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\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}

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\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}

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\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,
\[
\text{(a) counterclockwise,} \qquad \text{(b) clockwise.}
\]

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\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}

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\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}

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\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,
\[
\tau = 0.25\,\mathrm{s},
\qquad
I(0.25\,\mathrm{s}) = 3.79\,\mathrm{A},
\qquad
I_{\text{max}} = 6.0\,\mathrm{A},
\qquad
U_B(0.25\,\mathrm{s}) = 7.2\,\mathrm{J}.
\]

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\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,
\[
f = 450\,\mathrm{Hz},
\qquad
I_{\max} = 28.3\,\mathrm{mA},
\qquad
U_L\!\left(\frac{T}{4}\right) = 1.00\times 10^{-5}\,\mathrm{J},
\qquad
U_C(t_{1/2}) = 5.00\times 10^{-6}\,\mathrm{J}.
\]