\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}
