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