content(warnings): Add W17-W28, X8-X12, N7 — E&M misconceptions, cross-refs, notes

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

@@ -14,6 +14,8 @@ dI=\vec{J}\cdot d\vec{A}.
 \]
 Thus $\vec{J}$ links the local flow of charge to the total current through a cross section.}
 
+\wc{Electrons flow opposite to conventional current}{The \emph{conventional current} direction is defined as the direction positive charges would move (from higher to lower potential). In metal wires, the actual charge carriers are electrons, which move \emph{opposite} to the conventional current direction. This historical convention does not affect circuit analysis results.}
+
 \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

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

@@ -8,6 +8,8 @@ R = \frac{V}{I}.
 \]
 The SI unit of resistance is the \emph{ohm}, denoted $\Omega$, where $1\,\Omega = 1\,\mathrm{V/A}$.}
 
+\wc{Current is not ``used up'' by resistors}{Current $I$ is the same through every element in a series circuit. A resistor does not ``consume'' current --- it drops voltage (potential energy per charge). Think of the circuit as a closed loop: whatever current leaves the battery returns to it. What is ``used up'' is electrical potential energy (converted to heat), not charge.}
+
 \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}.

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

@@ -49,6 +49,8 @@ The assumed direction of each branch current must be declared before writing equ
 \end{enumerate}
 The number of independent equations needed equals the number of unknown branch currents.}
 
+Kirchhoff's rules generalize the equivalent-resistance method from Section 11.4. For single-battery circuits with pure series/parallel arrangements, equivalent resistance is faster. Kirchhoff's rules become necessary when multiple batteries or loops make reduction impossible.
+
 \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.}

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

@@ -16,6 +16,8 @@ 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 $I \to 0$.}
 
+\nt{The RC time constant $\tau=RC$ has units of seconds: $[\Omega]\cdot[\mathrm{F}]=[\mathrm{V/A}]\cdot[\mathrm{C/V}]=[\mathrm{C/A}]=[\mathrm{C/(C/s)}]=[\mathrm{s}]$. After one time constant during charging, the capacitor reaches $1-e^{-1}\approx 63.2\%$ of its final voltage. After five time constants, it reaches $99.3\%$, which is the practical definition of ``fully charged'' in circuit analysis.}
+
 \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,

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

@@ -24,6 +24,8 @@ where $\theta$ is the angle between the vectors $\vec{v}$ and $\vec{B}$ measured
 \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}}
 
+The magnetic force on a single charge (this section) generalizes to the magnetic force on a current-carrying wire (Section 12.3) by summing over all charge carriers. The Biot-Savart law (Section 12.4) gives the reverse: how currents produce the $\vec{B}$ field that exerts these forces.
+
 \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})
@@ -32,6 +34,8 @@ has magnitude $|q|vB\sin\theta$ and the correct directional behaviour. This is t
 
 \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.}
 
+\wc{A stationary charge produces zero magnetic field}{A point charge at rest produces only an \emph{electric} field. A magnetic field is produced only by \emph{moving} charges (currents) or changing electric fields (displacement current). A single stationary charge $q$ has $\vec{B}=\vec{0}$ everywhere.}
+
 \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
 \[
 \vec{F} = q\,\vec{E} + q\,\vec{v}\times\vec{B}.
@@ -49,6 +53,8 @@ P = \vec{F}_B\cdot\vec{v} = q\,(\vec{v}\times\vec{B})\cdot\vec{v} = 0.
 \]
 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 $(\vec{a}\times\vec{b})\cdot\vec{a}=0$.}
 
+\wc{Magnetic force does zero work --- always}{Because $\vec{F}_B=q\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$, the power $P=\vec{F}_B\cdot\vec{v}=0$. The magnetic force can change a particle's direction but never its speed or kinetic energy. Even in complex field configurations, the magnetic force contributes zero to the work integral $\int\vec{F}\cdot d\vec{r}$.}
+
 \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},

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

@@ -72,6 +72,8 @@ The sense of the circular rotation follows the same rule as cyclotron motion: co
 
 \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 $\alpha\to 90^\circ$.}
 
+\wc{Field lines are a visualization tool, not physical threads}{Magnetic field lines are a mathematical construct to visualize $\vec{B}$. The field is continuous and exists at every point --- field lines are just a convenient drawing. Charges do not follow field lines (except when $\vec{v}\parallel\vec{B}$). The density of lines indicates field strength, but the lines themselves have no physical substance.}
+
 \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)} \\

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

@@ -12,6 +12,8 @@ This subsection states Amp\`ere's law and shows how symmetry can reduce a diffic
 \]
 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.}
 
+\wc{Ampere's law is always true for steady currents}{Like Gauss's law, $\oint\vec{B}\cdot d\vec{\ell}=\mu_0 I_{\mathrm{enc}}$ holds for \emph{any} steady current distribution and \emph{any} closed path. Symmetry only makes it \emph{useful} for calculating $\vec{B}$. Without symmetry, you know the line integral but cannot extract $\vec{B}$ at each point.}
+
 \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,

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

@@ -18,6 +18,8 @@ where $d\vec{A}$ is the oriented area element (direction given by the right-hand
 \]
 The minus sign encodes Lenz's law: the induced current produces a magnetic field that opposes the change in flux that created it.}
 
+\wc{Faraday's law creates EMF, not necessarily current}{A changing magnetic flux induces an \emph{electromotive force} $\mathcal{E}=-d\Phi_B/dt$. Whether current flows depends on whether the loop is conducting and whether there is a complete circuit. A changing flux through an open loop or a broken ring creates EMF (a potential difference) but no current flows.}
+
 \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):
 \[

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

@@ -20,10 +20,14 @@ The operational procedure is:
 \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}}
 
+\wc{Lenz's law opposes the \emph{change} in flux, not the field itself}{The induced current creates a field that opposes the \emph{change} in magnetic flux, not the external field itself. If flux is \emph{increasing}, the induced field opposes the external field. If flux is \emph{decreasing}, the induced field is in the \emph{same} direction as the external field (to try to maintain it).}
+
 \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).}
 
+Lenz's law is the directional consequence of Faraday's law (Section 13.2) and is deeply connected to energy conservation. The same principle governs motional EMF in Section 13.4 and inductance in Section 13.5.
+
 \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}

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

@@ -100,6 +100,8 @@ Thus $P_{\text{mech}} = P_{\text{elec}}$, consistent with conservation of energy
 \]
 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.}
 
+The motional EMF $\mathcal{E}=B\ell v$ is consistent with Faraday's law $\mathcal{E}=-d\Phi_B/dt$ through the flux change perspective. Both derivations (force-on-charges and flux-change) appear in Sections 13.2 and 13.4 respectively and yield identical results.
+
 \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:
 \[
 \mathcal{E} = \int_{0}^{\ell} B\,(r\omega)\,dr = \frac{1}{2}\,B\,\omega\,\ell^{2}.\]}

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

@@ -12,6 +12,8 @@ Equivalently, from Faraday's law of induction, a changing current induces an EMF
 \]
 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 $1\;\mathrm{H} = 1\;\mathrm{V\!\cdot\!s/A} = 1\;\mathrm{kg\!\cdot\!m^2/(s^2\!\cdot\!A^2)}$.}
 
+\wc{An inductor opposes \emph{change} in current, not current itself}{An ideal inductor with steady (DC) current has zero voltage drop across it ($V=0$ when $dI/dt=0$). It only produces a back-EMF $\mathcal{E}=-L(dI/dt)$ when the current is \emph{changing}. A steady current passes through an ideal inductor as freely as through a wire.}
+
 \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

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

@@ -8,6 +8,8 @@ This subsection defines the electric field from source charges and shows how it
 \]
 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 $\mathrm{N/C}$.}
 
+\wc{Electric field fills all space (even if zero)}{The electric field $\vec{E}$ is defined at every point in space, regardless of whether a test charge is present. At points far from all source charges, the field magnitude may be very small or effectively zero, but the field itself extends throughout space. A test charge placed anywhere will experience $\vec{F}=q\vec{E}$ at that location.}
+
 \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,

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

@@ -10,8 +10,14 @@ This subsection states Gauss's law and shows how symmetry can reduce a difficult
 \]
 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.}
 
+Gauss's law is one of Maxwell's four equations (in electrostatic form). Its analogue for magnetism is Gauss's law for magnetic fields, $\oint\vec{B}\cdot d\vec{A}=0$ (Section 12.5), which reflects the absence of magnetic monopoles.
+
 \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.}
 
+\wc{Gauss's law is always true, not just for symmetric cases}{Gauss's law $\oint\vec{E}\cdot d\vec{A}=q_{\mathrm{enc}}/\varepsilon_0$ holds for \emph{any} charge distribution and \emph{any} closed surface. Symmetry only makes it \emph{useful} for calculating $\vec{E}$. Without symmetry, you know the total flux but cannot solve for the field at each point on the surface.}
+
+\wc{Zero net enclosed charge does not mean zero field}{A Gaussian surface enclosing zero net charge has zero \emph{net flux}, but the electric field on the surface need not be zero. External charges can produce nonzero field on the surface, with field lines entering on one side and leaving on the other, yielding zero net flux. $\oint\vec{E}\cdot d\vec{A}=0$ does \emph{not} imply $\vec{E}=0$.}
+
 \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