Lenz rule and induced electric field

Learning objectives

Outcome
Determine the direction of the induced current in a loop for a given change of magnetic flux by applying Lenz's rule and the right-hand rule.
Outcome
Calculate the magnitude of the induced emf and current using $|\mathcal{E}| = \left|\frac{d\Phi}{dt}\right|$ and $I = |\mathcal{E}|/R$ , relating the sign of $\mathcal{E}$ to Lenz's rule.
Outcome
Compute the magnitude of the induced electric field at a given distance from the symmetry axis using $\oint \vec{E}\cdot d\vec{l} = -\frac{d\Phi}{dt}$ and state its direction.
Outcome
Sketch the closed field lines of the induced electric field for a region of changing magnetic field and explain why this field is non-conservative.

Understand the idea

Lenz's Rule and the Induced (Vortex) Electric Field

Text

Faraday's law gives the magnitude of the induced emf, $|\mathcal{E}| = \left|\frac{d\Phi}{dt}\right|$ , but not its direction. Lenz's rule fixes the direction: the induced current always opposes the change of magnetic flux that produces it. Equivalently, the induced current creates its own magnetic field whose flux through the circuit counteracts the change $\Delta\Phi$ .

A changing magnetic flux is accompanied by an induced (vortex) electric field $\vec{E}$ whose field lines are closed and encircle the region of changing $\vec{B}$ . Quantitatively this is written as

\oint \vec{E}\cdot d\vec{l} = -\frac{d\Phi}{dt}.

Because the line integral of $\vec{E}$ around a closed loop need not vanish, the induced field is non-conservative: it cannot be represented as the gradient of a scalar potential, and its field lines are closed curves rather than lines running from positive to negative charge. The minus sign in the equation is Lenz's rule, and the field exists even in empty space, where there are no charges or currents to push.

Worked examples

Direct

Problem

A circular conducting loop lies horizontally. A bar magnet is brought toward it from above with its north pole facing down. Determine the direction of the induced current in the loop as seen from the magnet's side. Justify your answer with Lenz's rule.

Solution

Choose the positive normal to the loop along the magnet's field (downward from the north pole), so the flux is positive. As the magnet approaches, the magnitude of the flux increases: $\frac{d\Phi}{dt} > 0$ . By Lenz's rule the induced current must create a field that opposes this increase, so $\vec{B}_{\text{ind}}$ points upward, opposite to the magnet's field. Right-hand rule: thumb up, fingers curl anticlockwise when viewed from above. Therefore, viewed from the magnet's side, the induced current is counterclockwise.

Application

Problem

A circular loop of radius $r = 0.20\,\text{m}$ and resistance $R = 2.0\,\Omega$ lies in a uniform magnetic field perpendicular to its plane and directed out of the page. The field decreases from $0.60\,\text{T}$ to $0.20\,\text{T}$ in $\Delta t = 0.40\,\text{s}$ . Find the magnitude of the induced emf and the current, and state the current's direction.

Solution

Area of the loop: $A = \pi r^{2} = \pi(0.20\,\text{m})^{2} = 0.126\,\text{m}^{2}$ . Change of flux: $\Delta\Phi = \Delta B\cdot A = (-0.40\,\text{T})(0.126\,\text{m}^{2}) = -5.0\times 10^{-2}\,\text{Wb}$ . Magnitude of the emf: $|\mathcal{E}| = \left|\frac{\Delta\Phi}{\Delta t}\right| = \frac{5.0\times 10^{-2}\,\text{Wb}}{0.40\,\text{s}} = 0.126\,\text{V}$ (approximately $0.13\,\text{V}$ ). Current: $I = \frac{|\mathcal{E}|}{R} = \frac{0.126\,\text{V}}{2.0\,\Omega} = 6.3\times 10^{-2}\,\text{A} = 63\,\text{mA}$ . Direction: the flux out of the page is decreasing, so the induced current must create a field out of the page; such a field is produced by a counterclockwise current (viewed from the side from which $\vec{B}$ emerges).

Transfer

Problem

A uniform magnetic field $\vec{B}$ points out of the page and exists only inside a circle of radius $R = 0.30\,\text{m}$ . The field increases at the rate $\frac{dB}{dt} = 2.0\,\text{T/s}$ . Find the magnitude and direction of the induced electric field at distances $r_1 = 0.10\,\text{m}$ and $r_2 = 0.50\,\text{m}$ from the center.

Solution

By symmetry, $\vec{E}$ is tangential to a circle of radius $r$ and has constant magnitude on it, so $\oint \vec{E}\cdot d\vec{l} = E\cdot 2\pi r$ . For $r_1 < R$ : $E_1\cdot 2\pi r_1 = \pi r_1^{2}\frac{dB}{dt}$ , hence $E_1 = \frac{r_1}{2}\frac{dB}{dt} = \frac{0.10\,\text{m}}{2}\cdot 2.0\,\text{T/s} = 0.10\,\text{V/m}$ . For $r_2 > R$ the whole flux lies inside the loop: $E_2\cdot 2\pi r_2 = \pi R^{2}\frac{dB}{dt}$ , hence $E_2 = \frac{R^{2}}{2r_2}\frac{dB}{dt} = \frac{(0.30\,\text{m})^{2}}{2(0.50\,\text{m})}\cdot 2.0\,\text{T/s} = 0.18\,\text{V/m}$ . Direction: the field $\vec{B}$ out of the page is increasing, so by Lenz's rule the induced effect inside opposes it, meaning the $\vec{E}$ lines run clockwise (viewed from above).

Practice

Text

Use the PhET simulation Faraday's Electromagnetic Lab in the Magnet and Coil mode. The induced vortex field cannot be measured with household equipment (it requires precise instruments and is not tied to charges), so we observe it on a model.

Place the magnet with its north pole toward the coil and move it in slowly. Predict the current direction first using Lenz's rule, then check the sign of the galvanometer reading.
Move the magnet away; predict again and compare with the observation.
Reverse the magnet (south pole toward the coil) and repeat steps 1–2.
Increase the magnet's speed and note how the current magnitude changes; verify that $|\mathcal{E}| \propto \left|\frac{d\Phi}{dt}\right|$ .

Keep a table: motion (approach/recede), pole, flux change (increase/decrease), predicted current direction, observed direction. Finally, state in one sentence where the $\vec{E}$ lines point if the field $\vec{B}$ out of the page is increasing.

Visual reference

Purpose

Show two linked ideas: (1) the induced current is directed so that its magnetic field opposes the flux change (Lenz's rule), and (2) the field lines of the induced electric field are closed loops around the region of changing magnetic field.

Representation

A two-part diagram. Left: a magnet approaching a circular loop with its north pole; arrows show the magnet's field $\vec{B}$ (downward, into the loop), the induced field $\vec{B}_{\text{ind}}$ (upward, opposing), and circular arrows for the induced current (counterclockwise as seen from above). Right: a top view of a circle of radius $R$ with the field $\vec{B}$ out of the page (symbol $\odot$ ); closed concentric circles around the axis represent the $\vec{E}$ field lines, with tangential arrows pointing clockwise.

Real-world examples

An induction hob in the kitchen

Scenario

A $2.0~\text{kW}$ induction ring heats $1.0~\text{L}$ of water in a steel-based pan from $20~^\circ\text{C}$ to $100~^\circ\text{C}$ in about $3.5~\text{min}$ . Under the ceramic glass a coil carries an alternating current of frequency $25~\text{kHz}$ . The same pan made of aluminium barely warms on the same ring: the water in it stays cold.

Connection

The coil's alternating current changes the magnetic flux through the pan base, $\frac{d\Phi}{dt}\ne 0$ , and the induced vortex electric field drives closed eddy-current loops inside the metal. Lenz's rule fixes their direction: the magnetic field of these currents opposes the change of flux, so the loops close in a thin layer facing the coil and deposit heat there by Joule heating $P=I^2R$ . The $25~\text{kHz}$ frequency (rather than $50~\text{Hz}$ ) raises the induced emf; a steel base with resistance of order $0.1~\Omega$ turns the energy into heating, while aluminium has both low resistance and no ferromagnetism, so the effect is too weak to boil water.

Regenerative braking of an electric train

Scenario

An electric train of mass $3.0\times10^{5}~\text{kg}$ brakes from $20~\text{m/s}$ ( $72~\text{km/h}$ ) to a full stop in $20~\text{s}$ . Instead of only heating the brake pads, the traction motors are switched to generator mode and push about $60~\text{MJ}$ back into the overhead line — exactly the kinetic energy $\frac12 mv^2$ of this train.

Connection

While the wheels turn, the magnetic flux through the machine windings changes and an emf is induced in them; by Lenz's rule the current in the closed winding flows so that the magnetic force on it points against the rotation. A braking torque appears that does negative work on the train, and the energy goes to the grid instead of into heat. The vortex field obeys $\oint \vec E\cdot d\vec l=-\frac{d\Phi}{dt}$ : the faster the train, the larger $\frac{d\Phi}{dt}$ and the stronger the current and the braking force, so braking weakens as the train slows.

Wireless charging of a phone

Scenario

A $15~\text{W}$ charging pad fills a $4000~\text{mA}\cdot\text{h}$ phone battery (about $15~\text{W}\cdot\text{h}$ ) in roughly $1.5~\text{h}$ ; the flat coils are separated by a $5~\text{mm}$ plastic back, and the current in the pad alternates at about $150~\text{kHz}$ . Keys or a coin left on the pad get noticeably hot within a minute.

Connection

Lenz's rule explains both the power transfer and the hot coin: the induced current in the phone coil flows so that its field opposes the change of flux from the pad. That is why the pad starts drawing more current from the adapter (the reflected load), while in the receiver coil the induced emf $\mathcal{E}=-\frac{d\Phi}{dt}$ drives current into the battery. A coin has no closed circuit to charge: the vortex electric field swirls circular currents in it, and all the power is dissipated as $I^2R$ heat.