{12}/L = \mu_0 I_1 I_2/(2\pi d)$. The direction follows from the cross product: if both currents point upward, $\vec{B}_1$ at wire 2 points into the page, and $\vec{L}_2\times\vec{B}_1$ points toward wire 1 (attractive).}
+
+\thm{Magnetic dipole moment of a current loop}{A planar current loop carrying current $I$ and enclosing area $A$ has a \emph{magnetic dipole moment}
+\[
+\vec{\mu} = I\,A\,\hat{n},
+\]
+where $\hat{n}$ is the unit normal to the plane of the loop, determined by the right-hand rule: curl the fingers of your right hand in the direction of the current, and your thumb points along $\hat{n}$. The SI unit of $\mu$ is the ampere-square metre ($\mathrm{A\!\cdot\!m^2}$), equivalent to joule per tesla ($\mathrm{J/T}$). For a coil with $N$ turns, $\vec{\mu} = N I A\,\hat{n}$.}
+
+\nt{The magnetic dipole moment characterises the torque a current loop experiences in a uniform external field: $\vec{\tau} = \vec{\mu}\times\vec{B}$. It is also the quantity that determines the far-field of the loop -- at distances much greater than the loop size, a current loop produces the same magnetic field as a magnetic dipole. This is the quantum-mechanical basis of atomic magnetism (electron orbital and spin angular momenta give rise to magnetic dipole moments).}
+
+\mprop{Magnetic dipole properties}{For a planar current loop (or coil of $N$ turns) with area $A$ and current $I$:
+\begin{itemize}
+\item \textbf{Dipole moment:} $\mu = N I A$. The direction is given by the right-hand rule.
+\item \textbf{Torque in a uniform field:} $\vec{\tau} = \vec{\mu}\times\vec{B}$, with magnitude $\tau = \mu B \sin\theta$, where $\theta$ is the angle between $\vec{\mu}$ and $\vec{B}$.
+\item \textbf{Potential energy:} $U = -\vec{\mu}\cdot\vec{B} = -\mu B \cos\theta$. The minimum energy (stable equilibrium) occurs when $\vec{\mu}$ aligns with $\vec{B}$ ($\theta = 0$).
+\end{itemize}}
+
+\cor{Rectangular and circular loops}{For a rectangular loop of sides $a$ and $b$, $A = a\,b$. For a circular loop of radius $R$, $A = \pi R^2$. In both cases $\mu = I A$ for a single turn, and the dipole moment points along the axis of symmetry.}
+
+\ex{Illustrative example}{Two parallel wires are separated by $d = 8.0\,\mathrm{cm}$ and carry currents $I_1 = 3.0\,\mathrm{A}$ and $I_2 = 5.0\,\mathrm{A}$ in the same direction. The force per unit length is
+\[
+\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A})(3.0\,\mathrm{A})(5.0\,\mathrm{A})}{2\pi(0.080\,\mathrm{m})} = \frac{6\times 10^{-6}}{0.080}\,\mathrm{N/m} = 7.5\times 10^{-5}\,\mathrm{N/m},
+\]
+attractive. A circular loop of radius $5.0\,\mathrm{cm}$ carrying $2.0\,\mathrm{A}$ has dipole moment $\mu = I(\pi R^2) = (2.0)(\pi)(0.050)^2 = 1.57\times 10^{-2}\,\mathrm{A\!\cdot\!m^2}$.}
+
+\qs{Worked example}{A solenoid is $40.0\,\mathrm{cm}$ long, has $N = 600$ turns uniformly distributed along its length, and a circular cross-section of diameter $3.0\,\mathrm{cm}$. A steady current $I = 4.0\,\mathrm{A}$ flows through the wire.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the turn density $n$ of the solenoid,
+\item the magnitude of the magnetic field inside the solenoid,
+\item the direction of the magnetic field if the current, viewed from the left end, flows counterclockwise,
+\item the magnetic dipole moment of the solenoid, and
+\item the torque on the solenoid if it is placed in a uniform external magnetic field $B_{\text{ext}} = 0.15\,\mathrm{T}$ with its axis at $30^\circ$ to the field.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The turn density is the number of turns divided by the length:
+\[
+n = \frac{N}{L} = \frac{600}{0.400\,\mathrm{m}} = 1500\,\mathrm{turns/m}.
+\]
+
+\textbf{Part (b).} The magnetic field inside the solenoid is
+\[
+B = \mu_0\,n\,I.
+\]
+Substitute the values:
+\[
+B = \bigl(4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A}\bigr)\bigl(1500\,\mathrm{m^{-1}}\bigr)\bigl(4.0\,\mathrm{A}\bigr).
+\]
+Compute step by step:
+\[
+1500\times 4.0 = 6000,
+\]
+\[
+B = 4\pi\times 10^{-7}\times 6000 = 4\pi\times 6.0\times 10^{-4} = 24\pi\times 10^{-4}\,\mathrm{T}.
+\]
+Using $\pi \approx 3.1416$:
+\[
+B = 24\times 3.1416\times 10^{-4}\,\mathrm{T} = 7.54\times 10^{-3}\,\mathrm{T} = 7.54\,\mathrm{mT}.
+\]
+
+\textbf{Part (c).} The current, viewed from the left end, flows counterclockwise. By the right-hand rule: curl the fingers of your right hand counterclockwise (as seen from the left), and your thumb points to the right. Thus the magnetic field inside the solenoid points to the \textbf{right} (along the solenoid axis, away from the viewer of the left end).
+
+\textbf{Part (d).} The magnetic dipole moment of a solenoid is $\mu = N I A$, where $A$ is the cross-sectional area of one turn. The radius of the solenoid is
+\[
+R = \frac{3.0\,\mathrm{cm}}{2} = 1.5\,\mathrm{cm} = 0.015\,\mathrm{m}.
+\]
+The area is
+\[
+A = \pi R^2 = \pi (0.015\,\mathrm{m})^2 = \pi \times 2.25\times 10^{-4}\,\mathrm{m^2} = 7.07\times 10^{-4}\,\mathrm{m^2}.
+\]
+The dipole moment magnitude is
+\[
+\mu = N I A = (600)(4.0\,\mathrm{A})(7.07\times 10^{-4}\,\mathrm{m^2}).
+\]
+Compute:
+\[
+600\times 4.0 = 2400,
+\]
+\[
+\mu = 2400\times 7.07\times 10^{-4}\,\mathrm{A\!\cdot\!m^2} = 1.70\,\mathrm{A\!\cdot\!m^2}.
+\]
+The direction of $\vec{\mu}$ is along the axis to the right (same as $\vec{B}$ inside).
+
+\textbf{Part (e).} The torque on a magnetic dipole in a uniform field is $\vec{\tau} = \vec{\mu}\times\vec{B}_{\text{ext}}$. Its magnitude is
+\[
+\tau = \mu\,B_{\text{ext}}\,\sin\theta,
+\]
+where $\theta = 30^\circ$. Substituting:
+\[
+\tau = (1.70\,\mathrm{A\!\cdot\!m^2})(0.15\,\mathrm{T})\,\sin 30^\circ.
+\]
+Since $\sin 30^\circ = 0.5$:
+\[
+\tau = 1.70\times 0.15\times 0.5 = 0.1275\,\mathrm{N\!\cdot\!m} \approx 0.13\,\mathrm{N\!\cdot\!m}.
+\]
+
+\bigskip
+\textbf{Final answers:}
+\begin{enumerate}[label=(\alph*)]
+\item $n = 1500\,\mathrm{turns/m}$
+\item $B = 7.54\,\mathrm{mT}$ (or $7.5\times 10^{-3}\,\mathrm{T}$)
+\item To the right (along the solenoid axis)
+\item $\mu = 1.70\,\mathrm{A\!\cdot\!m^2}$
+\item $\tau = 0.13\,\mathrm{N\!\cdot\!m}$
+\end{enumerate}

concepts/em/u13/e13-1-magnetic-flux.tex

@@ -0,0 +1,111 @@
+\subsection{Magnetic Flux}
+
+Magnetic flux is a scalar quantity that measures the total magnetic field passing through a given surface. It is the magnetic analogue of electric flux and provides the natural framework for understanding Faraday's law of induction, which is the subject of the next section.
+
+\dfn{Magnetic flux}{Let $\vec{B}$ denote the magnetic-field vector and let $S$ be an oriented surface. The \emph{magnetic flux} through $S$ is the surface integral
+\[
+\Phi_B = \iint_S \vec{B}\cdot d\vec{A},
+\]
+where $d\vec{A}$ is the vector area element: its magnitude is the area of the infinitesimal surface patch, and its direction is normal to the surface. For an open surface the orientation (and hence the direction of $d\vec{A}$) must be specified; for a closed surface the convention is that $d\vec{A}$ points outward.
+
+The magnetic flux $\Phi_B$ is a scalar quantity. Its sign depends on the relative orientation of the surface normal and the field: positive when $\vec{B}$ has a component along the surface normal, negative when it opposes the normal, and zero when $\vec{B}$ is tangent to the surface.}
+
+\nt{Think of magnetic flux as the ``number of magnetic field lines'' passing through a surface. A larger surface catches more lines; tilting the surface reduces the effective area and hence the flux; reversing the surface normal flips the sign of the flux. This picture is qualitative, but it is extremely useful for building intuition about how $\Phi_B$ changes when the surface moves or rotates.}
+
+\thm{Magnetic flux through a flat surface in a uniform field}{When the magnetic field $\vec{B}$ is uniform (constant in magnitude and direction) and the surface is flat with area $A$ and unit normal $\hat{n}$, the surface integral simplifies to a single dot product:
+\[
+\Phi_B = \vec{B}\cdot\vec{A} = B\,A\,\cos\theta,
+\]
+where $\vec{A}=A\,\hat{n}$ is the area vector, $B=|\vec{B}|$, and $\theta$ is the angle between $\vec{B}$ and the area normal $\hat{n}$.}
+
+\nt{The angle $\theta$ is always measured between $\vec{B}$ and the \emph{surface normal}, \emph{not} between $\vec{B}$ and the surface itself. When $\vec{B}$ is perpendicular to the surface, $\theta=0^\circ$ and $\cos\theta=1$, giving maximum flux $\Phi_B = BA$. When $\vec{B}$ is parallel to the surface, $\theta=90^\circ$ and $\cos\theta=0$, giving zero flux. These are the two extremes that frequently appear on exams.}
+
+\mprop{Magnetic flux: key properties}{
+\begin{itemize}
+\item \textbf{SI unit:} The weber, $\mathrm{Wb}$, where $1\,\mathrm{Wb}=1\,\mathrm{T\!\cdot\!m^2}$.
+\item \textbf{Scalar:} $\Phi_B$ is a scalar. It can be positive, negative, or zero, depending on the choice of surface orientation.
+\item \textbf{Orientation dependence:} $\Phi_B = BA\cos\theta$, where $\theta$ is the angle between $\vec{B}$ and the surface normal. Flux is maximal when $\vec{B}\perp$ surface ($\theta=0^\circ$) and zero when $\vec{B}\parallel$ surface ($\theta=90^\circ$).
+\item \textbf{Comparison with electric flux:} The electric flux through a surface is $\Phi_E = \iint_S \vec{E}\cdot d\vec{A}$. The mathematical structure is identical; the only difference is that $\Phi_E$ can be non-zero for closed surfaces enclosing net charge ($\Phi_E = Q_{\text{enc}}/\varepsilon_0$, Gauss's law), while $\Phi_B$ is always zero through a closed surface.
+\item \textbf{Time dependence:} If any of $B$, $A$, or $\theta$ changes with time, the flux changes: $\Delta\Phi_B = \Phi_{B,f} - \Phi_{B,i}$. A changing magnetic flux is the fundamental ingredient behind electromagnetic induction (Faraday's law).
+\end{itemize}}
+
+\cor{Gauss's law for magnetism (qualitative)}{The net magnetic flux through any \emph{closed} surface is zero:
+\[
+\oiint_{\text{closed}} \vec{B}\cdot d\vec{A} = 0.
+\]
+This is Gauss's law for magnetism. It reflects the experimental fact that magnetic monopoles have never been observed: magnetic field lines always form closed loops, so every field line that enters a closed surface must also exit it. This stands in contrast to electric flux through a closed surface, which is proportional to the enclosed charge.}
+
+\ex{Illustrative example}{A square loop of side $0.20\,\mathrm{m}$ sits in a uniform magnetic field $B=0.30\,\mathrm{T}$. The field makes an angle $\theta=35^\circ$ with the loop's normal. The area is $A=(0.20\,\mathrm{m})^2=0.040\,\mathrm{m^2}$, so the flux is
+\[
+\Phi_B = B\,A\,\cos\theta = (0.30\,\mathrm{T})(0.040\,\mathrm{m^2})\cos 35^\circ = 0.012 \times 0.8192 = 9.83\times 10^{-3}\,\mathrm{Wb}.
+\]}
+
+\qs{Worked example}{A rectangular loop of wire has width $w=12.0\,\mathrm{cm}=0.120\,\mathrm{m}$ and height $h=8.00\,\mathrm{cm}=0.0800\,\mathrm{m}$. The loop is situated in a uniform magnetic field of magnitude $B=0.450\,\mathrm{T}$. The field is constant in space. The loop has area $A=w\,h$.
+
+Find the magnetic flux $\Phi_B$ through the loop for each of the following orientations:
+\begin{enumerate}[label=(\alph*)]
+\item The loop is in the $xy$-plane and the magnetic field points in the $+\hat{k}$ direction (perpendicular to the loop, with the area normal also taken as $+\hat{k}$).
+\item The loop is in the $xz$-plane and the magnetic field points in the $+\hat{k}$ direction (the field is parallel to the loop).
+\item The loop is in the $xy$-plane and the magnetic field lies in the $xz$-plane, making an angle $\theta=30.0^\circ$ below the $x$-axis. Take the area normal as $+\hat{k}$.
+\item The loop is in the $yz$-plane and the magnetic field points in the $+\hat{k}$ direction (the field is perpendicular to the loop, with the area normal taken as $+\hat{k}$).
+\item The loop is in the $xy$-plane. The magnetic field is $\vec{B}=(0.450\,\mathrm{T})\,\hat{\imath}+(0.300\,\mathrm{T})\,\hat{\jmath}$. Take the area normal as $+\hat{k}$.
+\end{enumerate}}
+
+\sol First compute the loop's area:
+\[
+A = w\,h = (0.120\,\mathrm{m})(0.0800\,\mathrm{m}) = 9.60\times 10^{-3}\,\mathrm{m^2}.
+\]
+The area vector is $\vec{A} = A\,\hat{n}$, where $\hat{n}$ is the unit normal determined by the loop's orientation. Since $\vec{B}$ is uniform, we use $\Phi_B = \vec{B}\cdot\vec{A} = B\,A\,\cos\theta$, where $\theta$ is the angle between $\vec{B}$ and $\hat{n}$.
+
+\textbf{(a)} The loop is in the $xy$-plane with normal $\hat{n}=+\hat{k}$, and $\vec{B} = B\,\hat{k}$. The angle between $\vec{B}$ and $\hat{n}$ is $\theta = 0^\circ$:
+\[
+\Phi_B = B\,A\,\cos 0^\circ = (0.450\,\mathrm{T})(9.60\times 10^{-3}\,\mathrm{m^2})(1).
+\]
+Compute:
+\[
+\Phi_B = 0.450 \times 9.60\times 10^{-3}\,\mathrm{Wb} = 4.32\times 10^{-3}\,\mathrm{Wb}.
+\]
+
+\textbf{(b)} The loop is in the $xz$-plane with normal $\hat{n}=+\hat{j}$ (by the right-hand rule, or equivalently, the area vector points along $y$). The field is $\vec{B}=B\,\hat{k}$. The angle between $\vec{B}$ and $\hat{n}$ is $\theta = 90^\circ$:
+\[
+\Phi_B = B\,A\,\cos 90^\circ = 0.
+\]
+No field lines pass through the loop; the field runs parallel to the plane of the loop.
+
+\textbf{(c)} The loop is in the $xy$-plane with normal $\hat{n}=+\hat{k}$. The field is $\vec{B}$ lying in the $xz$-plane at $30.0^\circ$ below the $x$-axis. We need the angle between $\vec{B}$ and $\hat{k}$. Since $\vec{B}$ is in the $xz$-plane and makes $30.0^\circ$ with the $x$-axis, the angle with the $z$-axis ($\hat{k}$) is $\theta = 90^\circ - 30.0^\circ = 60.0^\circ$:
+\[
+\Phi_B = B\,A\,\cos 60.0^\circ = (0.450\,\mathrm{T})(9.60\times 10^{-3}\,\mathrm{m^2})\,(0.500).
+\]
+Compute:
+\[
+\Phi_B = 0.450 \times 9.60\times 10^{-3} \times 0.500\,\mathrm{Wb} = 2.16\times 10^{-3}\,\mathrm{Wb}.
+\]
+
+\textbf{(d)} The loop is in the $yz$-plane with normal $\hat{n}=+\hat{k}$, and the field is $\vec{B}=B\,\hat{k}$. Again $\theta = 0^\circ$:
+\[
+\Phi_B = B\,A\,\cos 0^\circ = (0.450\,\mathrm{T})(9.60\times 10^{-3}\,\mathrm{m^2})(1).
+\]
+\[
+\Phi_B = 4.32\times 10^{-3}\,\mathrm{Wb}.
+\]
+The orientation of the loop in space does not matter; only the relative angle between $\vec{B}$ and the area normal matters. Since $\vec{B}$ is again perpendicular to the loop and aligned with the normal, the flux is the same as in part (a).
+
+\textbf{(e)} The loop is in the $xy$-plane with normal $\hat{n}=+\hat{k}$. The field is $\vec{B}=(0.450\,\mathrm{T})\,\hat{\imath}+(0.300\,\mathrm{T})\,\hat{\jmath}$. The area vector is $\vec{A} = A\,\hat{k}$. The flux is the dot product:
+\[
+\Phi_B = \vec{B}\cdot\vec{A} = \bigl[(0.450)\,\hat{\imath}+(0.300)\,\hat{\jmath}\bigr]\cdot\bigl[(9.60\times 10^{-3})\,\hat{k}\bigr].
+\]
+Since $\hat{\imath}\cdot\hat{k}=0$ and $\hat{\jmath}\cdot\hat{k}=0$:
+\[
+\Phi_B = 0.
+\]
+The field lies entirely in the $xy$-plane, parallel to the loop, so no flux passes through.
+
+\bigskip
+\textbf{Final answers:}
+\begin{enumerate}[label=(\alph*)]
+\item $\Phi_B = 4.32\times 10^{-3}\,\mathrm{Wb} = 4.32\,\mathrm{mWb}$
+\item $\Phi_B = 0$
+\item $\Phi_B = 2.16\times 10^{-3}\,\mathrm{Wb} = 2.16\,\mathrm{mWb}$
+\item $\Phi_B = 4.32\times 10^{-3}\,\mathrm{Wb} = 4.32\,\mathrm{mWb}$
+\item $\Phi_B = 0$
+\end{enumerate}

concepts/em/u13/e13-2-faraday.tex

@@ -0,0 +1,160 @@
+\subsection{Faraday's Law of Induction}
+
+This subsection states Faraday's law of induction, explains the meaning of the sign in the law via Lenz's law, and derives the motional-EMF formula $\mathcal{E}=B\ell v$ as a special case. It establishes the connection between Faraday's law and energy conservation.
+
+\dfn{Electromotive force (EMF)}{The \emph{electromotive force} (EMF) $\mathcal{E}$ around a closed loop $C$ is the work done per unit charge by the non-electrostatic forces that drive charge carriers around the loop. In terms of the total force per unit charge $\vec{f}_{\text{non-e}}$ (which may include magnetic Lorentz forces or induced electric fields),
+\[
+\mathcal{E}=\oint_C \vec{f}_{\text{non-e}}\cdot d\vec{\ell}.
+\]
+In a resistive loop of total resistance $R$, the induced current obeys $\mathcal{E}=IR$.}
+
+\thm{Faraday's law of induction}{Let $C$ be a closed conducting loop and $S$ any surface bounded by $C$. The magnetic flux through $S$ is
+\[
+\Phi_B=\int_S \vec{B}\cdot d\vec{A},
+\]
+where $d\vec{A}$ is the oriented area element (direction given by the right-hand rule from the chosen circulation direction around $C$). Faraday's law states that the induced electromotive force around the loop is
+\[
+\mathcal{E}=-\frac{d\Phi_B}{dt}.
+\]
+The minus sign encodes Lenz's law: the induced current produces a magnetic field that opposes the change in flux that created it.}
+
+\pf{Derivation from Faraday's law}{
+The EMF is the work per unit charge by the induced non-conservative field: $\mathcal{E}=\oint_C\vec{E}\cdot d\vec{\ell}$. When the magnetic flux $\Phi_B=\int_S\vec{B}\cdot d\vec{A}$ through the loop changes in time, a non-conservative electric field is induced with non-zero circulation. Energy conservation requires this circulation to equal the rate of flux change (with the minus sign from Lenz's law):
+\[
+\mathcal{E}=\oint_C\vec{E}\cdot d\vec{\ell}=-\frac{d\Phi_B}{dt},
+\]
+which is Faraday's law of induction.
+}
+
+\nt{The surface $S$ is not unique --- any surface bounded by the same loop $C$ gives the same flux because $\nabla\cdot\vec{B}=0$. The orientation of $d\vec{A}$ is fixed by the right-hand rule applied to the chosen direction of integration around $C$: if your fingers curl along the integration direction, your thumb points along $d\vec{A}$. The sign of $\Phi_B$ then tells you whether the field generally threads the loop in the ``positive'' direction (along $d\vec{A}$) or the ``negative'' direction (opposite to $d\vec{A}$). A negative $d\Phi_B/dt$ means the flux in the positive direction is decreasing, so $\mathcal{E}>0$ and the induced EMF drives current in the positive (integration) direction.}
+
+\nt{Faraday's law can be understood through two complementary mechanisms that both change the flux:
+\begin{itemize}
+\item \textbf{Transformer EMF:} The loop is stationary but the magnetic field $\vec{B}(t)$ changes with time. An induced non-conservative electric field $\vec{E}$ is created by $\nabla\times\vec{E}=-\partial\vec{B}/\partial t$, and this field drives the current.
+\item \textbf{Motional EMF:} The magnetic field is static but the loop moves, changes shape, or rotates. The magnetic component of the Lorentz force $\vec{F}=q\vec{v}\times\vec{B}$ on the charge carriers drives the current.
+\end{itemize}
+Both mechanisms give the same $\mathcal{E}=-d\Phi_B/dt$; the distinction is frame-dependent.}
+
+\mprop{Differential form (Maxwell--Faraday equation)}{A time-varying magnetic field produces a spatially non-conservative electric field. In differential form,
+\[
+\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}.
+\]
+By Stokes' theorem, for a stationary loop $C$ bounding surface $S$, this is equivalent to the integral form:
+\[
+\oint_C \vec{E}\cdot d\vec{\ell}=-\iint_S \frac{\partial\vec{B}}{\partial t}\cdot d\vec{A}=-\frac{d}{dt}\iint_S \vec{B}\cdot d\vec{A}=-\frac{d\Phi_B}{dt}.
+\]
+This equation describes the transformer-EMF mechanism and holds even in the absence of any physical conductor.}
+
+\mprop{EMF for a coil of $N$ turns}{If a tightly wound coil has $N$ identical turns and the same flux $\Phi_B$ passes through each turn, the total EMF is
+\[
+\mathcal{E}=-N\,\frac{d\Phi_B}{dt}.
+\]
+The product $N\Phi_B$ is called the \emph{flux linkage}. The factor $N$ appears because the EMFs of the individual turns add in series.}
+
+\mprop{Motional EMF}{When a straight conductor of length $\ell$ moves with velocity $\vec{v}$ perpendicular to a uniform magnetic field $\vec{B}$, the magnetic Lorentz force on the charge carriers produces an EMF
+\[
+\mathcal{E}=B\,\ell\,v.
+\]
+More generally, for a moving loop the motional EMF is $\mathcal{E}=\oint (\vec{v}\times\vec{B})\cdot d\vec{\ell}$, and the total EMF (including any transformer contribution) is $\mathcal{E}=-d\Phi_B/dt$.}
+
+\cor{Energy conservation and the sign of Faraday's law}{The minus sign in Faraday's law is required by conservation of energy. If the induced current reinforced the flux change instead of opposing it, a small perturbation would produce a positive feedback loop that creates energy from nothing. The minus sign ensures that mechanical work done to change the flux is converted to electrical energy (dissipated as Joule heat in the resistance of the loop).}
+
+\ex{Illustrative example}{A conducting rod of length $\ell=0.50\,\mathrm{m}$ slides at constant speed $v=4.0\,\mathrm{m/s}$ on frictionless rails in a uniform magnetic field $B=0.30\,\mathrm{T}$ perpendicular to the plane of the circuit. The rod and rails form a closed loop of total resistance $R=2.0\,\Omega$. The induced EMF is $\mathcal{E}=B\ell v=(0.30\,\mathrm{T})(0.50\,\mathrm{m})(4.0\,\mathrm{m/s})=0.60\,\mathrm{V}$, and the induced current is $I=\mathcal{E}/R=0.30\,\mathrm{A}$. The magnetic force on the rod is $F=BI\ell=0.036\,\mathrm{N}$, opposing the motion.}
+
+\qs{Worked example}{A rectangular conducting loop has horizontal width $L=0.400\,\mathrm{m}$ and vertical height $w=0.200\,\mathrm{m}$, and total resistance $R=5.00\,\Omega$. The loop lies in the $xy$-plane and is partially inside a region of uniform magnetic field $\vec{B}=0.750\,\mathrm{T}$ pointing in the $-\hat{k}$ direction (into the page). The field region is confined to the half-space $x<x_0$, and the loop is being pulled to the right with constant velocity $v=3.00\,\mathrm{m/s}$, as shown. At the instant shown, the right edge of the loop is outside the field region and the left edge is still inside it.
+
+\begin{enumerate}[label=(\alph*)]
+\item Find the magnitude and direction of the induced current in the loop.
+\item Find the magnitude and direction of the applied force required to maintain the constant velocity.
+\item Verify that the mechanical power input by the applied force equals the electrical power dissipated in the loop's resistance.
+\end{enumerate}}
+
+\sol
+\textbf{Geometry and sign conventions.} The loop lies in the $xy$-plane. The magnetic field is $\vec{B}=-(0.750\,\mathrm{T})\,\hat{k}$. Choose the integration direction around the loop to be clockwise (as viewed from above the $xy$-plane). By the right-hand rule, this makes the area vector point in the $-\hat{k}$ direction (into the page), so $d\vec{A}=dA\,(-\hat{k})$. The area vector and the field are in the same direction, so the flux $\Phi_B$ is positive.
+
+Let $x$ be the horizontal extent of the portion of the loop still inside the field region. At the instant shown, the loop is partially out, so $0<x<L$. The flux is
+\[
+\Phi_B=\vec{B}\cdot\vec{A}=BA=B\,w\,x,
+\]
+where $A=w\,x$ is the area of the loop inside the field. As the loop is pulled to the right, $x$ decreases, so $dx/dt=-v$.
+
+\textbf{(a) Induced current.} The rate of change of flux is
+\[
+\frac{d\Phi_B}{dt}=B\,w\,\frac{dx}{dt}=B\,w\,(-v)=-B\,w\,v.
+\]
+By Faraday's law, the EMF is
+\[
+\mathcal{E}=-\frac{d\Phi_B}{dt}=B\,w\,v.
+\]
+Substitute the values:
+\[
+\mathcal{E}=(0.750\,\mathrm{T})(0.200\,\mathrm{m})(3.00\,\mathrm{m/s}).
+\]
+Compute step by step:
+\[
+(0.750)(0.200)=0.150,\qquad (0.150)(3.00)=0.450,
+\]
+so
+\[
+\mathcal{E}=0.450\,\mathrm{V}.
+\]
+The EMF is positive, meaning the induced current flows in the chosen (clockwise) direction. Using Ohm's law:
+\[
+I=\frac{\mathcal{E}}{R}=\frac{0.450\,\mathrm{V}}{5.00\,\Omega}=0.0900\,\mathrm{A}.
+\]
+The current flows \textbf{clockwise} as viewed from above.
+
+\textbf{(b) Applied force.} The portion of the loop inside the field that carries current and experiences a magnetic force is the left (leading) vertical segment, of length $w=0.200\,\mathrm{m}$. The current in this segment flows downward (consistent with the clockwise circulation). The magnetic field points into the page ($-\hat{k}$). The magnetic force on this segment is
+\[
+\vec{F}_{\text{mag}}=I\vec{\ell}\times\vec{B}=Iw(-\hat{\jmath})\times(-B\,\hat{k})=IBw\,(\hat{\jmath}\times\hat{k})=IBw\,\hat{\imath}.
+\]
+The magnetic force points to the right (in the direction of motion). To maintain constant velocity, the net force must be zero, so the applied force must balance the magnetic force:
+\[
+\vec{F}_{\text{app}}+\vec{F}_{\text{mag}}=0\quad\Rightarrow\quad\vec{F}_{\text{app}}=-IBw\,\hat{\imath}.
+\]
+The applied force points to the \textbf{left} (opposite to the direction of motion) with magnitude
+\[
+F_{\text{app}}=IBw=BIw.
+\]
+Substitute the values:
+\[
+F_{\text{app}}=(0.750\,\mathrm{T})(0.0900\,\mathrm{A})(0.200\,\mathrm{m}).
+\]
+Compute:
+\[
+(0.750)(0.0900)=0.0675,\qquad(0.0675)(0.200)=0.0135,
+\]
+so
+\[
+F_{\text{app}}=0.0135\,\mathrm{N}.
+\]
+
+\textbf{(c) Energy conservation.} The electrical power dissipated in the loop is
+\[
+P_{\text{elec}}=I^2R=I\mathcal{E}=I(Bwv).
+\]
+Substitute:
+\[
+P_{\text{elec}}=(0.0900\,\mathrm{A})(0.450\,\mathrm{V})=0.0405\,\mathrm{W}.
+\]
+The mechanical power delivered by the applied force is
+\[
+P_{\text{mech}}=\vec{F}_{\text{app}}\cdot\vec{v}.
+\]
+Since $\vec{F}_{\text{app}}$ points left and $\vec{v}$ points right, $\vec{F}_{\text{app}}\cdot\vec{v}=-F_{\text{app}}v$.
+\[
+P_{\text{mech}}=-(0.0135\,\mathrm{N})(3.00\,\mathrm{m/s})=-0.0405\,\mathrm{W}.
+\]
+The mechanical power is negative because the applied force opposes the motion (you are holding back the loop). The magnetic force does positive work at rate $+0.0405\,\mathrm{W}$, which exactly equals the electrical power dissipated. Thus,
+\[
+|P_{\text{mech}}|=P_{\text{elec}}=0.0405\,\mathrm{W}.
+\]
+This verifies energy conservation: mechanical energy lost by the system equals electrical energy dissipated as heat.
+
+\bigskip
+\textbf{Final answers:}
+\begin{enumerate}[label=(\alph*)]
+\item $I=0.0900\,\mathrm{A}$, clockwise.
+\item $F_{\text{app}}=0.0135\,\mathrm{N}$, to the left.
+\item $P_{\text{mech}}=-0.0405\,\mathrm{W}$, $P_{\text{elec}}=0.0405\,\mathrm{W}$; magnitudes equal.
+\end{enumerate}

concepts/em/u13/e13-3-lenz.tex

@@ -0,0 +1,51 @@
+\subsection{Lenz's Law and Induced Current Direction}
+
+This subsection explains how the sign in Faraday's law encodes the direction of induced current, states Lenz's law in its operational form, and connects it to energy conservation.
+
+\dfn{Lenz's law and induced current direction}{Let a closed conducting loop bound an oriented surface $S$ with unit normal $\hat{n}$ determined by the right-hand rule from the chosen circulation direction. The magnetic flux through the loop is
+\[
+\Phi_B=\int_S \vec{B}\cdot d\vec{A}.
+\]
+Faraday's law gives the induced electromotive force around the loop:
+\[
+\mathcal{E}=-\frac{d\Phi_B}{dt}.
+\]
+\textbf{Lenz's law} states that the induced current flows in the direction that produces a magnetic field $\vec{B}_{\text{ind}}$ opposing the change in flux $\Phi_B$.
+
+The operational procedure is:
+\begin{enumerate}[label=(\arabic*)]
+\item Determine the direction of the external magnetic field $\vec{B}_{\text{ext}}$ through the loop.
+\item Determine whether $\Phi_B$ is increasing or decreasing.
+\item The induced field $\vec{B}_{\text{ind}}$ points in the same direction as $\vec{B}_{\text{ext}}$ if the flux is decreasing, and in the opposite direction if the flux is increasing.
+\item Use the right-hand rule on $\vec{B}_{\text{ind}}$ to find the induced current direction: curl the fingers of your right hand in the current direction; your thumb points along $\vec{B}_{\text{ind}}$.
+\end{enumerate}}
+
+\nt{The negative sign in Faraday's law \emph{is} Lenz's law written as an equation. If the sign were positive, the induced current would reinforce the flux change, producing more flux in the same direction, which would drive yet more current --- an energy-creating runaway. Lenz's law prevents this by ensuring the induced field opposes the change.}
+
+\thm{Lenz's law (energy-conservation form)}{The direction of induced current in any closed loop is always such that the magnetic force or torque on the loop opposes the motion or change that produced the induction. Equivalently, mechanical work must be done against the magnetic forces to sustain the change in flux; this work is converted to electrical energy (and ultimately to thermal energy in the resistance of the loop).}
+
+\pf{Lenz's law from energy conservation}{Suppose a magnet is pushed toward a conducting loop. The induced current creates a magnetic field $\vec{B}_{\text{ind}}$ that opposes the approaching magnet. An external agent must do positive work against the magnetic repulsion to keep the magnet moving. This work supplies the electrical energy dissipated as Joule heating in the loop.
+
+If Lenz's law were reversed --- if the induced field \emph{aided} the approaching magnet --- the magnet would accelerate toward the loop without any external work, increasing both the kinetic energy of the magnet and the electrical energy dissipated in the loop, with no energy input. This violates conservation of energy. Therefore, the minus sign in Faraday's law is required by energy conservation. \Qed}
+
+\cor{Flux-change sign convention}{When the flux through a loop is increasing ($d\Phi_B/dt>0$), the induced EMF is negative and the induced current flows in the direction that creates a field opposing the external field. When the flux is decreasing ($d\Phi_B/dt<0$), the induced EMF is positive and the induced current flows in the direction that reinforces the external field.}
+
+\ex{Illustrative example}{A circular loop of radius $R$ lies in a uniform magnetic field $\vec{B}$ pointing out of the page. The radius is shrunk at constant speed. Since the outward flux $\Phi_B=BR^2\pi$ is decreasing, the induced current must create an outward field to oppose the decrease. By the right-hand rule, this corresponds to a counterclockwise current.}
+
+\qs{Worked example}{A bar magnet with its north pole facing downward is released from rest above a horizontal copper ring. The magnet falls along the central axis of the ring.
+
+\begin{enumerate}[label=(\alph*)]
+\item As the north pole \emph{approaches} the ring from above, what is the direction of the induced current in the ring as viewed from above?
+\item Once the magnet has passed through and the south pole is \emph{leaving} the ring from below, what is the direction of the induced current in the ring as viewed from above?
+\end{enumerate}}
+
+\sol We view the ring from above (looking downward along the axis of the magnet). Let downward be the direction of the external field $\vec{B}_{\text{ext}}$ through the ring (the field lines emerge from the north pole and point downward through the ring while the north pole is above it, and continue downward through the ring while the south pole is below it).
+
+\noindent\textbf{Part (a):} As the north pole approaches the ring from above, the downward magnetic field through the ring becomes stronger. The downward flux $\Phi_B$ is therefore \emph{increasing}. By Lenz's law, the induced current must create an induced magnetic field $\vec{B}_{\text{ind}}$ that opposes this increase, so $\vec{B}_{\text{ind}}$ must point \emph{upward} (out of the page). Using the right-hand rule, with the thumb pointing upward, the fingers curl \emph{counterclockwise}. The induced current is \textbf{counterclockwise} as viewed from above.
+
+\noindent\textbf{Part (b):} As the south pole leaves the ring from below, the downward magnetic field through the ring becomes weaker. The downward flux $\Phi_B$ is therefore \emph{decreasing}. By Lenz's law, the induced current must create an induced magnetic field $\vec{B}_{\text{ind}}$ that opposes this decrease, so $\vec{B}_{\text{ind}}$ must point \emph{downward} (into the page) to supplement the collapsing field. Using the right-hand rule, with the thumb pointing downward, the fingers curl \emph{clockwise}. The induced current is \textbf{clockwise} as viewed from above.
+
+Therefore,
+\[
+\text{(a) counterclockwise,} \qquad \text{(b) clockwise.}
+\]

concepts/em/u13/e13-4-motional-emf.tex

@@ -0,0 +1,205 @@
+\subsection{Motional Electromotive Force}
+
+When a conductor moves through a magnetic field, the magnetic force on the charge carriers inside the conductor separates charge and produces an electromotive force. This effect, called \emph{motional emf}, is one of the two fundamental mechanisms of electromagnetic induction (the other being a time-varying magnetic field).
+
+\dfn{Motional emf}{When a conductor moves with velocity $\vec{v}$ through a magnetic field $\vec{B}$, each charge carrier of charge $q$ experiences the magnetic force
+\[
+\vec{F}_B = q\,\vec{v}\times\vec{B}.
+\]
+This force acts as a non-electrostatic force per unit charge, $\vec{v}\times\vec{B}$, which drives charges along the conductor. The \emph{motional emf} along a path from point $a$ to point $b$ within the conductor is the line integral
+\[
+\mathcal{E} = \int_{a}^{b} (\vec{v}\times\vec{B})\cdot d\vec{\ell},
+\]
+where $d\vec{\ell}$ is an infinitesimal displacement vector along the conductor from $a$ to $b$. The motional emf represents the work done per unit charge by the magnetic force as charges move along the conductor.}
+
+\nt{Even though the magnetic force itself does no net work on a moving charge (since $\vec{F}_B\perp\vec{v}$), the motional emf arises because the conductor's motion provides the energy transfer mechanism. The external agent pushing the conductor does mechanical work; the magnetic field acts as the intermediary that converts this work into electrical energy.}
+
+\nt{Inside the moving conductor (the ``source''), the emf drives current from lower potential to higher potential, just as a battery drives current from its negative to its positive terminal. The moving conductor thus acts as a battery with emf $\mathcal{E}$ and zero internal resistance (if the conductor is ideal).}
+
+\thm{Motional emf of a straight conductor in a uniform field}{Let a straight conductor of length $\ell$ move with constant velocity $\vec{v}$ through a uniform magnetic field $\vec{B}$.
+
+\begin{itemize}
+\item \textbf{General case:} The motional emf between the two ends of the conductor is
+\[
+\mathcal{E} = \int_{a}^{b} (\vec{v}\times\vec{B})\cdot d\vec{\ell},
+\]
+where the integral is taken along the conductor from end $a$ to end $b$.
+
+\item \textbf{Mutually perpendicular case:} When $\vec{v}$, $\vec{B}$, and the conductor are mutually perpendicular, the motional emf simplifies to
+\[
+\mathcal{E} = B\,\ell\,v.
+\]
+
+\item \textbf{Direction of emf:} The end of the conductor toward which positive charges are pushed by $\vec{v}\times\vec{B}$ is at higher potential. Use the right-hand rule for the cross product $\vec{v}\times\vec{B}$: point your fingers along $\vec{v}$, curl toward $\vec{B}$; your thumb points in the direction of $\vec{v}\times\vec{B}$, which is the direction positive charges move inside the conductor.
+\end{itemize}}
+
+\pf{Motional emf from the Lorentz force}{Consider a straight conducting bar of length $\ell$ moving with velocity $\vec{v}$ to the right through a uniform magnetic field $\vec{B}$ pointing out of the page. The bar is oriented vertically, perpendicular to both $\vec{v}$ and $\vec{B}$.
+
+Each conduction electron of charge $-e$ experiences a magnetic force
+\[
+\vec{F}_e = -e\,(\vec{v}\times\vec{B}).
+\]
+With $\vec{v}$ to the right ($+\hat{\imath}$) and $\vec{B}$ out of the page ($+\hat{k}$),
+\[
+\vec{v}\times\vec{B} = vB\,(\hat{\imath}\times\hat{k}) = -vB\,\hat{\jmath},
+\]
+so the force on electrons is $-e(-vB\,\hat{\jmath}) = evB\,\hat{\jmath}$, pointing \emph{upward} along the bar. Electrons accumulate at the top, leaving the bottom end positively charged.
+
+Charge separation continues until the resulting electrostatic field $\vec{E}_{\text{ind}}$ balances the magnetic force:
+\[
+-e\,\vec{E}_{\text{ind}} + (-e)(\vec{v}\times\vec{B}) = \vec{0}
+\quad\Rightarrow\quad
+\vec{E}_{\text{ind}} = -(\vec{v}\times\vec{B}).
+\]
+The motional emf is the line integral of the non-electrostatic force per unit charge along the bar. Equivalently, it equals the potential difference between the ends:
+\[
+\mathcal{E} = \int_{\text{bottom}}^{\text{top}} \vec{E}_{\text{ind}}\cdot d\vec{\ell} = E_{\text{ind}}\,\ell.
+\]
+Since $\vec{E}_{\text{ind}} = vB\,\hat{\jmath}$ and the bar extends along $\hat{\jmath}$, we obtain
+\[
+\mathcal{E} = vB\,\ell.
+\]
+More generally, without assuming perpendicularity,
+\[
+\mathcal{E} = \int_{a}^{b} (\vec{v}\times\vec{B})\cdot d\vec{\ell},
+\]
+which reduces to $B\ell v$ when $\vec{v}$, $\vec{B}$, and $d\vec{\ell}$ are mutually perpendicular. \Qed}
+
+\mprop{Motional emf in a circuit with resistance}{A conducting bar of length $\ell$ slides on frictionless conducting rails in a uniform magnetic field $B$, with the rails connected by a resistor $R$. The bar moves with instantaneous speed $v$ perpendicular to $B$. Then:
+\begin{enumerate}
+\item The motional emf is
+\[
+\mathcal{E} = B\,\ell\,v.
+\]
+\item The induced current in the circuit (magnitude) is
+\[
+I = \frac{\mathcal{E}}{R} = \frac{B\,\ell\,v}{R}.
+\]
+\item The magnetic force on the bar (magnitude) opposes the motion:
+\[
+F_{\text{mag}} = I\,\ell\,B = \frac{B^{2}\,\ell^{2}\,v}{R}.
+\]
+\item The external force needed to maintain constant velocity $v$ equals the magnetic force in magnitude:
+\[
+F_{\text{ext}} = \frac{B^{2}\,\ell^{2}\,v}{R}.
+\]
+\item Power dissipated in the resistor:
+\[
+P_{\text{elec}} = I^{2}R = \frac{B^{2}\,\ell^{2}\,v^{2}}{R}.
+\]
+\item Mechanical power delivered by the external agent:
+\[
+P_{\text{mech}} = F_{\text{ext}}\,v = \frac{B^{2}\,\ell^{2}\,v^{2}}{R}.
+\]
+\end{enumerate}
+Thus $P_{\text{mech}} = P_{\text{elec}}$, consistent with conservation of energy.}
+
+\cor{Motional emf and Faraday's law agree}{For a conducting bar sliding on rails, Faraday's law gives the induced emf as $\mathcal{E} = -\dfrac{d\Phi_B}{dt}$. If the bar moves a distance $dx$ in time $dt$, the area changes by $dA = \ell\,dx$ and the flux by $d\Phi_B = B\,\ell\,dx$, so
+\[
+\left\lvert\frac{d\Phi_B}{dt}\right\rvert = B\,\ell\,\frac{dx}{dt} = B\,\ell\,v.
+\]
+This is exactly the motional-emf result $\mathcal{E} = B\ell v$. Faraday's law provides the same emf through the ``flux-change'' perspective, while the motional-emf derivation provides it through the ``force-on-charges'' perspective.}
+
+\ex{Illustrative example}{A metal rod of length $\ell$ rotates with angular speed $\omega$ about one end in a uniform magnetic field $B$ perpendicular to the plane of rotation. Different points on the rod have different speeds $v(r) = r\,\omega$, so the emf must be computed by integration:
+\[
+\mathcal{E} = \int_{0}^{\ell} B\,(r\omega)\,dr = \frac{1}{2}\,B\,\omega\,\ell^{2}.\]}
+
+\qs{Worked example}{A conducting bar of length $\ell = 0.50\,\mathrm{m}$ slides on two horizontal, frictionless, conducting rails connected by a resistor $R = 4.0\,\Omega$ at the left end, as shown in the figure. A uniform magnetic field $\vec{B} = (0.30\,\mathrm{T})\,\hat{k}$ points straight upward (out of the horizontal plane of the rails). At a particular instant, the bar is moving to the right with velocity $v = 2.0\,\mathrm{m/s}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Find the motional emf induced in the circuit at this instant.
+\item Find the induced current: magnitude and direction (clockwise or counterclockwise as viewed from above).
+\item Find the magnitude and direction of the external force that must be applied to the bar to maintain this constant velocity.
+\item Find the power dissipated in the resistor. Verify that the mechanical power delivered by the external force equals the electrical power dissipated.
+\end{enumerate}}
+
+\textbf{Given quantities:}
+\begin{itemize}
+\item Bar length: $\ell = 0.50\,\mathrm{m}$
+\item Resistance: $R = 4.0\,\Omega$
+\item Magnetic field: $B = 0.30\,\mathrm{T}$ (out of page)
+\item Bar velocity: $v = 2.0\,\mathrm{m/s}$ (to the right)
+\item $\vec{v}$, $\vec{B}$, and the bar are mutually perpendicular.
+\end{itemize}
+
+\sol \textbf{(a) Motional emf.} Since $\vec{v}$, $\vec{B}$, and the bar are mutually perpendicular, the motional emf is
+\[
+\mathcal{E} = B\,\ell\,v.
+\]
+Substitute the given values:
+\[
+\mathcal{E} = (0.30\,\mathrm{T})(0.50\,\mathrm{m})(2.0\,\mathrm{m/s})
+= (0.30)(1.0)\,\mathrm{V} = 0.30\,\mathrm{V}.
+\]
+
+\textbf{(b) Induced current.} The magnitude of the induced current is given by Ohm's law:
+\[
+I = \frac{\mathcal{E}}{R} = \frac{0.30\,\mathrm{V}}{4.0\,\Omega}
+= 0.075\,\mathrm{A} = 75\,\mathrm{mA}.
+\]
+
+To find the direction, apply Lenz's law. The bar moves to the right, so the area of the loop increases, and the upward magnetic flux $\Phi_B$ through the loop is \emph{increasing}. Lenz's law requires the induced current to create a magnetic field $\vec{B}_{\text{ind}}$ that opposes this increase, so $\vec{B}_{\text{ind}}$ must point \emph{into} the page (downward). By the right-hand rule, a current that produces an into-the-page field flows \emph{clockwise} as viewed from above.
+
+Alternatively, use the force-on-charges argument: inside the moving bar, positive charge carriers experience $\vec{v}\times\vec{B}$, which points upward along the bar (since $\hat{\imath}\times\hat{k}=-\hat{\jmath}$ and positive charges move in the $+\hat{\jmath}$ direction if the bar is oriented in that direction). The upper end is thus at higher potential. Current flows from high to low potential through the external circuit (the resistor), i.e., clockwise.
+
+\[
+I = 0.075\,\mathrm{A},\quad \text{clockwise (as viewed from above)}.
+\]
+
+\textbf{(c) External force.} The induced current in the bar flows \emph{downward} (from the top rail to the bottom rail), i.e., in the $-\hat{\jmath}$ direction. Therefore, $\vec{\ell} = -\ell\,\hat{\jmath}$, and the magnetic force on the bar is
+\[
+\vec{F}_{\text{mag}} = I\,\vec{\ell}\times\vec{B} = I\,(-\ell\,\hat{\jmath})\times(B\,\hat{k})
+= -I\,\ell\,B\,(\hat{\jmath}\times\hat{k}) = -I\,\ell\,B\,\hat{\imath}.
+\]
+The magnetic force on the bar points to the \emph{left}, opposing the motion. Its magnitude is
+\[
+F_{\text{mag}} = I\,\ell\,B
+= (0.075\,\mathrm{A})(0.50\,\mathrm{m})(0.30\,\mathrm{T})
+= (0.075)(0.15)\,\mathrm{N} = 0.01125\,\mathrm{N}.
+\]
+To maintain constant velocity (zero net force), the external force must exactly balance the magnetic force:
+\[
+\vec{F}_{\text{ext}} = -\vec{F}_{\text{mag}} = +(0.01125\,\mathrm{N})\,\hat{\imath},
+\]
+i.e., $F_{\text{ext}} = 0.01125\,\mathrm{N}$ to the \emph{right}.
+
+Rounding to two significant figures (matching the given data):
+\[
+F_{\text{ext}} = 0.011\,\mathrm{N}\text{ to the right}.
+\]
+
+\textbf{(d) Power.} The electrical power dissipated in the resistor is
+\[
+P_{\text{elec}} = I^{2}R = (0.075\,\mathrm{A})^{2}(4.0\,\Omega)
+= (0.005625)(4.0)\,\mathrm{W} = 0.0225\,\mathrm{W}.
+\]
+Equivalently,
+\[
+P_{\text{elec}} = \frac{\mathcal{E}^{2}}{R}
+= \frac{(0.30\,\mathrm{V})^{2}}{4.0\,\Omega}
+= \frac{0.090}{4.0}\,\mathrm{W} = 0.0225\,\mathrm{W}.
+\]
+
+The mechanical power delivered by the external force is
+\[
+P_{\text{mech}} = F_{\text{ext}}\,v
+= (0.01125\,\mathrm{N})(2.0\,\mathrm{m/s})
+= 0.0225\,\mathrm{W}.
+\]
+
+Since $P_{\text{mech}} = P_{\text{elec}} = 0.0225\,\mathrm{W}$, mechanical energy is fully converted to electrical energy (Joule heating), as expected from energy conservation.
+
+\[
+P_{\text{elec}} = 0.0225\,\mathrm{W} = 22.5\,\mathrm{mW},
+\qquad
+P_{\text{mech}} = 0.0225\,\mathrm{W}.
+\]
+
+\bigskip
+\noindent\textbf{Summary of results:}
+\begin{enumerate}[label=(\alph*)]
+\item $\mathcal{E} = 0.30\,\mathrm{V}$
+\item $I = 0.075\,\mathrm{A} = 75\,\mathrm{mA}$, clockwise
+\item $F_{\text{ext}} = 0.011\,\mathrm{N}$ to the right
+\item $P_{\text{elec}} = 0.0225\,\mathrm{W}$, verified $P_{\text{mech}} = P_{\text{elec}}$
+\end{enumerate}

concepts/em/u13/e13-5-inductance.tex

@@ -0,0 +1,216 @@
+\subsection{Inductance and Magnetic Energy Storage}
+
+This subsection covers self-inductance, mutual inductance, and the energy stored in magnetic fields. When the current through a conductor changes, the magnetic flux it produces also changes, inducing an electromotive force (EMF) that opposes that change --- a phenomenon governed by Faraday's law and Lenz's law. The energy required to establish the current is stored in the magnetic field and can be recovered when the current decreases.
+
+\dfn{Self-inductance}{The \emph{self-inductance} $L$ of a circuit (or coil) quantifies the magnetic flux linkage per unit current. If a current $I$ through a coil of $N$ turns produces a magnetic flux $\Phi_B$ through each turn, the total flux linkage is $N\Phi_B$. The self-inductance is defined by
+\[
+L = \frac{N\,\Phi_B}{I}.
+\]
+Equivalently, from Faraday's law of induction, a changing current induces an EMF
+\[
+\mathcal{E} = -\,L\,\frac{dI}{dt},
+\]
+where the negative sign reflects Lenz's law: the induced EMF opposes the change in current. The SI unit of inductance is the henry (H), where $1\;\mathrm{H} = 1\;\mathrm{V\!\cdot\!s/A} = 1\;\mathrm{kg\!\cdot\!m^2/(s^2\!\cdot\!A^2)}$.}
+
+\nt{Inductance is purely a geometric property. For a fixed geometry (and no ferromagnetic material near the coil), $L$ is constant and independent of the current. The larger the coil, the more turns, and the greater the flux linkage per unit current, the larger the inductance. A coil with $L = 1\;\mathrm{H}$ and $dI/dt = 1\;\mathrm{A/s}$ develops a $1\;\mathrm{V}$ back EMF.}
+
+\thm{Solenoid self-inductance}{For an ideal long solenoid of length $\ell \gg R$, total turns $N$, cross-sectional area $A = \pi R^2$, and vacuum permeability $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A}$, the magnetic field inside is $B = \mu_0\,n\,I$ where $n = N/\ell$. The flux through each turn is $\Phi_B = B\,A = \mu_0 N I A/\ell$. Therefore
+\[
+L = \frac{N\,\Phi_B}{I} = \frac{\mu_0\,N^2\,A}{\ell}.
+\]}
+
+\nt{The solenoid inductance scales as $N^2$ because each additional turn both increases the flux linkage ($N$ factor) and increases the magnetic field produced per unit current ($N$ factor through $n = N/\ell$). The inductance is proportional to the cross-sectional area $A$ and inversely proportional to the length $\ell$.}
+
+\dfn{Mutual inductance}{Consider two coils placed near each other. If a current $I_1$ in coil 1 produces a magnetic flux $\Phi_{21}$ through each turn of coil 2 (which has $N_2$ turns), the \emph{mutual inductance} $M_{21}$ is defined as
+\[
+M_{21} = \frac{N_2\,\Phi_{21}}{I_1}.
+\]
+By Faraday's law, a changing current in coil 1 induces an EMF in coil 2:
+\[
+\mathcal{E}_2 = -\,M_{21}\,\frac{dI_1}{dt}.
+\]
+Mutual inductance is symmetric: $M_{21} = M_{12} = M$. The SI unit is the henry (H).}
+
+\nt{Mutual inductance is the operating principle of transformers. The amount of flux from coil 1 that threads coil 2 depends on their relative orientation, separation, and the presence of magnetic materials. When the coils are perfectly coupled (all flux from one threads the other), $M$ reaches its maximum value. In practice, the coupling is characterized by the coefficient $k = M/\sqrt{L_1 L_2}$, where $0 \leq k \leq 1$.}
+
+\thm{Magnetic energy and energy density}{When a current $I$ flows through an inductor of inductance $L$, the magnetic field stores energy. The total energy stored is
+\[
+U = \frac{1}{2}\,L\,I^2.
+\]
+Inside an ideal solenoid, the magnetic field is uniform with magnitude $B = \mu_0 n I$. The energy can be expressed in terms of $B$ by noting that the energy is distributed throughout the volume $V = A\,\ell$. The \emph{magnetic energy density} (energy per unit volume) in vacuum is
+\[
+u_B = \frac{U}{V} = \frac{B^2}{2\,\mu_0}.
+\]}
+
+\pf{Magnetic energy from power}{The power delivered to an inductor by an external source to drive current $I$ against the back EMF $\mathcal{E} = -L\,dI/dt$ is
+\[
+P = -\,\mathcal{E}\,I = L\,I\,\frac{dI}{dt}.
+\]
+The rate of energy storage is $dU/dt = P$, so
+\[
+\frac{dU}{dt} = L\,I\,\frac{dI}{dt}.
+\]
+Integrating with respect to time as the current rises from $0$ to $I$:
+\[
+U = \int_0^I L\,i\;di = \frac{1}{2}\,L\,I^2.
+\]}
+
+\pf{Energy density of the B field (solenoid derivation)}{Consider an ideal solenoid with $N$ turns, length $\ell$, cross-sectional area $A$, carrying current $I$. Its inductance is $L = \mu_0 N^2 A/\ell$, and the stored energy is
+\[
+U = \frac{1}{2}\,L\,I^2 = \frac{1}{2}\,\frac{\mu_0\,N^2\,A}{\ell}\;I^2.
+\]
+The magnetic field inside is $B = \mu_0 N I/\ell$, so $I = B\,\ell/(\mu_0 N)$. Substituting:
+\[
+U = \frac{1}{2}\,\frac{\mu_0 N^2 A}{\ell}\,\left(\frac{B\,\ell}{\mu_0 N}\right)^{\!2} = \frac{1}{2}\,\frac{\mu_0 N^2 A}{\ell}\,\frac{B^2\,\ell^2}{\mu_0^2 N^2} = \frac{B^2}{2\,\mu_0}\;A\,\ell.
+\]
+The volume is $V = A\,\ell$, so the energy density is
+\[
+u_B = \frac{U}{V} = \frac{B^2}{2\,\mu_0}.
+\]
+This result is general: the energy density $u_B = B^2/(2\mu_0)$ holds at any point in space in vacuum wherever a magnetic field $B$ exists.}
+
+\nt{The magnetic energy density $u_B = B^2/(2\mu_0)$ is the magnetic analogue of the electric energy density $u_E = \tfrac{1}{2}\varepsilon_0 E^2$. In both cases, energy is stored in the field itself, distributed throughout space. This field-energy viewpoint is essential in electrodynamics: changing fields carry energy via the Poynting vector $\vec{S} = \vec{E}\times\vec{B}/\mu_0$.}
+
+\mprop{Key inductance and magnetic-energy formulas}{
+\begin{itemize}
+\item \textbf{Self-inductance definition:} $L = N\,\Phi_B/I = -\,\mathcal{E}/(dI/dt)$.
+\item \textbf{Solenoid inductance:} $L = \mu_0 N^2 A/\ell$.
+\item \textbf{Mutual inductance:} $M = N_2\,\Phi_{21}/I_1$.
+\item \textbf{Magnetic energy:} $U = \tfrac{1}{2} L I^2$.
+\item \textbf{Magnetic energy density (vacuum):} $u_B = B^2/(2\mu_0)$.
+\end{itemize}}
+
+\cor{Units check}{Energy density $u_B$ has units of joules per cubic metre ($\mathrm{J/m^3}$). Since $1\;\mathrm{J} = 1\;\mathrm{N\!\cdot\!m}$ and $1\;\mathrm{N} = 1\;\mathrm{T\!\cdot\!A\!\cdot\!m}$, we have $B^2/\mu_0$ in units of $(\mathrm{T})^2/(\mathrm{T\!\cdot\!m/A}) = \mathrm{T\!\cdot\!A/m} = (\mathrm{N/(A\!\cdot\!m)})\!\cdot\!(\mathrm{A/m}) = \mathrm{N/m^2} = \mathrm{J/m^3}$, as expected.}
+
+\ex{Illustrative example}{A toroidal solenoid has mean circumference $0.40\,\mathrm{m}$, cross-sectional area $2.0\times 10^{-3}\,\mathrm{m^2}$, and $N = 500$ turns. Its inductance is $L = \mu_0 N^2 A/\ell = (4\pi\times 10^{-7})(500)^2(2.0\times 10^{-3})/(0.40) = 1.57\times 10^{-3}\,\mathrm{H} = 1.57\,\mathrm{mH}$. At $I = 3.0\,\mathrm{A}$, the stored energy is $U = \tfrac{1}{2}(1.57\times 10^{-3})(3.0)^2 = 7.07\times 10^{-3}\,\mathrm{J} = 7.1\,\mathrm{mJ}$.}
+
+\qs{Worked example}{An ideal solenoid is $50.0\,\mathrm{cm}$ long and has $N = 1200$ turns uniformly distributed along its length. Its circular cross-section has radius $R = 2.0\,\mathrm{cm}$. A steady current $I = 5.0\,\mathrm{A}$ flows through the wire. Take $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A}$.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the self-inductance $L$ of the solenoid,
+\item the magnetic energy $U$ stored in the solenoid,
+\item the magnitude of the magnetic field $B$ inside the solenoid, and
+\item the magnetic energy density $u_B$ inside the solenoid.
+Verify that $U = u_B\,V$, where $V$ is the interior volume of the solenoid.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The cross-sectional area of the solenoid is
+\[
+A = \pi R^2 = \pi\,(0.020\,\mathrm{m})^2 = \pi\times 4.0\times 10^{-4}\,\mathrm{m^2} = 1.26\times 10^{-3}\,\mathrm{m^2}.
+\]
+The length is $\ell = 0.500\,\mathrm{m}$. Using the solenoid inductance formula:
+\[
+L = \frac{\mu_0\,N^2\,A}{\ell}.
+\]
+Substitute the values:
+\[
+L = \frac{(4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A})(1200)^2(1.26\times 10^{-3}\,\mathrm{m^2})}{0.500\,\mathrm{m}}.
+\]
+Compute step by step:
+\[
+(1200)^2 = 1.44\times 10^{6},
+\]
+\[
+\mu_0\,(1200)^2 = (4\pi\times 10^{-7})(1.44\times 10^{6}) = 4\pi\times 0.144 = 1.810\times 10^{-6}\,\mathrm{T\!\cdot\!m^2/A}.
+\]
+Then
+\[
+L = \frac{(1.810\times 10^{-6})(1.26\times 10^{-3})}{0.500}\,\mathrm{H} = \frac{2.28\times 10^{-9}}{0.500}\,\mathrm{H} = 4.56\times 10^{-3}\,\mathrm{H}.
+\]
+More precisely, carrying $\pi$ through:
+\[
+L = \frac{4\pi\times 10^{-7}\times 1.44\times 10^{6}\times \pi\times 4.0\times 10^{-4}}{0.500} = \frac{4\pi^2\times 5.76\times 10^{-5}}{0.500} = 8\pi^2\times 5.76\times 10^{-5}.
+\]
+\[
+\pi^2 \approx 9.87,\qquad L = 8\times 9.87\times 5.76\times 10^{-5}\,\mathrm{H} = 4.55\times 10^{-3}\,\mathrm{H}.
+\]
+So
+\[
+L = 4.55\times 10^{-3}\,\mathrm{H} = 4.55\,\mathrm{mH}.
+\]
+
+\textbf{Part (b).} The stored magnetic energy is
+\[
+U = \frac{1}{2}\,L\,I^2.
+\]
+Substitute:
+\[
+U = \frac{1}{2}\,(4.55\times 10^{-3}\,\mathrm{H})\,(5.0\,\mathrm{A})^2.
+\]
+Since $(5.0)^2 = 25.0$:
+\[
+U = \frac{1}{2}\,(4.55\times 10^{-3})(25.0)\,\mathrm{J} = \frac{1}{2}\,(0.114)\,\mathrm{J} = 5.68\times 10^{-2}\,\mathrm{J}.
+\]
+So
+\[
+U = 5.68\times 10^{-2}\,\mathrm{J} = 56.8\,\mathrm{mJ}.
+\]
+
+\textbf{Part (c).} The magnetic field inside the solenoid is
+\[
+B = \mu_0\,n\,I,
+\]
+where $n = N/\ell$ is the turn density:
+\[
+n = \frac{1200}{0.500\,\mathrm{m}} = 2400\,\mathrm{turns/m}.
+\]
+Substitute:
+\[
+B = (4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A})(2400\,\mathrm{m^{-1}})(5.0\,\mathrm{A}).
+\]
+Compute:
+\[
+2400\times 5.0 = 1.20\times 10^{4},
+\]
+\[
+B = 4\pi\times 10^{-7}\times 1.20\times 10^{4} = 4.80\pi\times 10^{-3}\,\mathrm{T}.
+\]
+Using $\pi \approx 3.1416$:
+\[
+B = 4.80\times 3.1416\times 10^{-3}\,\mathrm{T} = 1.51\times 10^{-2}\,\mathrm{T}.
+\]
+So
+\[
+B = 1.51\times 10^{-2}\,\mathrm{T} = 15.1\,\mathrm{mT}.
+\]
+
+\textbf{Part (d).} The magnetic energy density is
+\[
+u_B = \frac{B^2}{2\,\mu_0}.
+\]
+Substitute $B = 1.51\times 10^{-2}\,\mathrm{T}$:
+\[
+B^2 = (1.51\times 10^{-2})^2 = 2.28\times 10^{-4}\,\mathrm{T^2}.
+\]
+Then
+\[
+u_B = \frac{2.28\times 10^{-4}}{2(4\pi\times 10^{-7})}\,\mathrm{J/m^3} = \frac{2.28\times 10^{-4}}{2.51\times 10^{-6}}\,\mathrm{J/m^3} = 9.09\times 10^{1}\,\mathrm{J/m^3}.
+\]
+More precisely:
+\[
+u_B = \frac{2.283\times 10^{-4}}{2.513\times 10^{-6}}\,\mathrm{J/m^3} = 90.9\,\mathrm{J/m^3}.
+\]
+So
+\[
+u_B = 90.9\,\mathrm{J/m^3}.
+\]
+
+Now verify $U = u_B\,V$. The interior volume of the solenoid is
+\[
+V = A\,\ell = (\pi\times 4.0\times 10^{-4}\,\mathrm{m^2})(0.500\,\mathrm{m}) = 6.28\times 10^{-4}\,\mathrm{m^3}.
+\]
+Then
+\[
+u_B\,V = (90.9\,\mathrm{J/m^3})(6.28\times 10^{-4}\,\mathrm{m^3}) = 5.71\times 10^{-2}\,\mathrm{J}.
+\]
+Using more precise intermediate values: $B = 1.508\times 10^{-2}\,\mathrm{T}$, $B^2 = 2.274\times 10^{-4}$, $u_B = 90.6\,\mathrm{J/m^3}$, $V = 6.283\times 10^{-4}\,\mathrm{m^3}$, giving $u_B\,V = 5.69\times 10^{-2}\,\mathrm{J}$, which matches $U = 5.68\times 10^{-2}\,\mathrm{J}$ within rounding. The relation $U = u_B V$ holds, confirming that the energy is uniformly distributed throughout the solenoid volume at density $u_B = B^2/(2\mu_0)$.
+
+\bigskip
+\textbf{Final answers:}
+\begin{enumerate}[label=(\alph*)]
+\item $L = 4.55\times 10^{-3}\,\mathrm{H} = 4.55\,\mathrm{mH}$
+\item $U = 5.68\times 10^{-2}\,\mathrm{J} = 56.8\,\mathrm{mJ}$
+\item $B = 1.51\times 10^{-2}\,\mathrm{T} = 15.1\,\mathrm{mT}$
+\item $u_B = 90.9\,\mathrm{J/m^3}$, and $U = u_B\,V$ verified
+\end{enumerate}

concepts/em/u13/e13-6-lr-circuits.tex

@@ -0,0 +1,195 @@
+\subsection{LR Circuits and Transients}
+
+This subsection introduces the series LR circuit, derives the time-dependent current during both the growth and decay phases, and discusses the energy stored in the inductor's magnetic field. The inductor opposes changes in current through a self-induced back EMF, producing an exponential transient with characteristic time constant $\tau = L/R$.
+
+\dfn{LR circuit and self-induced EMF}{Consider a series circuit consisting of a battery with constant EMF $\mathcal{E}$, a resistor of resistance $R$, an inductor of inductance $L$, and a switch, all connected in a single closed loop. When current $I$ flows through the inductor, any change in current induces a back EMF across the inductor given by
+\[
+\mathcal{E}_L = -L\,\frac{dI}{dt}.
+\]
+By Kirchhoff's loop rule, the sum of potential differences around the loop is zero. Traversing the loop in the direction of the current:
+\[
+\mathcal{E} - IR - L\,\frac{dI}{dt} = 0.
+\]
+This first-order differential equation governs the time evolution of the current $I(t)$. The inductor's back EMF opposes any change in current, analogous to how inertia opposes changes in velocity in mechanics.}
+
+\nt{The inductor acts as a ``magnetic inertia'' element: just as a mass cannot change its velocity instantaneously, current through an inductor cannot change instantaneously. At the instant the switch is closed, the current is zero and the full battery EMF appears across the inductor. After a long time, the current reaches a steady value and the inductor behaves as a short circuit (zero voltage drop).}
+
+\mprop{Time constant of an LR circuit}{The \emph{time constant} $\tau$ characterizes how quickly the current changes in an LR circuit:
+\[
+\tau = \frac{L}{R}.
+\]
+The SI unit of inductance is the henry (H) and the SI unit of resistance is the ohm ($\Omega$). Since $1\,\mathrm{H} = 1\,\mathrm{V\!\cdot\!s/A}$ and $1\,\Omega = 1\,\mathrm{V/A}$, the ratio $L/R$ has units of seconds, confirming that $\tau$ is a characteristic time. At $t = \tau$, the current during growth has reached $(1 - e^{-1}) \approx 63.2\%$ of its maximum value.}
+
+\thm{Current growth in a series LR circuit}{Consider a series LR circuit with battery EMF $\mathcal{E}$, resistance $R$, and inductance $L$. The switch is closed at $t = 0$, with the initial current $I(0) = 0$. The current as a function of time is
+\[
+I(t) = \frac{\mathcal{E}}{R}\,\bigl(1 - e^{-t/\tau}\bigr),
+\]
+where $\tau = L/R$. The maximum (steady-state) current is
+\[
+I_{\text{max}} = \frac{\mathcal{E}}{R}.
+\]}
+
+\pf{Current growth derivation}{We solve the differential equation obtained from Kirchhoff's loop rule:
+\[
+\mathcal{E} - IR - L\,\frac{dI}{dt} = 0.
+\]
+Rearrange to isolate the time derivative:
+\[
+L\,\frac{dI}{dt} = \mathcal{E} - IR.
+\]
+Separate variables, bringing all $I$ terms to one side:
+\[
+\frac{dI}{\mathcal{E} - IR} = \frac{dt}{L}.
+\]
+Integrate both sides. On the left, substitute $u = \mathcal{E} - IR$, so $du = -R\,dI$:
+\[
+\int_{0}^{I(t)} \frac{dI}{\mathcal{E} - IR} = \int_{0}^{t} \frac{dt}{L}.
+\]
+The left-hand integral is
+\[
+-\frac{1}{R}\,\ln\Bigl(\mathcal{E} - IR\Bigr)\,\bigg|_{0}^{\,I(t)}
+= -\frac{1}{R}\,\ln\!\left(\frac{\mathcal{E} - IR(t)}{\mathcal{E}}\right).
+\]
+The right-hand integral is $t/L$. Thus,
+\[
+-\frac{1}{R}\,\ln\!\left(\frac{\mathcal{E} - IR(t)}{\mathcal{E}}\right) = \frac{t}{L}.
+\]
+Multiply by $-R$:
+\[
+\ln\!\left(\frac{\mathcal{E} - IR(t)}{\mathcal{E}}\right) = -\frac{R}{L}\,t.
+\]
+Exponentiate both sides:
+\[
+\frac{\mathcal{E} - IR(t)}{\mathcal{E}} = e^{-Rt/L}.
+\]
+Solve for $I(t)$:
+\[
+I(t) = \frac{\mathcal{E}}{R}\,\bigl(1 - e^{-Rt/L}\bigr).
+\]
+Since $\tau = L/R$, this is equivalent to $I(t) = (\mathcal{E}/R)\,(1 - e^{-t/\tau})$.
+In the limit $t \to \infty$, the exponential vanishes and $I \to \mathcal{E}/R$. \Qed}
+
+\thm{Current decay in a series LR circuit}{Consider a series LR circuit carrying an initial current $I_0$ (established in steady state with a battery that is then disconnected, leaving only $R$ and $L$ in a closed loop). With $I(0) = I_0$, the current as a function of time is
+\[
+I(t) = I_0\,e^{-t/\tau},
+\]
+where $\tau = L/R$.}
+
+\pf{Current decay derivation}{With the battery removed, Kirchhoff's loop rule gives
+\[
+IR + L\,\frac{dI}{dt} = 0.
+\]
+Separate variables:
+\[
+\frac{dI}{I} = -\frac{R}{L}\,dt.
+\]
+Integrate both sides from $t = 0$ to $t$:
+\[
+\int_{I_0}^{I(t)} \frac{dI}{I} = -\frac{R}{L}\,\int_{0}^{t} dt.
+\]
+The left-hand side gives $\ln(I/I_0)$ and the right-hand side gives $-(R/L)\,t$. Thus,
+\[
+\ln\!\left(\frac{I(t)}{I_0}\right) = -\frac{R}{L}\,t,
+\]
+and exponentiating,
+\[
+I(t) = I_0\,e^{-Rt/L} = I_0\,e^{-t/\tau}.
+\]
+In the limit $t \to \infty$, $I \to 0$. \Qed}
+
+\cor{Back EMF across the inductor (growth phase)}{Differentiating the growth current gives the self-induced EMF across the inductor:
+\[
+\mathcal{E}_L = -L\,\frac{dI}{dt}
+= -L\,\frac{\mathcal{E}}{R}\,\frac{R}{L}\,e^{-t/\tau}
+= -\mathcal{E}\,e^{-t/\tau}.
+\]
+At $t = 0$, $\mathcal{E}_L = -\mathcal{E}$ (the full battery EMF opposes the change). As $t \to \infty$, $\mathcal{E}_L \to 0$ (the inductor becomes a short circuit). The magnitude of the back EMF decays exponentially with the same time constant $\tau$.}
+
+\cor{Voltage across the resistor (growth phase)}{The voltage across the resistor is
+\[
+V_R = IR = \mathcal{E}\,\bigl(1 - e^{-t/\tau}\bigr).
+\]
+At $t = 0$, $V_R = 0$. As $t \to \infty$, $V_R \to \mathcal{E}$, so the full battery EMF appears across the resistor in steady state.}
+
+\mprop{Energy stored in an inductor}{When a current $I$ flows through an inductor of inductance $L$, energy is stored in the magnetic field:
+\[
+U_B = \frac{1}{2}\,L\,I^{2}.
+\]
+The rate at which energy is stored is
+\[
+\frac{dU_B}{dt} = L\,I\,\frac{dI}{dt}.
+\]
+The SI unit of energy is the joule (J). During current growth in an LR circuit, the battery supplies energy at rate $\mathcal{E}I$, part of which is dissipated as Joule heating $I^2R$ in the resistor and part is stored in the inductor's magnetic field.}
+
+\nt{At $t = \tau = L/R$, the current reaches $I(\tau) = (\mathcal{E}/R)(1 - e^{-1}) \approx 0.632\,(\mathcal{E}/R)$. The inductor EMF has dropped to $e^{-1} \approx 36.8\%$ of its initial value. After $5\tau$, the current is within $1\%$ of its steady-state value and the circuit is effectively in steady state.}
+
+\ex{Illustrative example}{A series LR circuit with $\mathcal{E} = 12\,\mathrm{V}$, $R = 6.0\,\Omega$, and $L = 3.0\,\mathrm{H}$ has time constant $\tau = L/R = 0.50\,\mathrm{s}$. The maximum current is $I_{\text{max}} = \mathcal{E}/R = 2.0\,\mathrm{A}$.
+
+\begin{enumerate}[label=(\arabic*)]
+\item At $t = 0$, $I = 0$ and $\mathcal{E}_L = -12\,\mathrm{V}$.
+\item At $t = \tau = 0.50\,\mathrm{s}$, $I = 2.0(1 - e^{-1}) = 1.26\,\mathrm{A}$ and $\mathcal{E}_L = -12\,e^{-1} = -4.41\,\mathrm{V}$.
+\item At $t = 2\tau = 1.0\,\mathrm{s}$, $I = 2.0(1 - e^{-2}) = 1.73\,\mathrm{A}$ and $\mathcal{E}_L = -12\,e^{-2} = -1.62\,\mathrm{V}$.
+\item At $t \to \infty$, $I = 2.0\,\mathrm{A}$ and $\mathcal{E}_L = 0\,\mathrm{V}$.
+\end{enumerate}}
+
+\qs{Worked example}{A series LR circuit consists of a battery with emf $\mathcal{E} = 24\,\mathrm{V}$, a resistor with resistance $R = 4.0\,\Omega$, and an inductor with inductance $L = 1.0\,\mathrm{H}$, all connected in series with a switch. The switch is closed at time $t = 0$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Find the time constant $\tau$ of the circuit.
+\item Find the current at $t = 0.25\,\mathrm{s}$.
+\item Find the maximum (steady-state) current.
+\item Find the energy stored in the inductor at $t = 0.25\,\mathrm{s}$.
+\end{enumerate}}
+
+\sol
+\noindent\textbf{Part (a):} The time constant of a series LR circuit is
+\[
+\tau = \frac{L}{R}.
+\]
+Substituting the given values:
+\[
+\tau = \frac{1.0\,\mathrm{H}}{4.0\,\Omega} = 0.25\,\mathrm{s}.
+\]
+
+\noindent\textbf{Part (b):} During the growth phase, the current is
+\[
+I(t) = \frac{\mathcal{E}}{R}\,\bigl(1 - e^{-t/\tau}\bigr).
+\]
+At $t = 0.25\,\mathrm{s}$ and with $\tau = 0.25\,\mathrm{s}$:
+\[
+I(0.25) = \frac{24\,\mathrm{V}}{4.0\,\Omega}\,\Bigl(1 - e^{-0.25/0.25}\Bigr)
+= 6.0\,\mathrm{A}\,\bigl(1 - e^{-1}\bigr).
+\]
+Using $e^{-1} \approx 0.3679$:
+\[
+I(0.25) = 6.0\,\mathrm{A} \times (1 - 0.3679)
+= 6.0\,\mathrm{A} \times 0.6321
+= 3.79\,\mathrm{A}.
+\]
+
+\noindent\textbf{Part (c):} The maximum (steady-state) current is obtained as $t \to \infty$, when the exponential term vanishes:
+\[
+I_{\text{max}} = \frac{\mathcal{E}}{R} = \frac{24\,\mathrm{V}}{4.0\,\Omega} = 6.0\,\mathrm{A}.
+\]
+
+\noindent\textbf{Part (d):} The energy stored in the inductor's magnetic field is
+\[
+U_B = \frac{1}{2}\,L\,I^{2}.
+\]
+Using the current from part (b):
+\[
+U_B = \frac{1}{2}\,(1.0\,\mathrm{H})\,(3.79\,\mathrm{A})^{2}
+= 0.5 \times 14.36\,\mathrm{J}
+= 7.2\,\mathrm{J}.
+\]
+
+\noindent Therefore,
+\[
+\tau = 0.25\,\mathrm{s},
+\qquad
+I(0.25\,\mathrm{s}) = 3.79\,\mathrm{A},
+\qquad
+I_{\text{max}} = 6.0\,\mathrm{A},
+\qquad
+U_B(0.25\,\mathrm{s}) = 7.2\,\mathrm{J}.
+\]

concepts/em/u13/e13-7-lc-circuits.tex

@@ -0,0 +1,272 @@
+\subsection{LC Oscillations}
+
+This subsection introduces the LC circuit -- a capacitor and inductor connected in a closed loop with no resistance -- derives the second-order differential equation governing charge evolution, identifies the sinusoidal oscillation of charge and current, defines the natural angular frequency $\omega = 1/\sqrt{LC}$, and analyses the continuous energy exchange between the capacitor and inductor.
+
+\dfn{LC circuit}{An \emph{LC circuit} consists of an inductor of inductance $L$ and a capacitor of capacitance $C$ connected in a single closed loop. No resistor is present (ideal components). If the capacitor is initially charged and then connected to the inductor, the charge on the capacitor and the current through the inductor oscillate sinusoidally in time.}
+
+\nt{Think of an LC circuit as the electrical analogue of a frictionless spring-mass system. The capacitor stores energy in its electric field (just as a spring stores potential energy), and the inductor stores energy in its magnetic field (just as a moving mass has kinetic energy). The charge oscillates between the two plates of the capacitor while the current alternates direction through the inductor, and the total energy remains constant.}
+
+\thm{Oscillation of charge and current}{In a series LC circuit, let $L$ be the inductance and $C$ the capacitance. If at $t = 0$ the capacitor carries charge $q(0) = Q_{\max}$ and the current is $I(0) = 0$, then the charge on the capacitor at time $t$ is
+\[
+q(t) = Q_{\max}\,\cos(\omega t),
+\]
+and the current through the inductor is
+\[
+I(t) = \frac{dq}{dt} = -\,\omega\,Q_{\max}\,\sin(\omega t).
+\]
+Here $\omega = \dfrac{1}{\sqrt{LC}}$ is the \emph{angular frequency} of oscillation (in rad/s), $Q_{\max}$ is the maximum charge on the capacitor, and the phase constant is $\phi = 0$ for these initial conditions. The period of oscillation is
+\[
+T = \frac{2\pi}{\omega} = 2\pi\sqrt{LC}.
+\]
+The maximum current (current amplitude) is
+\[
+I_{\max} = \omega\,Q_{\max} = \frac{Q_{\max}}{\sqrt{LC}}.
+\]}
+
+\pf{Derivation of LC oscillations}{Apply Kirchhoff's voltage law around the loop. Going around, the voltage drop across the capacitor is $q/C$ and the voltage drop across the inductor is $L\,(dI/dt)$. Since the sum of voltage drops around a closed loop is zero:
+\[
+\frac{q}{C} + L\,\frac{dI}{dt} = 0.
+\]
+The current is the rate of change of charge on the capacitor. When the capacitor discharges, charge leaves the plate, so $I = -\,dq/dt$. (This sign convention is consistent: as $q$ decreases, $dq/dt < 0$ and $I > 0$, meaning positive current flows off the positively charged plate.) Substituting:
+\[
+\frac{q}{C} + L\,\frac{d}{dt}\!\left(-\,\frac{dq}{dt}\right) = 0,
+\]
+which simplifies to
+\[
+\frac{d^2q}{dt^2} + \frac{1}{LC}\,q = 0.
+\]
+This is the \emph{simple harmonic oscillator equation} (second-order linear ODE with constant coefficients). Comparing with the mechanical oscillator equation $\ddot{x} + (\kappa/m)\,x = 0$, the angular frequency is
+\[
+\omega = \frac{1}{\sqrt{LC}}.
+\]
+The general solution is
+\[
+q(t) = A\,\cos(\omega t) + B\,\sin(\omega t) = Q_{\max}\,\cos(\omega t + \phi),
+\]
+where $Q_{\max} = \sqrt{A^2 + B^2}$ is the amplitude and $\phi = \tan^{-1}(-B/A)$ is the phase constant. With $q(0) = Q_{\max}$ and $I(0) = 0$:
+\[
+q(0) = A = Q_{\max}, \qquad
+I(0) = -\omega B = 0 \;\Rightarrow\; B = 0.
+\]
+Thus $\phi = 0$ and
+\[
+q(t) = Q_{\max}\,\cos(\omega t).
+\]
+The current is
+\[
+I(t) = \frac{dq}{dt} = -\,\omega\,Q_{\max}\,\sin(\omega t).
+\]
+The maximum current occurs when $|\sin(\omega t)| = 1$:
+\[
+I_{\max} = \omega\,Q_{\max} = \frac{Q_{\max}}{\sqrt{LC}}.
+\]}
+
+\nt{The initial conditions determine the amplitude $Q_{\max}$ and phase $\phi$. If the capacitor is initially charged to $q(0) = q_0$ with $I(0) = 0$, then $Q_{\max} = q_0$ and $\phi = 0$. If the current has an initial value $I(0) = I_0$ while $q(0) = 0$, then $Q_{\max} = I_0/\omega$ and $\phi = \pi/2$.}
+
+\thm{Energy conservation in an LC circuit}{The energy stored in the capacitor at time $t$ is
+\[
+U_C(t) = \frac{q(t)^2}{2C} = \frac{Q_{\max}^2}{2C}\,\cos^2(\omega t),
+\]
+and the energy stored in the inductor is
+\[
+U_L(t) = \frac{1}{2}\,L\,I(t)^2 = \frac{1}{2}\,L\,\omega^2\,Q_{\max}^2\,\sin^2(\omega t).
+\]
+Using $\omega^2 = 1/(LC)$, the inductor energy becomes
+\[
+U_L(t) = \frac{Q_{\max}^2}{2C}\,\sin^2(\omega t).
+\]
+The total energy is therefore
+\[
+U_{\text{total}} = U_C(t) + U_L(t) = \frac{Q_{\max}^2}{2C}\,\bigl(\cos^2(\omega t) + \sin^2(\omega t)\bigr) = \frac{Q_{\max}^2}{2C}.
+\]
+The total energy is \emph{constant} and does not depend on time.}
+
+\pf{Energy conservation}{The energy stored in a capacitor with charge $q$ is $U_C = q^2/(2C)$, and the energy stored in an inductor with current $I$ is $U_L = \tfrac{1}{2}LI^2$. Substituting the oscillation formulas:
+\[
+U_C(t) = \frac{Q_{\max}^2}{2C}\,\cos^2(\omega t),
+\qquad
+U_L(t) = \frac{1}{2}\,L\,\bigl(-\,\omega\,Q_{\max}\,\sin(\omega t)\bigr)^{\!2}
+= \frac{1}{2}\,L\,\omega^2\,Q_{\max}^2\,\sin^2(\omega t).
+\]
+Since $\omega^2 = 1/(LC)$:
+\[
+U_L(t) = \frac{1}{2}\,L\cdot\frac{1}{LC}\cdot Q_{\max}^2\,\sin^2(\omega t)
+= \frac{Q_{\max}^2}{2C}\,\sin^2(\omega t).
+\]
+Therefore,
+\[
+U_{\text{total}} = U_C + U_L = \frac{Q_{\max}^2}{2C}\,\bigl(\cos^2(\omega t) + \sin^2(\omega t)\bigr) = \frac{Q_{\max}^2}{2C}.
+\]
+This is time-independent, confirming energy conservation.}
+
+\mprop{Energy exchange in an LC circuit}{The total energy is
+\[
+U = \frac{Q_{\max}^2}{2C} = \frac{1}{2}\,L\,I_{\max}^{\,2},
+\]
+where we used $I_{\max} = \omega Q_{\max} = Q_{\max}/\sqrt{LC}$, so $\tfrac{1}{2}LI_{\max}^{\,2} = \tfrac{1}{2}L\,(Q_{\max}^2/LC) = Q_{\max}^2/(2C)$.
+
+The energy is purely capacitive at times $t$ when $\cos(\omega t) = \pm 1$ (i.e., $t = 0,\; T/2,\; T,\; \ldots$):
+\[
+U_C = \frac{Q_{\max}^2}{2C} = U_{\text{total}},
+\qquad U_L = 0.
+\]
+The energy is purely inductive at times $t$ when $\sin(\omega t) = \pm 1$ (i.e., $t = T/4,\; 3T/4,\; \ldots$):
+\[
+U_L = \frac{Q_{\max}^2}{2C} = U_{\text{total}},
+\qquad U_C = 0.
+\]
+At all other times, the energy is shared between the two elements.}
+
+\cor{Analogy to spring-mass simple harmonic motion}{The LC circuit is directly analogous to a frictionless spring-mass oscillator:
+
+\begin{center}
+\begin{tabular}{c|c|c}
+ & Spring--mass system & LC circuit \\ \hline
+Displacement & $x(t) = A\,\cos(\omega t + \phi)$ & Charge $q(t) = Q_{\max}\,\cos(\omega t + \phi)$ \\
+Velocity & $v(t) = -\,\omega A\,\sin(\omega t + \phi)$ & Current $I(t) = -\,\omega Q_{\max}\,\sin(\omega t + \phi)$ \\
+Mass & $m$ & Inductance $L$ \\
+Spring constant & $k$ & Inverse capacitance $1/C$ \\
+Angular frequency & $\omega = \sqrt{k/m}$ & $\omega = 1/\sqrt{LC}$ \\
+Potential energy & $\tfrac{1}{2}kx^2$ & $\tfrac{1}{2}q^2/C$ \\
+Kinetic energy & $\tfrac{1}{2}mv^2$ & $\tfrac{1}{2}LI^2$ \\
+Total energy & $\tfrac{1}{2}kA^2$ & $\dfrac{Q_{\max}^2}{2C} = \tfrac{1}{2}LI_{\max}^{\,2}$ \\
+\end{tabular}
+\end{center}
+This analogy is useful for building physical intuition about LC circuits.}
+
+\ex{Illustrative example}{An LC circuit has $L = 40\,\mathrm{mH}$ and $C = 2.0\,\mathrm{\mu F}$, initially charged to $Q_{\max} = 5.0\,\mathrm{\mu C}$. The angular frequency is $\omega = 1/\sqrt{LC} = 1/\sqrt{(40\times 10^{-3})(2.0\times 10^{-6})} = 1/\sqrt{8.0\times 10^{-8}} = 1/(8.94\times 10^{-4}) = 112\,\mathrm{rad/s}$. The total energy is $U = Q_{\max}^2/(2C) = (5.0\times 10^{-6})^2/(2\cdot 2.0\times 10^{-6}) = 6.25\times 10^{-6}\,\mathrm{J} = 6.25\,\mathrm{\mu J}$. The maximum current is $I_{\max} = \omega Q_{\max} = 112 \times 5.0\times 10^{-6} = 5.6\times 10^{-4}\,\mathrm{A} = 0.56\,\mathrm{mA}$.}
+
+\qs{Worked example}{An LC circuit consists of an ideal inductor with inductance
+\[
+L = 25.0\,\mathrm{mH} = 25.0 \times 10^{-3}\,\mathrm{H},
+\]
+and an ideal capacitor with capacitance
+\[
+C = 5.00\,\mathrm{\mu F} = 5.00 \times 10^{-6}\,\mathrm{F}.
+\]
+At $t = 0$, the capacitor carries its maximum charge
+\[
+Q_{\max} = 10.0\,\mathrm{\mu C} = 1.00 \times 10^{-5}\,\mathrm{C},
+\]
+and the current is $I(0) = 0$. The phase constant is therefore $\phi = 0$, and
+\[
+q(t) = Q_{\max}\,\cos(\omega t),
+\qquad
+I(t) = -\,\omega\,Q_{\max}\,\sin(\omega t),
+\]
+with $\omega = 1/\sqrt{LC}$.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the oscillation frequency $f$ of the circuit,
+\item the maximum current $I_{\max}$,
+\item the energy $U_L$ stored in the inductor at time $t = T/4$ (one-quarter of the oscillation period), and
+\item the energy $U_C$ stored in the capacitor at time $t_{1/2}$, the first instant at which the energy stored in the capacitor equals one-half of the total energy.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The angular frequency of oscillation is
+\[
+\omega = \frac{1}{\sqrt{LC}}.
+\]
+Substitute $L = 25.0\times 10^{-3}\,\mathrm{H}$ and $C = 5.00\times 10^{-6}\,\mathrm{F}$:
+\[
+LC = (25.0\times 10^{-3})(5.00\times 10^{-6}) = 1.25\times 10^{-7}\,\mathrm{H\cdot F}.
+\]
+Thus
+\[
+\omega = \frac{1}{\sqrt{1.25\times 10^{-7}}}
+= \frac{1}{3.536\times 10^{-4}}
+= 2828\,\mathrm{rad/s}.
+\]
+The oscillation frequency is
+\[
+f = \frac{\omega}{2\pi}
+= \frac{2828}{2\pi}\,\mathrm{Hz}
+= 450\,\mathrm{Hz}.
+\]
+Rounded to three significant figures:
+\[
+f = 450\,\mathrm{Hz}.
+\]
+(Equivalently, $4.50\times 10^2\,\mathrm{Hz}$.)
+
+\textbf{Part (b).} The maximum current is
+\[
+I_{\max} = \omega\,Q_{\max}.
+\]
+Substitute $\omega = 2828\,\mathrm{rad/s}$ and $Q_{\max} = 1.00\times 10^{-5}\,\mathrm{C}$:
+\[
+I_{\max} = (2828)\,(1.00\times 10^{-5})\,\mathrm{A}
+= 2.83\times 10^{-2}\,\mathrm{A}
+= 28.3\,\mathrm{mA}.
+\]
+
+\textbf{Part (c).} At $t = T/4$, the angular argument is $\omega t = (2\pi/T)(T/4) = \pi/2$. Therefore:
+\[
+\cos\!\left(\frac{\pi}{2}\right) = 0,
+\qquad
+\sin\!\left(\frac{\pi}{2}\right) = 1.
+\]
+The charge and current at this instant are:
+\[
+q\!\left(\frac{T}{4}\right) = Q_{\max}\,\cos\!\left(\frac{\pi}{2}\right) = 0,
+\qquad
+I\!\left(\frac{T}{4}\right) = -\,\omega\,Q_{\max}\,\sin\!\left(\frac{\pi}{2}\right)
+= -\,\omega\,Q_{\max}
+= -\,I_{\max}.
+\]
+The energy stored in the capacitor is $U_C = q^2/(2C) = 0$, so \emph{all} the energy is in the inductor:
+\[
+U_L\!\left(\frac{T}{4}\right) = U_{\text{total}} = \frac{Q_{\max}^2}{2C}.
+\]
+Substitute the values:
+\[
+U_L\!\left(\frac{T}{4}\right)
+= \frac{(1.00\times 10^{-5}\,\mathrm{C})^2}{2\,(5.00\times 10^{-6}\,\mathrm{F})}
+= \frac{1.00\times 10^{-10}}{1.00\times 10^{-5}}\,\mathrm{J}
+= 1.00\times 10^{-5}\,\mathrm{J}.
+\]
+
+\textbf{Part (d).} We seek the first time $t_{1/2}$ such that the capacitor energy is half the total:
+\[
+U_C = \frac{q^2}{2C} = \frac{1}{2}\,U_{\text{total}}
+= \frac{1}{2}\cdot\frac{Q_{\max}^2}{2C}
+= \frac{Q_{\max}^2}{4C}.
+\]
+Multiply both sides by $2C$:
+\[
+q^2 = \frac{Q_{\max}^2}{2},
+\qquad
+q = \frac{Q_{\max}}{\sqrt{2}}.
+\]
+Now use $q(t) = Q_{\max}\,\cos(\omega t)$:
+\[
+Q_{\max}\,\cos(\omega t_{1/2}) = \frac{Q_{\max}}{\sqrt{2}},
+\qquad
+\cos(\omega t_{1/2}) = \frac{1}{\sqrt{2}}.
+\]
+The first solution (smallest positive angle) is $\omega t_{1/2} = \pi/4$. We will not need the numerical value of $t_{1/2}$ to find the energy. The capacitor energy at this instant is
+\[
+U_C = \frac{1}{2}\,U_{\text{total}}
+= \frac{1}{2}\cdot\frac{Q_{\max}^2}{2C}
+= \frac{Q_{\max}^2}{4C}.
+\]
+Substitute the values:
+\[
+U_C = \frac{(1.00\times 10^{-5}\,\mathrm{C})^2}{4\,(5.00\times 10^{-6}\,\mathrm{F})}
+= \frac{1.00\times 10^{-10}}{2.00\times 10^{-5}}\,\mathrm{J}
+= 5.00\times 10^{-6}\,\mathrm{J}.
+\]
+
+\textbf{Check.} At $t = T/4$, the charge is zero and the energy is entirely in the inductor: $U_L = 1.00\times 10^{-5}\,\mathrm{J} = U_{\text{total}}$, which checks out. At $t = t_{1/2}$, the capacitor holds half the total energy, so $U_C = 5.00\times 10^{-6}\,\mathrm{J}$, and the inductor holds the other half: $U_L = U_{\text{total}} - U_C = 5.00\times 10^{-6}\,\mathrm{J}$, as expected.
+
+Therefore,
+\[
+f = 450\,\mathrm{Hz},
+\qquad
+I_{\max} = 28.3\,\mathrm{mA},
+\qquad
+U_L\!\left(\frac{T}{4}\right) = 1.00\times 10^{-5}\,\mathrm{J},
+\qquad
+U_C(t_{1/2}) = 5.00\times 10^{-6}\,\mathrm{J}.
+\]

concepts/em/u8/e8-1-charge-conservation.tex

@@ -0,0 +1,142 @@
+\subsection{Charge Conservation and Charging Processes}
+
+This subsection treats electric charge as a conserved quantity and introduces the AP-level charging processes of friction, contact, and induction as charge-bookkeeping ideas.
+
+\dfn{Charge conservation and basic charging processes}{Let $q$ denote the net charge of an object or subsystem, measured in coulombs, and let
+\[
+Q_{\mathrm{tot}}=\sum_i q_i
+\]
+denote the total charge of a chosen system.
+
+Charge conservation states that for an isolated system,
+\[
+Q_{\mathrm{tot},f}=Q_{\mathrm{tot},i}.
+\]
+
+In ordinary AP charging problems, objects become charged because charge is redistributed:
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item \emph{Charging by friction}: rubbing two materials can transfer electrons from one object to the other.
+\item \emph{Charging by contact}: when objects touch, charge can transfer between them before they separate.
+\item \emph{Charging by induction}: a nearby charged object causes charge separation in another object; with grounding, charge can enter or leave so the object may be left with a net charge after the process.
+\end{enumerate}
+
+In each case, the total charge of the full isolated system remains constant.}
+
+\nt{The sign of charge is bookkeeping. A positive net charge means an electron deficit, while a negative net charge means an electron excess. In ordinary friction, contact, and induction processes, charge is transferred from one place to another; it is not created from nothing. If one part of an isolated system gains $+\Delta q$, the rest of the system must change by $-\Delta q$. When grounding is involved, include the Earth in the system because it can supply or receive the transferred charge.}
+
+\mprop{Practical bookkeeping relations for isolated systems and simple sharing}{Consider a chosen system of objects with initial charges $q_{1,i},q_{2,i},\dots$ and final charges $q_{1,f},q_{2,f},\dots$.
+
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item For any isolated system,
+\[
+\sum_k q_{k,f}=\sum_k q_{k,i}
+\qquad\text{or}\qquad
+\Delta Q_{\mathrm{tot}}=0.
+\]
+
+\item For charge transfer between two objects $A$ and $B$ within an isolated system,
+\[
+\Delta q_A+\Delta q_B=0,
+\]
+so
+\[
+q_{A,f}-q_{A,i}=-(q_{B,f}-q_{B,i}).
+\]
+
+\item If two identical small conducting spheres with initial charges $q_{A,i}$ and $q_{B,i}$ are touched together and then separated, symmetry gives equal final charges:
+\[
+q_{A,f}=q_{B,f}=\frac{q_{A,i}+q_{B,i}}{2}.
+\]
+
+\item If a neutral object is charged by induction while connected to ground, then charge conservation must be applied to the combined object-Earth system. If the object starts neutral and ends with charge $q_f$, then the Earth changes by
+\[
+\Delta q_{\mathrm{Earth}}=-q_f.
+\]
+\end{enumerate}}
+
+\qs{Worked AP-style problem}{Three identical small conducting spheres $A$, $B$, and $C$ are far apart initially. Let their initial charges be
+\[
+q_{A,i}=+8.0\,\mathrm{nC},
+\qquad
+q_{B,i}=-2.0\,\mathrm{nC},
+\qquad
+q_{C,i}=0.
+\]
+
+First, sphere $A$ is touched to sphere $B$ and then separated. Next, sphere $B$ is touched to sphere $C$ and then separated.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the charge on each sphere after the first contact,
+\item the final charge on each sphere after the second contact, and
+\item the number of electrons transferred during the second contact.
+\end{enumerate}
+
+Take the elementary charge magnitude to be $e=1.60\times 10^{-19}\,\mathrm{C}$.}
+
+\sol Because the spheres are identical, whenever two of them touch and then separate, they share the total charge equally.
+
+For the first contact, apply charge conservation to spheres $A$ and $B$:
+\[
+q_{A,i}+q_{B,i}=(+8.0\,\mathrm{nC})+(-2.0\,\mathrm{nC})=+6.0\,\mathrm{nC}.
+\]
+
+Since the spheres are identical, after they separate each has half of this total charge:
+\[
+q_{A}=q_{B}=\frac{+6.0\,\mathrm{nC}}{2}=+3.0\,\mathrm{nC}.
+\]
+
+So after the first contact,
+\[
+q_{A}=+3.0\,\mathrm{nC},
+\qquad
+q_{B}=+3.0\,\mathrm{nC},
+\qquad
+q_{C}=0.
+\]
+
+Now sphere $B$ touches sphere $C$. Just before this second contact, their total charge is
+\[
+q_{B}+q_{C}=(+3.0\,\mathrm{nC})+0=+3.0\,\mathrm{nC}.
+\]
+
+Again they are identical, so after separation they share this total equally:
+\[
+q_{B,f}=q_{C,f}=\frac{+3.0\,\mathrm{nC}}{2}=+1.5\,\mathrm{nC}.
+\]
+
+Sphere $A$ is not involved in the second contact, so its charge stays
+\[
+q_{A,f}=+3.0\,\mathrm{nC}.
+\]
+
+Therefore the final charges are
+\[
+q_{A,f}=+3.0\,\mathrm{nC},
+\qquad
+q_{B,f}=+1.5\,\mathrm{nC},
+\qquad
+q_{C,f}=+1.5\,\mathrm{nC}.
+\]
+
+To find the number of electrons transferred during the second contact, compute the magnitude of the charge change on either sphere. Sphere $B$ changes from $+3.0\,\mathrm{nC}$ to $+1.5\,\mathrm{nC}$, so
+\[
+\Delta q_B=(+1.5-3.0)\,\mathrm{nC}=-1.5\,\mathrm{nC}.
+\]
+
+The negative change means sphere $B$ gained electrons. Equivalently, sphere $C$ changed from $0$ to $+1.5\,\mathrm{nC}$, so sphere $C$ lost electrons. The magnitude of transferred charge is
+\[
+|\Delta q|=1.5\times 10^{-9}\,\mathrm{C}.
+\]
+
+Let $N$ denote the number of electrons transferred. Then
+\[
+N=\frac{|\Delta q|}{e}=\frac{1.5\times 10^{-9}}{1.60\times 10^{-19}}=9.375\times 10^9.
+\]
+
+To two significant figures,
+\[
+N\approx 9.4\times 10^9\text{ electrons}.
+\]
+
+These electrons moved from sphere $C$ to sphere $B$. This direction makes sense because sphere $B$ became less positive while sphere $C$ became more positive.

concepts/em/u8/e8-2-coulomb-superposition.tex

@@ -0,0 +1,149 @@
+\subsection{Coulomb's Law and Superposition}
+
+This subsection gives the electrostatic force between point charges and shows how forces from multiple source charges combine by vector addition.
+
+\dfn{Point charges, separation vector, and superposition}{Let point charges $q_1,q_2,\dots,q_N$ be located at position vectors $\vec{r}_1,\vec{r}_2,\dots,\vec{r}_N$ in an inertial frame. For two distinct charges $q_i$ and $q_j$, define the separation vector from $q_i$ to $q_j$ by
+\[
+\vec{r}_{ij}=\vec{r}_j-\vec{r}_i,
+\]
+let $r_{ij}=|\vec{r}_{ij}|$ be the separation distance, and let $\hat{r}_{ij}=\vec{r}_{ij}/r_{ij}$ be the corresponding unit vector.
+
+A \emph{point charge} is an idealized charged object whose size is negligible compared with the distances of interest. The \emph{superposition principle} states that when several source charges act on a chosen charge, the net electric force is the vector sum of the individual forces exerted by each source charge separately.}
+
+\thm{Coulomb's law in vector form and force superposition}{Let $k=\dfrac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}$. For two point charges $q_i$ and $q_j$ with separation vector $\vec{r}_{ij}\neq \vec{0}$, the electric force on $q_j$ due to $q_i$ is
+\[
+\vec{F}_{i\to j}=k\frac{q_i q_j}{r_{ij}^2}\hat{r}_{ij}
+=k\frac{q_i q_j}{r_{ij}^3}\vec{r}_{ij}.
+\]
+Its magnitude is
+\[
+F_{i\to j}=k\frac{|q_i q_j|}{r_{ij}^2}.
+\]
+Thus the force is proportional to the product of the charges, inversely proportional to the square of the separation distance, and directed along the line joining the charges. If $N$ source charges act on $q_j$, then
+\[
+\vec{F}_{\mathrm{net},j}=\sum_{i\ne j}\vec{F}_{i\to j}
+=\sum_{i\ne j} k\frac{q_i q_j}{|\vec{r}_j-\vec{r}_i|^3}(\vec{r}_j-\vec{r}_i).
+\]}
+
+\pf{Why the vector law has this form}{For two point charges separated by distance $r_{ij}$, Coulomb's law gives the force magnitude
+\[
+F_{i\to j}=k\frac{|q_i q_j|}{r_{ij}^2}.
+\]
+The force must lie along the line connecting the charges, so its direction is either $+\hat{r}_{ij}$ or $-\hat{r}_{ij}$. If $q_i q_j>0$, the charges have the same sign and repel, so the force on $q_j$ points away from $q_i$, which is $+\hat{r}_{ij}$. If $q_i q_j<0$, the charges have opposite signs and attract, so the force on $q_j$ points toward $q_i$, which is $-\hat{r}_{ij}$. Writing the force as
+\[
+\vec{F}_{i\to j}=k\frac{q_i q_j}{r_{ij}^2}\hat{r}_{ij}
+\]
+captures both cases automatically through the sign of $q_i q_j$. Because force is a vector, multiple electric forces combine by ordinary vector addition, giving the superposition formula.}
+
+\cor{Collinear charges on the $x$-axis}{Let fixed source charges $q_1,\dots,q_N$ lie on the $x$-axis at coordinates $x_1,\dots,x_N$. Let a test charge $q$ be at coordinate $x$, with $x\neq x_i$ for all $i$. Then the net force on the test charge is purely along the $x$-axis:
+\[
+\vec{F}_{\mathrm{net}}=kq\left(\sum_{i=1}^N q_i\frac{x-x_i}{|x-x_i|^3}\right)\hat{\imath}.
+\]
+So in one dimension, Coulomb superposition reduces to an algebraic sum of signed $x$-components. In particular, if two equal source charges $+Q$ are placed at $x=-a$ and $x=+a$, then at the midpoint $x=0$ their forces cancel, so $\vec{F}_{\mathrm{net}}=\vec{0}$ on any test charge placed there.}
+
+\qs{Worked AP-style problem}{In an $xy$-plane, let $q_1=+4.0\,\mu\mathrm{C}$ be fixed at $\vec{r}_1=\vec{0}$, let $q_2=-2.0\,\mu\mathrm{C}$ be fixed at $\vec{r}_2=(0.30\,\mathrm{m})\hat{\imath}$, and let $q_3=+1.5\,\mu\mathrm{C}$ be located at $\vec{r}_3=(0.40\,\mathrm{m})\hat{\jmath}$. Let $k=8.99\times 10^9\,\mathrm{N\,m^2/C^2}$.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the force $\vec{F}_{1\to 3}$ on $q_3$ due to $q_1$,
+\item the force $\vec{F}_{2\to 3}$ on $q_3$ due to $q_2$, and
+\item the net force $\vec{F}_{\mathrm{net},3}$ on $q_3$, including its magnitude and direction measured counterclockwise from the positive $x$-axis.
+\end{enumerate}}
+
+\sol First find the separation vectors to the charge $q_3$.
+
+For the force due to $q_1$,
+\[
+\vec{r}_{13}=\vec{r}_3-\vec{r}_1=(0.40\,\mathrm{m})\hat{\jmath},
+\qquad
+r_{13}=0.40\,\mathrm{m}.
+\]
+For the force due to $q_2$,
+\[
+\vec{r}_{23}=\vec{r}_3-\vec{r}_2=(-0.30\hat{\imath}+0.40\hat{\jmath})\,\mathrm{m},
+\qquad
+r_{23}=\sqrt{(0.30)^2+(0.40)^2}\,\mathrm{m}=0.50\,\mathrm{m}.
+\]
+
+For part (a), use Coulomb's law. Since $q_1$ and $q_3$ are both positive, the force on $q_3$ is repulsive and points away from $q_1$, which is in the $+\hat{\jmath}$ direction:
+\[
+\vec{F}_{1\to 3}=k\frac{q_1 q_3}{r_{13}^3}\vec{r}_{13}.
+\]
+Its magnitude is
+\[
+F_{1\to 3}=k\frac{|q_1 q_3|}{r_{13}^2}
+=(8.99\times 10^9)\frac{(4.0\times 10^{-6})(1.5\times 10^{-6})}{(0.40)^2}\,\mathrm{N}.
+\]
+So
+\[
+F_{1\to 3}=0.337\,\mathrm{N},
+\]
+and therefore
+\[
+\vec{F}_{1\to 3}=(0.337\,\mathrm{N})\hat{\jmath}.
+\]
+
+For part (b), $q_2$ is negative and $q_3$ is positive, so the force on $q_3$ is attractive and points from $q_3$ toward $q_2$. Let $\hat{u}_{3\to 2}$ denote the unit vector from $q_3$ to $q_2$. Then
+\[
+\hat{u}_{3\to 2}=\frac{(0.30\hat{\imath}-0.40\hat{\jmath})\,\mathrm{m}}{0.50\,\mathrm{m}}
+=0.60\hat{\imath}-0.80\hat{\jmath}.
+\]
+The magnitude is
+\[
+F_{2\to 3}=k\frac{|q_2 q_3|}{r_{23}^2}
+=(8.99\times 10^9)\frac{(2.0\times 10^{-6})(1.5\times 10^{-6})}{(0.50)^2}\,\mathrm{N}.
+\]
+Thus
+\[
+F_{2\to 3}=0.108\,\mathrm{N}.
+\]
+So the vector force is
+\[
+\vec{F}_{2\to 3}=F_{2\to 3}\hat{u}_{3\to 2}
+=(0.108)(0.60\hat{\imath}-0.80\hat{\jmath})\,\mathrm{N}.
+\]
+Therefore,
+\[
+\vec{F}_{2\to 3}=(0.0647\hat{\imath}-0.0863\hat{\jmath})\,\mathrm{N}.
+\]
+
+For part (c), add the forces componentwise:
+\[
+\vec{F}_{\mathrm{net},3}=\vec{F}_{1\to 3}+\vec{F}_{2\to 3}.
+\]
+So
+\[
+\vec{F}_{\mathrm{net},3}=(0.0647\hat{\imath}+0.2507\hat{\jmath})\,\mathrm{N}.
+\]
+Its magnitude is
+\[
+|\vec{F}_{\mathrm{net},3}|=\sqrt{(0.0647)^2+(0.2507)^2}\,\mathrm{N}=0.259\,\mathrm{N}.
+\]
+Let $\theta$ denote the direction measured counterclockwise from the positive $x$-axis. Then
+\[
+\tan\theta=\frac{0.2507}{0.0647}=3.88,
+\]
+so
+\[
+\theta=\tan^{-1}(3.88)=75.5^\circ.
+\]
+
+Therefore,
+\[
+\vec{F}_{1\to 3}=(0.337\,\mathrm{N})\hat{\jmath},
+\qquad
+\vec{F}_{2\to 3}=(0.0647\hat{\imath}-0.0863\hat{\jmath})\,\mathrm{N},
+\]
+and the net force on $q_3$ is
+\[
+\vec{F}_{\mathrm{net},3}=(0.0647\hat{\imath}+0.2507\hat{\jmath})\,\mathrm{N},
+\]
+with magnitude
+\[
+|\vec{F}_{\mathrm{net},3}|=0.259\,\mathrm{N}
+\]
+at angle
+\[
+\theta=75.5^\circ
+\]
+above the positive $x$-axis.

concepts/em/u8/e8-3-electric-field.tex

@@ -0,0 +1,156 @@
+\subsection{Electric Field as Force per Unit Charge}
+
+This subsection defines the electric field from source charges and shows how it determines the force on any charge placed at a point.
+
+\dfn{Electric field, source charges, and test charges}{Let source charges create an electrostatic interaction in space. Let a field point have position vector $\vec{r}$, and let a small positive test charge $q_0$ be placed at that point. If the electric force on the test charge is $\vec{F}$, then the \emph{electric field} at that point is defined by
+\[
+\vec{E}(\vec{r})=\frac{\vec{F}}{q_0}.
+\]
+The source charges are the charges that produce the field. The test charge is a charge used only to probe the field at a location. Because the factor of $q_0$ divides out, the field depends on the source-charge configuration and the location $\vec{r}$, not on the particular test charge used to measure it. By convention, the direction of $\vec{E}$ is the direction of the force on a positive test charge. The SI units of electric field are $\mathrm{N/C}$.}
+
+\thm{Point-charge field law and force relation}{Let a point source charge $Q$ be fixed at position vector $\vec{r}_Q$. Let the field point have position vector $\vec{r}$, and define
+\[
+\vec{R}=\vec{r}-\vec{r}_Q,
+\qquad
+R=|\vec{R}|,
+\qquad
+\hat{R}=\frac{\vec{R}}{R},
+\]
+with $\vec{r}\neq \vec{r}_Q$. Then the electric field due to the point charge $Q$ is
+\[
+\vec{E}(\vec{r})=k\frac{Q}{R^2}\hat{R}=k\frac{Q}{R^3}\vec{R},
+\]
+where
+\[
+k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
+\]
+If any charge $q$ is placed at that field point, the electric force on that charge is
+\[
+\vec{F}=q\vec{E}.
+\]}
+
+\pf{Derivation from Coulomb's law}{Let a positive test charge $q_0$ be placed at the field point. Coulomb's law gives the force on the test charge due to the source charge $Q$ as
+\[
+\vec{F}=k\frac{Qq_0}{R^3}\vec{R}.
+\]
+Now divide by $q_0$ and use the definition of electric field:
+\[
+\vec{E}(\vec{r})=\frac{\vec{F}}{q_0}=k\frac{Q}{R^3}\vec{R}=k\frac{Q}{R^2}\hat{R}.
+\]
+This shows that the field is determined entirely by the source charge and geometry. Once $\vec{E}$ is known at a point, the force on any charge $q$ placed there is obtained by multiplying by $q$, so $\vec{F}=q\vec{E}$.}
+
+\cor{Direction and superposition of electric fields}{Let point source charges $Q_1,Q_2,\dots,Q_N$ be fixed at position vectors $\vec{r}_1,\vec{r}_2,\dots,\vec{r}_N$. Let the field point have position vector $\vec{r}$, with $\vec{r}\neq \vec{r}_i$ for all $i$. Then the net electric field is the vector sum of the individual fields:
+\[
+\vec{E}_{\mathrm{net}}(\vec{r})=\sum_{i=1}^N k\frac{Q_i}{|\vec{r}-\vec{r}_i|^3}(\vec{r}-\vec{r}_i).
+\]
+For a single source charge, the field points radially away from the charge if $Q_i>0$ and radially toward the charge if $Q_i<0$.}
+
+\qs{Worked AP-style problem}{Two point source charges lie on the $x$-axis. Let
+\[
+q_1=+3.0\,\mu\mathrm{C}
+\qquad\text{at}\qquad
+\vec{r}_1=(-0.20\,\mathrm{m})\hat{\imath},
+\]
+and let
+\[
+q_2=-2.0\,\mu\mathrm{C}
+\qquad\text{at}\qquad
+\vec{r}_2=(+0.30\,\mathrm{m})\hat{\imath}.
+\]
+Let point $P$ be at the origin, so $\vec{r}_P=\vec{0}$. Take $k=8.99\times 10^9\,\mathrm{N\,m^2/C^2}$.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the electric field $\vec{E}_1$ at $P$ due to $q_1$,
+\item the electric field $\vec{E}_2$ at $P$ due to $q_2$,
+\item the net electric field $\vec{E}_{\mathrm{net}}$ at $P$, and
+\item the electric force on a charge $q=-4.0\,\mathrm{nC}$ placed at $P$.
+\end{enumerate}}
+
+\sol First find the displacement vectors from each source charge to the field point $P$.
+
+For $q_1$,
+\[
+\vec{R}_1=\vec{r}_P-\vec{r}_1=(0.20\,\mathrm{m})\hat{\imath},
+\qquad
+R_1=0.20\,\mathrm{m}.
+\]
+For $q_2$,
+\[
+\vec{R}_2=\vec{r}_P-\vec{r}_2=(-0.30\,\mathrm{m})\hat{\imath},
+\qquad
+R_2=0.30\,\mathrm{m}.
+\]
+
+For part (a), use the point-charge field law:
+\[
+\vec{E}_1=k\frac{q_1}{R_1^3}\vec{R}_1.
+\]
+Because $q_1$ is positive, the field points away from $q_1$, which at the origin is in the $+\hat{\imath}$ direction. Its magnitude is
+\[
+E_1=k\frac{|q_1|}{R_1^2}=(8.99\times 10^9)\frac{3.0\times 10^{-6}}{(0.20)^2}\,\mathrm{N/C}.
+\]
+So
+\[
+E_1=6.74\times 10^5\,\mathrm{N/C},
+\]
+and therefore
+\[
+\vec{E}_1=(6.74\times 10^5\,\mathrm{N/C})\hat{\imath}.
+\]
+
+For part (b),
+\[
+\vec{E}_2=k\frac{q_2}{R_2^3}\vec{R}_2.
+\]
+Here $q_2$ is negative, so the field points toward $q_2$. Since $q_2$ is to the right of the origin, the field at the origin is again in the $+\hat{\imath}$ direction. Its magnitude is
+\[
+E_2=k\frac{|q_2|}{R_2^2}=(8.99\times 10^9)\frac{2.0\times 10^{-6}}{(0.30)^2}\,\mathrm{N/C}.
+\]
+Thus,
+\[
+E_2=2.00\times 10^5\,\mathrm{N/C},
+\]
+so
+\[
+\vec{E}_2=(2.00\times 10^5\,\mathrm{N/C})\hat{\imath}.
+\]
+
+For part (c), add the fields as vectors:
+\[
+\vec{E}_{\mathrm{net}}=\vec{E}_1+\vec{E}_2.
+\]
+Since both fields point in the same direction,
+\[
+\vec{E}_{\mathrm{net}}=(6.74\times 10^5+2.00\times 10^5)\hat{\imath}\,\mathrm{N/C}.
+\]
+Therefore,
+\[
+\vec{E}_{\mathrm{net}}=(8.74\times 10^5\,\mathrm{N/C})\hat{\imath}.
+\]
+
+For part (d), the force on a charge placed at $P$ is
+\[
+\vec{F}=q\vec{E}_{\mathrm{net}}.
+\]
+Substitute $q=-4.0\times 10^{-9}\,\mathrm{C}$:
+\[
+\vec{F}=(-4.0\times 10^{-9})(8.74\times 10^5)\hat{\imath}\,\mathrm{N}.
+\]
+So
+\[
+\vec{F}=(-3.50\times 10^{-3}\,\mathrm{N})\hat{\imath}.
+\]
+The negative sign means the force points in the $-\hat{\imath}$ direction. Its magnitude is $3.50\times 10^{-3}\,\mathrm{N}$.
+
+Therefore,
+\[
+\vec{E}_1=(6.74\times 10^5\,\mathrm{N/C})\hat{\imath},
+\qquad
+\vec{E}_2=(2.00\times 10^5\,\mathrm{N/C})\hat{\imath},
+\]
+\[
+\vec{E}_{\mathrm{net}}=(8.74\times 10^5\,\mathrm{N/C})\hat{\imath},
+\qquad
+\vec{F}=(-3.50\times 10^{-3}\,\mathrm{N})\hat{\imath}.
+\]

concepts/em/u8/e8-4-continuous-distributions.tex

@@ -0,0 +1,156 @@
+\subsection{Fields of Continuous Charge Distributions}
+
+This subsection extends electric-field superposition from point charges to rods, arcs, rings, surfaces, and volumes by replacing discrete sums with integrals over charge elements.
+
+\dfn{Charge densities and differential field contribution}{Let the field point have position vector $\vec{r}$. Let a small source element at position vector $\vec{r}'$ carry charge $dq$. Define
+\[
+\vec{R}=\vec{r}-\vec{r}',
+\qquad
+R=|\vec{R}|,
+\qquad
+\hat{R}=\frac{\vec{R}}{R}.
+\]
+For a continuous distribution, the charge element is written as
+\[
+dq=\lambda\,dl
+\qquad\text{(line charge)},
+\qquad
+dq=\sigma\,dA
+\qquad\text{(surface charge)},
+\qquad
+dq=\rho\,dV
+\qquad\text{(volume charge)},
+\]
+where $\lambda$ is linear charge density in $\mathrm{C/m}$, $\sigma$ is surface charge density in $\mathrm{C/m^2}$, and $\rho$ is volume charge density in $\mathrm{C/m^3}$. The electric-field contribution of the source element at the field point is
+\[
+d\vec{E}=k\frac{dq}{R^2}\hat{R}=k\frac{dq}{R^3}\vec{R},
+\]
+where
+\[
+k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
+\]
+}
+
+\thm{Continuous superposition integral for electric field}{For a static continuous charge distribution, the net electric field at the field point $\vec{r}$ is the vector integral
+\[
+\vec{E}(\vec{r})=\int d\vec{E}
+=k\int \frac{1}{R^3}\vec{R}\,dq.
+\]
+Equivalently,
+\[
+\vec{E}(\vec{r})
+=k\int \frac{1}{R^3}\vec{R}\,\lambda\,dl,
+\qquad
+\vec{E}(\vec{r})
+=k\int \frac{1}{R^3}\vec{R}\,\sigma\,dA,
+\qquad
+\vec{E}(\vec{r})
+=k\int \frac{1}{R^3}\vec{R}\,\rho\,dV,
+\]
+depending on the geometry. In practice, write $d\vec{E}$ for one source element, resolve it into components, use symmetry to identify any canceling components, and integrate only the nonzero component(s).}
+
+\ex{Illustrative example}{A uniformly charged semicircular arc of radius $R$ lies above the $x$-axis and is centered at the origin. Let its total charge be $Q>0$. Find the electric field at the center.
+
+The arc length is $\pi R$, so the linear charge density is
+\[
+\lambda=\frac{Q}{\pi R}.
+\]
+Let $\theta$ denote the polar angle of a source element, measured from the positive $x$-axis, with $0\le \theta \le \pi$. Then
+\[
+dq=\lambda R\,d\theta.
+\]
+Each source element is distance $R$ from the center, so
+\[
+dE=k\frac{dq}{R^2}.
+\]
+By symmetry, the $x$-components cancel. The $y$-components all point downward, so
+\[
+dE_y=-dE\sin\theta=-k\frac{dq}{R^2}\sin\theta.
+\]
+Substitute $dq=\lambda R\,d\theta$:
+\[
+dE_y=-k\frac{\lambda}{R}\sin\theta\,d\theta.
+\]
+Integrate from $0$ to $\pi$:
+\[
+E_y=-k\frac{\lambda}{R}\int_0^\pi \sin\theta\,d\theta
+=-k\frac{\lambda}{R}(2).
+\]
+Therefore,
+\[
+\vec{E}=-\frac{2kQ}{\pi R^2}\hat{\jmath}.
+\]
+The field points downward because the positive charges on the upper arc repel a positive test charge at the center.}
+
+\nt{For continuous distributions, the hardest step is usually not the integral but the geometry. Start with $d\vec{E}$ from one source element, then ask which components cancel by symmetry. On a ring, sideways components cancel and only the axial component survives. On a symmetric finite line, horizontal components cancel at the perpendicular bisector and only the perpendicular component survives. Also choose $dq$ to match the object's dimension: use $dq=\lambda\,dl$ for rods and arcs, $dq=\sigma\,dA$ for sheets, and $dq=\rho\,dV$ for three-dimensional charge distributions.}
+
+\qs{Worked AP-style problem}{A thin ring of radius $a$ is centered at the origin and lies in the $yz$-plane. The ring carries total charge $Q$ distributed uniformly around its circumference. Let point $P$ lie on the ring's axis at position
+\[
+\vec{r}_P=x\hat{\imath},
+\]
+where $x>0$. Let
+\[
+k=\frac{1}{4\pi\varepsilon_0}.
+\]
+Find the electric field $\vec{E}$ at point $P$ in terms of $Q$, $a$, $x$, and $k$.}
+
+\sol Let $\lambda$ denote the ring's linear charge density. Since the ring circumference is $2\pi a$,
+\[
+\lambda=\frac{Q}{2\pi a}.
+\]
+
+Choose a small source element $dq$ on the ring. Let $\vec{R}$ denote the displacement vector from that source element to point $P$. Every source element on the ring is the same distance from $P$, so
+\[
+R=\sqrt{x^2+a^2}.
+\]
+The magnitude of the field due to $dq$ is therefore
+\[
+dE=k\frac{dq}{R^2}=k\frac{dq}{x^2+a^2}.
+\]
+
+Now use symmetry. For each source element on the ring, there is an opposite element whose field contribution has the same magnitude. Their components in the $y$- and $z$-directions cancel, while their components along the axis add. Therefore the net field must point along $\hat{\imath}$, so we only need the $x$-component of $d\vec{E}$.
+
+Let $\phi$ denote the angle between $\vec{R}$ and the positive $x$-axis. Then
+\[
+\cos\phi=\frac{x}{R}=\frac{x}{\sqrt{x^2+a^2}}.
+\]
+So the axial component of the differential field is
+\[
+dE_x=dE\cos\phi
+=\left(k\frac{dq}{R^2}\right)\left(\frac{x}{R}\right)
+=k\frac{x\,dq}{R^3}.
+\]
+Since $R=\sqrt{x^2+a^2}$ is constant over the ring,
+\[
+dE_x=k\frac{x\,dq}{(x^2+a^2)^{3/2}}.
+\]
+
+Integrate all the way around the ring:
+\[
+E_x=\int dE_x
+=k\frac{x}{(x^2+a^2)^{3/2}}\int dq.
+\]
+But
+\[
+\int dq=Q,
+\]
+so
+\[
+E_x=k\frac{Qx}{(x^2+a^2)^{3/2}}.
+\]
+
+Therefore,
+\[
+\vec{E}=k\frac{Qx}{(x^2+a^2)^{3/2}}\hat{\imath}.
+\]
+If $Q>0$, the field points in the $+\hat{\imath}$ direction, and if $Q<0$, it points in the $-\hat{\imath}$ direction.
+
+This result also passes two quick checks. If $x=0$, then
+\[
+\vec{E}=\vec{0},
+\]
+which matches the symmetry at the ring's center. If $x\gg a$, then $x^2+a^2\approx x^2$, so
+\[
+\vec{E}\approx k\frac{Q}{x^2}\hat{\imath},
+\]
+which is the field of a point charge $Q$ far away from the ring.

concepts/em/u8/e8-5-electric-flux.tex

@@ -0,0 +1,136 @@
+\subsection{Electric Flux}
+
+This subsection introduces electric flux as a signed measure of how much electric field passes through an oriented surface.
+
+\dfn{Area vector and electric flux}{Let $S$ be a surface broken into small area elements of scalar area $dA$. Let $\hat{n}$ denote a chosen unit normal to a surface element. The corresponding \emph{area vector element} is
+\[
+d\vec{A}=\hat{n}\,dA.
+\]
+For an open surface, either choice of normal may be used, but the choice must be kept consistent across the surface. For a closed surface, the standard choice is the outward normal.
+
+Let $\vec{E}$ denote the electric field at each point of the surface. The \emph{electric flux} through the oriented surface is the scalar
+\[
+\Phi_E=\iint_S \vec{E}\cdot d\vec{A}.
+\]
+Its SI units are $\mathrm{N\cdot m^2/C}$.}
+
+\thm{Surface-integral form and uniform-field special case}{Let $S$ be an oriented surface with area vector element $d\vec{A}=\hat{n}\,dA$, and let $\vec{E}$ be the electric field on that surface. Then the electric flux through $S$ is
+\[
+\Phi_E=\iint_S \vec{E}\cdot d\vec{A}.
+\]
+This integral adds the component of $\vec{E}$ perpendicular to the surface over the entire surface.
+
+If the surface is flat with area $A$, the field is uniform over it, and $\vec{A}=A\hat{n}$ denotes the surface's area vector, then
+\[
+\Phi_E=\vec{E}\cdot \vec{A}=EA\cos\theta,
+\]
+where $E=|\vec{E}|$, $\theta$ is the angle between $\vec{E}$ and $\vec{A}$, and $A=|\vec{A}|$. For a closed surface, the same formula is applied piece by piece using outward area vectors on all patches of the surface.}
+
+\ex{Illustrative example}{Let a uniform electric field be
+\[
+\vec{E}=(300\,\mathrm{N/C})\hat{\imath}.
+\]
+Let a flat surface have area
+\[
+A=0.20\,\mathrm{m^2},
+\]
+and let its area vector make an angle $\theta=60^\circ$ with $\vec{E}$.
+
+Then the flux is
+\[
+\Phi_E=EA\cos\theta=(300)(0.20)\cos 60^\circ\,\mathrm{N\cdot m^2/C}.
+\]
+So
+\[
+\Phi_E=30\,\mathrm{N\cdot m^2/C}.
+\]
+Because the angle is acute, the flux is positive.}
+
+\nt{Electric flux is not the same thing as electric field magnitude. Flux depends on both the field and the oriented surface. Reversing the chosen normal reverses the sign of $\Phi_E$. A positive flux means the field points generally in the same direction as the chosen area vector, while a negative flux means it points generally opposite that direction. If the field is parallel to the surface, then it is perpendicular to $d\vec{A}$ and the flux is zero even if $|\vec{E}|$ is large.}
+
+\qs{Worked AP-style problem}{A cube of side length $L=0.20\,\mathrm{m}$ is placed in a uniform electric field
+\[
+\vec{E}=(500\,\mathrm{N/C})\hat{\imath}.
+\]
+Let the cube's faces be aligned with the coordinate axes. Let the outward area vector of the right face be in the $+\hat{\imath}$ direction, and let the outward area vector of the left face be in the $-\hat{\imath}$ direction.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the electric flux through the right face,
+\item the electric flux through the left face,
+\item the electric flux through any one of the four remaining faces, and
+\item the net electric flux through the entire closed cube.
+\end{enumerate}}
+
+\sol Let the area of one face be $A$. Since each face is a square of side length $L$,
+\[
+A=L^2=(0.20\,\mathrm{m})^2=0.040\,\mathrm{m^2}.
+\]
+
+For each face of the cube, use
+\[
+\Phi_E=\vec{E}\cdot \vec{A}=EA\cos\theta,
+\]
+where $\theta$ is the angle between the electric field and that face's outward area vector.
+
+For part (a), the right face has outward area vector in the $+\hat{\imath}$ direction, the same direction as $\vec{E}$. Thus
+\[
+\theta=0^\circ.
+\]
+So
+\[
+\Phi_{E,\mathrm{right}}=EA\cos 0^\circ=(500)(0.040)(1)\,\mathrm{N\cdot m^2/C}.
+\]
+Therefore,
+\[
+\Phi_{E,\mathrm{right}}=20\,\mathrm{N\cdot m^2/C}.
+\]
+
+For part (b), the left face has outward area vector in the $-\hat{\imath}$ direction, opposite the field. Thus
+\[
+\theta=180^\circ.
+\]
+So
+\[
+\Phi_{E,\mathrm{left}}=EA\cos 180^\circ=(500)(0.040)(-1)\,\mathrm{N\cdot m^2/C}.
+\]
+Therefore,
+\[
+\Phi_{E,\mathrm{left}}=-20\,\mathrm{N\cdot m^2/C}.
+\]
+
+For part (c), on any of the other four faces, the outward area vector is perpendicular to $\vec{E}$. Thus
+\[
+\theta=90^\circ,
+\]
+so
+\[
+\Phi_E=EA\cos 90^\circ=0.
+\]
+Therefore, the flux through each of those four faces is
+\[
+0\,\mathrm{N\cdot m^2/C}.
+\]
+
+For part (d), add the fluxes from all six faces:
+\[
+\Phi_{E,\mathrm{net}}=\Phi_{E,\mathrm{right}}+\Phi_{E,\mathrm{left}}+4(0).
+\]
+So
+\[
+\Phi_{E,\mathrm{net}}=20+(-20)=0\,\mathrm{N\cdot m^2/C}.
+\]
+
+Therefore,
+\[
+\Phi_{E,\mathrm{right}}=20\,\mathrm{N\cdot m^2/C},
+\qquad
+\Phi_{E,\mathrm{left}}=-20\,\mathrm{N\cdot m^2/C},
+\]
+\[
+\Phi_E=0\,\mathrm{N\cdot m^2/C}\text{ for each of the other four faces},
+\]
+and the net flux through the closed cube is
+\[
+\Phi_{E,\mathrm{net}}=0.
+\]

concepts/em/u8/e8-6-gauss-law.tex

@@ -0,0 +1,100 @@
+\subsection{Gauss's Law and Symmetry Reduction}
+
+This subsection states Gauss's law and shows how symmetry can reduce a difficult flux integral to simple algebra when the charge distribution is highly symmetric.
+
+\dfn{Gaussian surface and enclosed charge}{Let $S$ be any closed imaginary surface in space, and let $d\vec{A}$ denote an outward-pointing area element on that surface. The surface $S$ is called a \emph{Gaussian surface}. The \emph{enclosed charge} $q_{\mathrm{enc}}$ is the algebraic sum of all charges contained inside $S$. Charges outside $S$ can affect the electric field on the surface, but they do not contribute to $q_{\mathrm{enc}}$.}
+
+\thm{Gauss's law and when symmetry makes it useful}{Let $S$ be any closed surface with outward area element $d\vec{A}$, and let $q_{\mathrm{enc}}$ be the net charge enclosed by $S$. Then Gauss's law states
+\[
+\oint_S \vec{E}\cdot d\vec{A}=\frac{q_{\mathrm{enc}}}{\varepsilon_0}.
+\]
+This law is always true. It becomes a practical method for solving for the electric field when the charge distribution has enough symmetry that one can choose a Gaussian surface for which the magnitude $E=|\vec{E}|$ is constant on each flux-contributing part of the surface and the angle between $\vec{E}$ and $d\vec{A}$ is everywhere $0^\circ$, $180^\circ$, or $90^\circ$. Then the flux integral reduces to algebraic terms such as $EA$, $-EA$, or $0$. Common useful cases are spherical, cylindrical, and planar symmetry.}
+
+\nt{Gauss's law is always true, but it is not always useful for finding $\vec{E}$. In a general asymmetric charge distribution, knowing only the total flux through a closed surface does not tell you the field at each point on that surface. Also, zero net enclosed charge implies zero \emph{net flux}, not necessarily zero field everywhere. The main strategy is therefore: first identify strong symmetry, then choose a Gaussian surface matched to that symmetry.}
+
+\pf{Why symmetry reduces the integral}{Let a point charge $Q$ be at the center of a spherical Gaussian surface of radius $r$. By spherical symmetry, the electric field is radial and has the same magnitude $E(r)$ at every point on the sphere. The outward area element $d\vec{A}$ is also radial, so
+\[
+\vec{E}\cdot d\vec{A}=E(r)\,dA
+\]
+everywhere on the surface. Therefore,
+\[
+\oint_S \vec{E}\cdot d\vec{A}=E(r)\oint_S dA=E(r)(4\pi r^2).
+\]
+Since the enclosed charge is $q_{\mathrm{enc}}=Q$, Gauss's law gives
+\[
+E(r)(4\pi r^2)=\frac{Q}{\varepsilon_0},
+\qquad
+E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}.
+\]
+The law itself is general, but the symmetry is what allowed $E(r)$ to be pulled outside the integral.}
+
+\qs{Worked AP-style problem}{A very long straight wire carries a uniform positive linear charge density
+\[
+\lambda=3.0\times 10^{-6}\,\mathrm{C/m}.
+\]
+Let point $P$ be at perpendicular distance
+\[
+r=0.20\,\mathrm{m}
+\]
+from the wire. Choose a cylindrical Gaussian surface of radius $r$ and length $L$ coaxial with the wire.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the enclosed charge $q_{\mathrm{enc}}$ for that Gaussian surface,
+\item the electric flux through the curved side and through the two flat end caps, and
+\item the magnitude and direction of the electric field at $P$.
+\end{enumerate}}
+
+\sol Let the cylinder have radius $r$ and length $L$. Because the wire has uniform linear charge density $\lambda$, the charge enclosed by the Gaussian surface is
+\[
+q_{\mathrm{enc}}=\lambda L.
+\]
+
+By cylindrical symmetry, the electric field due to the long wire points radially outward from the wire and has the same magnitude $E(r)$ everywhere on the curved side of the Gaussian cylinder. Let $\hat{s}$ denote the outward radial unit vector from the wire.
+
+On the curved side, the area element $d\vec{A}$ also points radially outward, so $\vec{E}$ is parallel to $d\vec{A}$. Thus,
+\[
+\vec{E}\cdot d\vec{A}=E(r)\,dA
+\]
+on the curved surface.
+
+On each flat end cap, the area element $d\vec{A}$ points along the axis of the wire, while $\vec{E}$ points perpendicular to that axis. Therefore,
+\[
+\vec{E}\cdot d\vec{A}=0
+\]
+on both end caps, so the flux through each end cap is zero.
+
+The total flux is therefore entirely through the curved side:
+\[
+\oint_S \vec{E}\cdot d\vec{A}=E(r)\bigl(2\pi rL\bigr).
+\]
+Apply Gauss's law:
+\[
+E(r)(2\pi rL)=\frac{\lambda L}{\varepsilon_0}.
+\]
+Cancel $L$:
+\[
+E(r)=\frac{\lambda}{2\pi \varepsilon_0 r}.
+\]
+Now substitute $\lambda=3.0\times 10^{-6}\,\mathrm{C/m}$, $r=0.20\,\mathrm{m}$, and $\varepsilon_0=8.85\times 10^{-12}\,\mathrm{C^2/(N\,m^2)}$:
+\[
+E(r)=\frac{3.0\times 10^{-6}}{2\pi(8.85\times 10^{-12})(0.20)}\,\mathrm{N/C}.
+\]
+This gives
+\[
+E(r)=2.7\times 10^5\,\mathrm{N/C}.
+\]
+
+So the fluxes are
+\[
+\Phi_{\mathrm{curved}}=E(r)(2\pi rL)=\frac{\lambda L}{\varepsilon_0},
+\qquad
+\Phi_{\mathrm{cap\ 1}}=0,
+\qquad
+\Phi_{\mathrm{cap\ 2}}=0,
+\]
+and the electric field at $P$ is
+\[
+\vec{E}(P)=(2.7\times 10^5\,\mathrm{N/C})\hat{s},
+\]
+where $\hat{s}$ points radially away from the positively charged wire.

concepts/em/u9/e9-1-electric-potential-energy.tex

@@ -0,0 +1,153 @@
+\subsection{Electric Potential Energy}
+
+This subsection introduces electric potential energy as an energy of a charge configuration and relates it to work done by electric forces.
+
+\dfn{Electric potential energy of a system}{Let a system contain interacting charges. Let $U$ denote the \emph{electric potential energy} of the system, measured in joules. Electric potential energy is a property of the \emph{configuration of the system}, not of one charge by itself. For any two configurations,
+\[
+\Delta U=U_f-U_i,
+\]
+and if the electric force does work $W_{\mathrm{elec}}$ on the system, then
+\[
+W_{\mathrm{elec}}=-\Delta U.
+\]
+Thus, when the electric force does positive work, the system loses electric potential energy, and when an external agent slowly assembles a configuration against the electric force, the system gains electric potential energy.}
+
+\thm{Point-charge pair formula and work relations}{Let two point charges $q_1$ and $q_2$ be separated by distance $r$, and choose the reference value
+\[
+U(\infty)=0.
+\]
+Then the electric potential energy of the two-charge system is
+\[
+U(r)=k\frac{q_1q_2}{r},
+\]
+where
+\[
+k=\frac{1}{4\pi\varepsilon_0}=8.99\times10^9\,\mathrm{N\,m^2/C^2}.
+\]
+If the separation changes from $r_i$ to $r_f$, then
+\[
+\Delta U=U_f-U_i=kq_1q_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right).
+\]
+The work done by the electric force is
+\[
+W_{\mathrm{elec}}=-\Delta U,
+\]
+and for a slow external rearrangement of the charges,
+\[
+W_{\mathrm{ext}}=\Delta U.
+\]}
+
+\nt{Because $U=kq_1q_2/r$, the sign of $U$ depends on the signs of the two charges. If $q_1q_2>0$, then $U>0$ and positive external work is required to bring the like charges closer together. If $q_1q_2<0$, then $U<0$ and the electric force itself tends to pull the unlike charges together. The important viewpoint is that $U$ belongs to the pair of charges as a system. It is not correct to say that a single isolated charge ``has'' electric potential energy by itself.}
+
+\pf{Derivation from quasistatic assembly}{Let charge $q_1$ be fixed, and let charge $q_2$ be brought slowly from infinity to a final separation $r$. Let $x$ denote the instantaneous separation during the motion, with outward radial unit vector $\hat{r}$. The electric force on $q_2$ is
+\[
+\vec{F}_{\mathrm{elec}}=k\frac{q_1q_2}{x^2}\hat{r}.
+\]
+For a quasistatic move, the external force balances the electric force, so
+\[
+\vec{F}_{\mathrm{ext}}=-\vec{F}_{\mathrm{elec}}.
+\]
+The external work done in assembling the pair is the increase in potential energy:
+\[
+U(r)-U(\infty)=W_{\mathrm{ext}}=\int_{\infty}^{r}\vec{F}_{\mathrm{ext}}\cdot d\vec{r}.
+\]
+Since $d\vec{r}=\hat{r}\,dx$,
+\[
+U(r)-0=-\int_{\infty}^{r}k\frac{q_1q_2}{x^2}\,dx
+=-kq_1q_2\left[-\frac{1}{x}\right]_{\infty}^{r}
+=k\frac{q_1q_2}{r}.
+\]
+This also gives
+\[
+\Delta U=kq_1q_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right),
+\]
+and because electric potential energy is defined for a conservative force, the electric-force work satisfies $W_{\mathrm{elec}}=-\Delta U$.}
+
+\qs{Worked AP-style problem}{Two point charges form a system. Let
+\[
+q_1=+2.0\,\mu\mathrm{C},
+\qquad
+q_2=-3.0\,\mu\mathrm{C}.
+\]
+Initially the charges are separated by
+\[
+r_i=0.50\,\mathrm{m},
+\]
+and they are moved slowly until their final separation is
+\[
+r_f=0.20\,\mathrm{m}.
+\]
+Take
+\[
+k=8.99\times10^9\,\mathrm{N\,m^2/C^2}.
+\]
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the initial electric potential energy $U_i$,
+\item the final electric potential energy $U_f$,
+\item the change in electric potential energy $\Delta U$, and
+\item the work done by the external agent and by the electric force during the slow move.
+\end{enumerate}}
+
+\sol Use
+\[
+U=k\frac{q_1q_2}{r}.
+\]
+Because the charges have opposite signs, the product $q_1q_2$ is negative:
+\[
+q_1q_2=(2.0\times10^{-6}\,\mathrm{C})(-3.0\times10^{-6}\,\mathrm{C})=-6.0\times10^{-12}\,\mathrm{C}^2.
+\]
+
+For part (a), the initial potential energy is
+\[
+U_i=k\frac{q_1q_2}{r_i}
+=(8.99\times10^9)\frac{-6.0\times10^{-12}}{0.50}\,\mathrm{J}.
+\]
+So
+\[
+U_i=-1.08\times10^{-1}\,\mathrm{J}=-0.108\,\mathrm{J}.
+\]
+
+For part (b), the final potential energy is
+\[
+U_f=k\frac{q_1q_2}{r_f}
+=(8.99\times10^9)\frac{-6.0\times10^{-12}}{0.20}\,\mathrm{J}.
+\]
+Thus,
+\[
+U_f=-2.70\times10^{-1}\,\mathrm{J}=-0.270\,\mathrm{J}.
+\]
+
+For part (c),
+\[
+\Delta U=U_f-U_i=(-0.270)-(-0.108)\,\mathrm{J}.
+\]
+Therefore,
+\[
+\Delta U=-0.162\,\mathrm{J}.
+\]
+
+For part (d), because the move is slow,
+\[
+W_{\mathrm{ext}}=\Delta U=-0.162\,\mathrm{J}.
+\]
+The negative sign means the external agent removes energy from the system rather than supplying it. The electric force does the opposite amount of work:
+\[
+W_{\mathrm{elec}}=-\Delta U=+0.162\,\mathrm{J}.
+\]
+
+This result makes physical sense. Unlike charges attract, so when they are brought closer together, the system energy becomes more negative.
+
+Therefore,
+\[
+U_i=-0.108\,\mathrm{J},
+\qquad
+U_f=-0.270\,\mathrm{J},
+\]
+\[
+\Delta U=-0.162\,\mathrm{J},
+\qquad
+W_{\mathrm{ext}}=-0.162\,\mathrm{J},
+\qquad
+W_{\mathrm{elec}}=+0.162\,\mathrm{J}.
+\]

concepts/em/u9/e9-2-potential-voltage.tex

@@ -0,0 +1,145 @@
+\subsection{Electric Potential and Voltage}
+
+This subsection defines electric potential as electric potential energy per unit charge, interprets voltage as a potential difference, and emphasizes that potentials from multiple source charges add as scalars.
+
+\dfn{Electric potential and voltage}{Let $U$ denote the electric potential energy of a system containing a chosen test charge $q\neq 0$ at a specified point in an electrostatic field. The \emph{electric potential} $V$ at that point is defined by
+\[
+V=\frac{U}{q}.
+\]
+If points $A$ and $B$ have potentials $V_A$ and $V_B$, then the \emph{potential difference} from $A$ to $B$ is
+\[
+\Delta V=V_B-V_A=\frac{\Delta U}{q},
+\]
+where
+\[
+\Delta U=U_B-U_A.
+\]
+In common AP usage, \emph{voltage} usually means this potential difference between two points.}
+
+\nt{Electric potential is a scalar, not a vector, so it has magnitude and sign but no direction. A positive source charge produces positive potential, and a negative source charge produces negative potential, when the reference value is chosen as $V=0$ at infinity. Only potential differences are directly physical, so the zero of potential is a convenient reference choice rather than an absolute requirement.}
+
+\mprop{Point-charge potential and scalar superposition}{Let fixed point charges $Q_1,Q_2,\dots,Q_N$ be located at position vectors $\vec{r}_1,\vec{r}_2,\dots,\vec{r}_N$. Let point $P$ have position vector $\vec{r}$, with $\vec{r}\ne \vec{r}_i$ for all $i$. For each source charge, define the separation distance from $Q_i$ to $P$ by
+\[
+R_i=|\vec{r}-\vec{r}_i|.
+\]
+Let
+\[
+k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
+\]
+Choosing the reference value $V(\infty)=0$, the electric potential at $P$ due to one point charge $Q_i$ is
+\[
+V_i(P)=k\frac{Q_i}{R_i}.
+\]
+Because potential is a scalar, the total potential at $P$ due to all the source charges is
+\[
+V(P)=\sum_{i=1}^N V_i(P)=k\sum_{i=1}^N \frac{Q_i}{R_i}.
+\]
+If points $A$ and $B$ have potentials $V_A$ and $V_B$, then the voltage between them is
+\[
+\Delta V_{A\to B}=V_B-V_A.
+\]
+For a charge $q$ moved quasistatically from $A$ to $B$ by an external agent,
+\[
+\Delta U=q\Delta V,
+\qquad
+W_{\mathrm{ext}}=\Delta U=q\Delta V.
+\]}
+
+\qs{Worked AP-style problem}{Three fixed point charges lie in an $xy$-plane. Let
+\[
+Q_1=+4.0\,\mu\mathrm{C}\quad\text{at}\quad \vec{r}_1=\vec{0},
+\]
+\[
+Q_2=-2.0\,\mu\mathrm{C}\quad\text{at}\quad \vec{r}_2=(0.30\,\mathrm{m})\hat{\imath},
+\]
+and
+\[
+Q_3=+3.0\,\mu\mathrm{C}\quad\text{at}\quad \vec{r}_3=(0.40\,\mathrm{m})\hat{\jmath}.
+\]
+Let point $P$ be at
+\[
+\vec{r}_P=(0.30\hat{\imath}+0.40\hat{\jmath})\,\mathrm{m}.
+\]
+Take
+\[
+k=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
+\]
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the electric potential $V(P)$ relative to infinity, and
+\item the external work required to bring a test charge
+\[
+q_0=+2.0\,\mathrm{nC}
+\]
+from infinity to point $P$.
+\end{enumerate}}
+
+\sol First find the distances from each source charge to point $P$.
+
+From $Q_1$ at the origin to $P$,
+\[
+R_1=|\vec{r}_P-\vec{r}_1|=\sqrt{(0.30)^2+(0.40)^2}\,\mathrm{m}=0.50\,\mathrm{m}.
+\]
+From $Q_2$ at $(0.30\,\mathrm{m})\hat{\imath}$ to $P$,
+\[
+R_2=0.40\,\mathrm{m}.
+\]
+From $Q_3$ at $(0.40\,\mathrm{m})\hat{\jmath}$ to $P$,
+\[
+R_3=0.30\,\mathrm{m}.
+\]
+
+Because electric potential is a scalar, add the contributions algebraically:
+\[
+V(P)=k\left(\frac{Q_1}{R_1}+\frac{Q_2}{R_2}+\frac{Q_3}{R_3}\right).
+\]
+Substitute the given values:
+\[
+V(P)=8.99\times 10^9\left(\frac{4.0\times 10^{-6}}{0.50}+\frac{-2.0\times 10^{-6}}{0.40}+\frac{3.0\times 10^{-6}}{0.30}\right)\mathrm{V}.
+\]
+Evaluate each term inside the parentheses:
+\[
+\frac{4.0\times 10^{-6}}{0.50}=8.0\times 10^{-6},
+\qquad
+\frac{-2.0\times 10^{-6}}{0.40}=-5.0\times 10^{-6},
+\qquad
+\frac{3.0\times 10^{-6}}{0.30}=1.0\times 10^{-5}.
+\]
+So
+\[
+V(P)=8.99\times 10^9\left(1.3\times 10^{-5}\right)\mathrm{V}.
+\]
+Therefore,
+\[
+V(P)=1.17\times 10^5\,\mathrm{V}.
+\]
+
+For part (b), the potential at infinity is the chosen reference value,
+\[
+V(\infty)=0,
+\]
+so the potential difference from infinity to $P$ is
+\[
+\Delta V=V(P)-V(\infty)=1.17\times 10^5\,\mathrm{V}.
+\]
+The external work required to bring the test charge slowly from infinity to $P$ is the change in electric potential energy:
+\[
+W_{\mathrm{ext}}=\Delta U=q_0\Delta V.
+\]
+Substitute $q_0=2.0\times 10^{-9}\,\mathrm{C}$:
+\[
+W_{\mathrm{ext}}=(2.0\times 10^{-9})(1.17\times 10^5)\,\mathrm{J}.
+\]
+Thus,
+\[
+W_{\mathrm{ext}}=2.34\times 10^{-4}\,\mathrm{J}.
+\]
+
+So the electric potential at point $P$ is
+\[
+V(P)=1.17\times 10^5\,\mathrm{V},
+\]
+and the required external work to bring the positive test charge from infinity to $P$ is
+\[
+W_{\mathrm{ext}}=2.34\times 10^{-4}\,\mathrm{J}.
+\]

concepts/em/u9/e9-3-field-potential.tex

@@ -0,0 +1,159 @@
+\subsection{The Field-Potential Relation}
+
+This subsection connects electric field and electric potential, both globally through a line integral and locally through the slope or gradient of the potential.
+
+\dfn{Potential difference from the electric field}{Let $A$ and $B$ be two points in an electrostatic region, let $C$ be any path from $A$ to $B$, and let $d\vec{\ell}$ denote an infinitesimal displacement along that path. If the electric field is $\vec{E}$, then the infinitesimal potential change is
+\[
+dV=-\vec{E}\cdot d\vec{\ell}.
+\]
+Integrating from $A$ to $B$ gives the potential difference
+\[
+\Delta V=V_B-V_A=-\int_A^B \vec{E}\cdot d\vec{\ell}.
+\]
+For electrostatics, this value is independent of the path because the electric field is conservative.}
+
+\thm{Global and local field-potential relations}{Let $V(\vec{r})$ denote the electric potential at position vector $\vec{r}$, and let $\vec{E}(\vec{r})$ denote the electric field there. Then in electrostatics,
+\[
+\Delta V=V_B-V_A=-\int_A^B \vec{E}\cdot d\vec{\ell}.
+\]
+Locally,
+\[
+dV=-\vec{E}\cdot d\vec{\ell}.
+\]
+Since also
+\[
+dV=\nabla V\cdot d\vec{\ell},
+\]
+comparison gives the vector relation
+\[
+\vec{E}=-\nabla V.
+\]
+For one-dimensional motion along the $x$-axis,
+\[
+E_x=-\frac{dV}{dx}.
+\]
+Thus the $x$-component of the electric field is the negative slope of the potential graph.}
+
+\pf{Derivation from work per unit charge}{Let a charge $q$ move through an infinitesimal displacement $d\vec{\ell}$ in an electric field $\vec{E}$. The electric force is
+\[
+\vec{F}=q\vec{E},
+\]
+so the infinitesimal work done by the field is
+\[
+dW=\vec{F}\cdot d\vec{\ell}=q\vec{E}\cdot d\vec{\ell}.
+\]
+Electric potential difference is potential-energy change per unit charge, so
+\[
+dV=\frac{dU}{q}.
+\]
+Because the work done by the electric field decreases electric potential energy,
+\[
+dU=-dW.
+\]
+Therefore,
+\[
+dV=\frac{-dW}{q}=-\vec{E}\cdot d\vec{\ell}.
+\]
+Integrating between two points gives
+\[
+\Delta V=-\int \vec{E}\cdot d\vec{\ell}.
+\]
+Comparing this with the differential identity $dV=\nabla V\cdot d\vec{\ell}$ yields $\vec{E}=-\nabla V$.}
+
+\cor{Useful special cases}{Let $x$ denote position along the $x$-axis.
+
+\begin{enumerate}[label=(\alph*)]
+    \item In one dimension,
+    \[
+    E_x=-\frac{dV}{dx}.
+    \]
+    So where a graph of $V$ versus $x$ slopes downward, $E_x$ is positive, and where it slopes upward, $E_x$ is negative.
+
+    \item If the electric field is uniform and parallel to the displacement, so $\vec{E}=E\hat{u}$ and $\Delta \vec{\ell}=\Delta s\hat{u}$, then
+    \[
+    \Delta V=-E\Delta s.
+    \]
+    In particular, between parallel plates with nearly uniform field magnitude $E$ and separation $d$ measured in the field direction,
+    \[
+    |\Delta V|=Ed.
+    \]
+\end{enumerate}}
+
+\qs{Worked AP-style problem}{Along the $x$-axis, the electric potential is
+\[
+V(x)=120-40x+5x^2,
+\]
+where $V$ is in volts and $x$ is in meters.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+    \item the electric-field component $E_x(x)$,
+    \item the electric field at $x=2.0\,\mathrm{m}$,
+    \item the potential difference $\Delta V=V(4.0\,\mathrm{m})-V(1.0\,\mathrm{m})$, and
+    \item the work done by the electric field on a charge $q=+2.0\,\mu\mathrm{C}$ moving from $x=1.0\,\mathrm{m}$ to $x=4.0\,\mathrm{m}$.
+\end{enumerate}}
+
+\sol For part (a), use the one-dimensional field-potential relation:
+\[
+E_x=-\frac{dV}{dx}.
+\]
+Differentiate the given potential function:
+\[
+\frac{dV}{dx}=\frac{d}{dx}(120-40x+5x^2)=-40+10x.
+\]
+Therefore,
+\[
+E_x=-( -40+10x)=40-10x.
+\]
+So the field as a function of position is
+\[
+E_x(x)=40-10x
+\]
+in units of $\mathrm{N/C}$.
+
+For part (b), substitute $x=2.0\,\mathrm{m}$:
+\[
+E_x(2.0)=40-10(2.0)=20\,\mathrm{N/C}.
+\]
+Since this value is positive, the electric field points in the $+x$ direction:
+\[
+\vec{E}(2.0\,\mathrm{m})=(20\,\mathrm{N/C})\hat{\imath}.
+\]
+
+For part (c), first evaluate the potential at each position:
+\[
+V(4.0)=120-40(4.0)+5(4.0)^2=120-160+80=40\,\mathrm{V},
+\]
+and
+\[
+V(1.0)=120-40(1.0)+5(1.0)^2=120-40+5=85\,\mathrm{V}.
+\]
+Thus,
+\[
+\Delta V=V(4.0)-V(1.0)=40-85=-45\,\mathrm{V}.
+\]
+
+For part (d), the work done by the electric field is related to potential difference by
+\[
+W_{\text{field}}=-q\Delta V.
+\]
+Substitute $q=+2.0\times 10^{-6}\,\mathrm{C}$ and $\Delta V=-45\,\mathrm{V}$:
+\[
+W_{\text{field}}=-(2.0\times 10^{-6})(-45)\,\mathrm{J}=9.0\times 10^{-5}\,\mathrm{J}.
+\]
+So the field does positive work:
+\[
+W_{\text{field}}=9.0\times 10^{-5}\,\mathrm{J}=90\,\mu\mathrm{J}.
+\]
+
+Therefore,
+\[
+E_x(x)=40-10x,
+\qquad
+\vec{E}(2.0\,\mathrm{m})=(20\,\mathrm{N/C})\hat{\imath},
+\]
+\[
+\Delta V=-45\,\mathrm{V},
+\qquad
+W_{\text{field}}=9.0\times 10^{-5}\,\mathrm{J}.
+\]

concepts/em/u9/e9-4-equipotentials.tex

@@ -0,0 +1,157 @@
+\subsection{Equipotentials and Energy Conservation for Moving Charges}
+
+This subsection explains how equipotential curves or surfaces encode the direction of $\vec{E}$ and how potential differences determine changes in kinetic and electric potential energy for moving charges.
+
+\dfn{Equipotentials and the energy-change relation}{Let $V(\vec{r})$ denote the electric potential at position vector $\vec{r}$. An \emph{equipotential} is a curve in two dimensions or a surface in three dimensions on which the potential has one constant value:
+\[
+V(\vec{r})=\text{constant}.
+\]
+If a charge $q$ moves from point $A$ to point $B$, define
+\[
+\Delta V=V_B-V_A
+\]
+and
+\[
+\Delta U=U_B-U_A.
+\]
+Then the change in electric potential energy is
+\[
+\Delta U=q\Delta V.
+\]
+Thus potential difference tells how the electric potential energy of a chosen charge changes between two points.}
+
+\thm{Equipotentials, no work, and kinetic-energy change}{Let $d\vec{\ell}$ denote an infinitesimal displacement in an electrostatic field $\vec{E}$. Then
+\[
+dV=-\vec{E}\cdot d\vec{\ell}.
+\]
+If the displacement is along an equipotential, then $dV=0$, so
+\[
+\vec{E}\cdot d\vec{\ell}=0.
+\]
+Therefore the electric field is perpendicular to an equipotential, and the electric force does no work on a charge moved along an equipotential:
+\[
+W_{\mathrm{elec}}=q\int \vec{E}\cdot d\vec{\ell}=0.
+\]
+For any motion of a charge $q$ from $A$ to $B$ in electrostatics,
+\[
+\Delta U=q\Delta V.
+\]
+If only the electric force does work, conservation of mechanical energy gives
+\[
+K_i+U_i=K_f+U_f,
+\]
+so
+\[
+\Delta K=K_f-K_i=-\Delta U=-q\Delta V.
+\]
+Hence a charge speeds up when its electric potential energy decreases.}
+
+\ex{Illustrative example}{Points $A$ and $B$ lie on the same equipotential,
+\[
+V_A=V_B=120\,\mathrm{V}.
+\]
+Let a proton of charge
+\[
+q=+e=+1.60\times 10^{-19}\,\mathrm{C}
+\]
+move from $A$ to $B$.
+
+Because the two points are on the same equipotential,
+\[
+\Delta V=V_B-V_A=0.
+\]
+So the change in electric potential energy is
+\[
+\Delta U=q\Delta V=0,
+\]
+and the work done by the electric field is also zero. The field may still be present, but along that displacement it is perpendicular to the motion.}
+
+\nt{Be careful to distinguish \emph{lower potential} from \emph{lower potential energy}. Since $\Delta U=q\Delta V$, a positive charge has lower potential energy at lower potential, but a negative charge has lower potential energy at \emph{higher} potential. Thus a positive charge released from rest tends to speed up toward lower $V$, whereas an electron released from rest tends to speed up toward higher $V$. In both cases the rule is the same: the charge moves spontaneously in the direction that makes $U$ decrease and $K$ increase.}
+
+\qs{Worked AP-style problem}{Two large parallel plates create a uniform electrostatic region. Let point $A$ be near the negative plate and point $B$ be near the positive plate. The potentials are
+\[
+V_A=100\,\mathrm{V},
+\qquad
+V_B=400\,\mathrm{V}.
+\]
+An electron is released from rest at point $A$ and moves to point $B$. Let the electron charge be
+\[
+q=-1.60\times 10^{-19}\,\mathrm{C}
+\]
+and the electron mass be
+\[
+m_e=9.11\times 10^{-31}\,\mathrm{kg}.
+\]
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the potential difference $\Delta V=V_B-V_A$,
+\item the change in electric potential energy $\Delta U$,
+\item the change in kinetic energy $\Delta K$, and
+\item the electron's speed at point $B$.
+\end{enumerate}}
+
+\sol First compute the potential difference:
+\[
+\Delta V=V_B-V_A=400\,\mathrm{V}-100\,\mathrm{V}=300\,\mathrm{V}.
+\]
+
+For part (b), use the relation
+\[
+\Delta U=q\Delta V.
+\]
+Substitute the electron charge and the potential difference:
+\[
+\Delta U=(-1.60\times 10^{-19}\,\mathrm{C})(300\,\mathrm{V}).
+\]
+Since $1\,\mathrm{V}=1\,\mathrm{J/C}$,
+\[
+\Delta U=-4.80\times 10^{-17}\,\mathrm{J}.
+\]
+
+For part (c), only the electric force does work, so
+\[
+\Delta K=-\Delta U.
+\]
+Therefore,
+\[
+\Delta K=+4.80\times 10^{-17}\,\mathrm{J}.
+\]
+
+This sign makes sense. The electron moves toward higher potential, but because its charge is negative, that motion lowers its electric potential energy and increases its kinetic energy.
+
+For part (d), the electron starts from rest, so
+\[
+K_i=0.
+\]
+Thus
+\[
+K_f=\Delta K=4.80\times 10^{-17}\,\mathrm{J}.
+\]
+Use the kinetic-energy formula
+\[
+K_f=\frac12 m_e v^2.
+\]
+Solve for the speed $v$:
+\[
+v=\sqrt{\frac{2K_f}{m_e}}.
+\]
+Substitute the values:
+\[
+v=\sqrt{\frac{2(4.80\times 10^{-17}\,\mathrm{J})}{9.11\times 10^{-31}\,\mathrm{kg}}}.
+\]
+This gives
+\[
+v=1.03\times 10^7\,\mathrm{m/s}.
+\]
+
+Therefore,
+\[
+\Delta V=300\,\mathrm{V},
+\qquad
+\Delta U=-4.80\times 10^{-17}\,\mathrm{J},
+\]
+\[
+\Delta K=+4.80\times 10^{-17}\,\mathrm{J},
+\qquad
+v=1.03\times 10^7\,\mathrm{m/s}.
+\]