Electromagnetic Induction

Lessons

Senior 6 · Physics

Electromagnetic Induction

4 lessons

Lesson 2 of 4

Lenz's Law and Motional Electromotive Force

25 min

Why a generator is hard to turn

Spin the wheel of a bicycle with its dynamo pressed against the tyre, first with the lamp disconnected and then with it connected. With the lamp lit, the wheel is noticeably harder to turn. The dynamo is not rubbing harder; the current it produces is pushing back. Faraday's law told you how big an induced emf is. This lesson answers the other half of the question: which way does it act? The answer, Lenz's law, turns out to be the law of conservation of energy in disguise.

Lenz's law

Look back at Experiment 1.1. Pushing the N pole in gave one deflection; pulling it out gave the opposite one. Lenz's law predicts which is which: the induced current always flows so as to oppose the change that produces it.

Figure 2.1: Solid arcs are the fronts of the turns (nearer to you), dashed arcs the backs. A current flowing anticlockwise when you look at a face makes that face a north pole; clockwise makes it a south pole.

In (a) the flux through the coil, towards the right, is increasing. The coil opposes this by making its own field towards the LEFT inside it: its near face becomes a N pole, which repels the approaching N pole. In (b) the flux is decreasing, so the coil tries to keep it: its near face becomes a S pole, which attracts the retreating N pole. In both cases you must do work against a magnetic force to move the magnet.

Definition

Lenz's law: the direction of an induced emf is such that the current it drives produces effects that oppose the change causing it. This is the meaning of the minus sign in Faraday's law, $\varepsilon = -N\frac{d\Phi}{dt}$ .

Important

Lenz's law and the conservation of energy. Suppose the induced current instead HELPED the change: an approaching N pole would meet a S face and be pulled in faster, which would increase the current, which would pull harder still. The magnet would gain kinetic energy and the circuit would gain electrical energy with no work done: energy from nothing. Because the induced current opposes the motion, the mechanical work you do is exactly what becomes electrical energy (and finally heat in the circuit). Lenz's law is required by the conservation of energy.

Experiment 2.1: finding the direction of an induced current

Apparatus: the coil, centre-zero galvanometer and magnet of Experiment 1.1; a 1.5 V cell; a resistor of about $10\,\mathrm{k\Omega}$ .

Calibrate the meter. Connect the cell, the resistor and the galvanometer in series for a moment. Note which way the pointer moves when conventional current enters the "+" (red) terminal.
Trace the direction in which the wire is wound round the coil, and connect the coil to the galvanometer.
Push the N pole in, as in Figure 2.1(a). From the deflection, deduce the direction of the current in the turns, and use the clock rule to find the polarity of the near face. It should be N.
Repeat for pulling the N pole out, and for the S pole.

Warning

Safety. Never connect the galvanometer directly across the cell: always include the large resistor, or the current will damage the meter, and connect it only for a moment. Handle strong magnets carefully so they do not snap together on your fingers, and keep them away from phones and cards.

Worked example 1: a loop beside a wire

A long straight wire lies along the bottom edge of a page, carrying a current to the right that is INCREASING. A small loop lies on the page just above the wire. Which way does the induced current flow round the loop?

By the right-hand grip rule, above a wire carrying current to the right the field points OUT of the page. The current is rising, so the outward flux through the loop is increasing. The loop opposes this by making a field INTO the page inside it, which needs a clockwise current (as seen by the reader).

Motional emf: a conductor moving through a field

An emf is also induced in a conductor that moves across a magnetic field, even when the field is steady. In Figure 2.2 a rod of length $L$ slides at speed $v$ along rails in a uniform field $B$ into the page.

Figure 2.2: The induced current is anticlockwise, and the force on the rod opposes its motion.

Derivation 1: forces on the charges

Each free charge $q$ in the rod moves with the rod at velocity $v$ , so it feels a magnetic force $F = q v B$ along the rod (S5). Charges pile up at the ends until the electric field $E$ they create balances this: $q E = q v B$ , so $E = v B$ . The potential difference across the rod is then

\varepsilon = EL = BLv

Derivation 2: rate of change of flux

In a time $\Delta t$ the rod moves $v\Delta t$ , so the area of the circuit grows by $Lv\Delta t$ and the flux by $BLv\Delta t$ :

\varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{BLv\,\Delta t}{\Delta t} = BLv

Both routes give the same answer. If the rod moves at an angle $\theta$ to the field lines, only the perpendicular component counts: $\varepsilon = BLv\sin\theta$ .

Direction and energy

Fleming's right-hand rule gives the direction of the induced current: hold the thumb, first finger and second finger of the RIGHT hand at right angles; point the thumb along the Motion and the First finger along the Field; the seCond finger then points along the induced Current. In Figure 2.2 this gives a current up the rod. That current in the field feels a force $F = B I L$ to the LEFT, opposing the motion (Lenz's law), so an agent must push the rod to keep it moving. The mechanical power supplied, $F v$ , equals the electrical power $\varepsilon I$ :

Fv = (BIL)v = I(BLv) = \varepsilon I

Definition

Motional emf is the emf induced in a conductor moving through a magnetic field: $\varepsilon = BLv$ for a straight conductor of length $L$ moving at speed $v$ at right angles to both itself and a uniform field $B$ .

Worked example 2: the sliding rod

In Figure 2.2, $L = 0.50\,\mathrm{m}$ , $v = 4.0\,\mathrm{m\,s^{-1}}$ , $B = 0.20\,\mathrm{T}$ and the total resistance is $2.0\,\Omega$ . Find the emf, the current, the force needed to keep the rod moving at constant speed and the power dissipated.

\varepsilon = BLv = 0.20 \times 0.50 \times 4.0 = 0.40\,\mathrm{V}

I = \frac{0.40}{2.0} = 0.20\,\mathrm{A}

F = BIL = 0.20 \times 0.20 \times 0.50 = 0.020\,\mathrm{N}

P = Fv = 0.020 \times 4.0 = 0.080\,\mathrm{W}

Check: $I^2 R = 0.20^2 \times 2.0 = 0.080\,\mathrm{W}$ . The mechanical work becomes heat in the resistor.

Worked example 3: a rotating rod

A metal rod of length $L$ rotates at angular velocity $\omega$ about one end, in a plane perpendicular to a uniform field $B$ . Show that the emf between its ends is $\frac{1}{2}B\omega L^2$ , and evaluate it for $L = 0.50\,\mathrm{m}$ , $\omega = 20\,\mathrm{rad\,s^{-1}}$ , $B = 0.10\,\mathrm{T}$ .

In one revolution (time $\frac{2\pi}{\omega}$ ) the rod sweeps a circle of area $\pi L^2$ , cutting a flux $B\pi L^2$ :

\varepsilon = \frac{B\pi L^2}{\frac{2\pi}{\omega}} = \tfrac{1}{2}B\omega L^2 = \tfrac{1}{2} \times 0.10 \times 20 \times 0.50^2 = 0.25\,\mathrm{V}

Worked example 4: an aircraft wing

An aircraft with a wingspan of 40 m flies horizontally at $250\,\mathrm{m\,s^{-1}}$ where the vertical component of the Earth's field is $4.0 \times 10^{-5}\,\mathrm{T}$ (illustrative value). Find the emf between the wingtips.

\varepsilon = BLv = 4.0 \times 10^{-5} \times 40 \times 250 = 0.40\,\mathrm{V}

No current flows and no meter on board can show this emf, because any wire joining the wingtips moves with the wing and has the same emf induced in it.

Warning

Common mistakes. (1) Saying the induced current opposes the FIELD: it opposes the CHANGE in flux. A decreasing flux is "supported", not opposed. (2) Using the left-hand (motor) rule for an induced current: use the right-hand rule, or Lenz's law. (3) Forgetting that $\varepsilon = BLv$ needs $B$ , $L$ and $v$ mutually perpendicular. (4) Thinking no force is needed to keep the rod moving at constant speed once a current flows: the force $B I L$ opposes it.

Tip

For direction problems use three steps: (1) which way is the flux, and is it increasing or decreasing? (2) the induced field opposes an increase, or supports a decrease; (3) the right-hand grip rule turns that field into a current direction.

Lenz's law with a copper pipe (MIT Physics Instructional Resources Lab). A magnet falls very slowly through a copper pipe: watch closely, and explain with Lenz's law why the induced currents in the pipe hold it back.

✦

Practice

State Lenz's law and explain why it follows from the conservation of energy.
The S pole of a magnet is pulled AWAY from the face of a coil. What polarity does that face acquire, and which way does the current flow as seen from the magnet?
A rod 0.30 m long moves at $5.0\,\mathrm{m\,s^{-1}}$ at right angles to a 0.50 T field. Find the emf between its ends.
The rod of question 3 slides on rails in a circuit of total resistance $1.5\,\Omega$ . Find the current, the force needed to keep it moving steadily and the power this force supplies.
A helicopter blade 1.2 m long rotates at $30\,\mathrm{rad\,s^{-1}}$ in a region where the field component along its axis is $5.0 \times 10^{-5}\,\mathrm{T}$ . Find the emf between the hub and the tip.
A magnet is dropped through a vertical coil. Explain why it falls more slowly when the coil is connected to a resistor than when the coil's ends are left unconnected.
Keza says, "The induced current in a coil always opposes the magnetic field of the magnet." Evaluate her statement, with an example.

Show answersHide answers

The induced current flows so as to oppose the change causing it. If it helped the change, the motion would speed itself up and produce electrical energy without any work being done, which would break the conservation of energy.
Moving a S pole away decreases the flux; the coil opposes this by attracting the magnet, so the near face becomes a N pole. Seen from the magnet the current is anticlockwise.
$\varepsilon = BLv = 0.50 \times 0.30 \times 5.0 = 0.75\,\mathrm{V}$
$I = \frac{0.75}{1.5} = 0.50\,\mathrm{A}$ ; $F = BIL = 0.50 \times 0.50 \times 0.30 = 0.075\,\mathrm{N}$ ; $P = Fv = 0.075 \times 5.0 = 0.375\,\mathrm{W}$ ( $= \varepsilon I$ ).
$\varepsilon = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 5.0 \times 10^{-5} \times 30 \times 1.2^2 = 1.1 \times 10^{-3}\,\mathrm{V}$
With the circuit closed, a current flows; by Lenz's law it produces a field that repels the magnet as it approaches and attracts it as it leaves, so it slows the fall. The magnet's lost kinetic energy becomes heat in the resistor. With the ends open, an emf is induced but no current flows, so there is no opposing force.
Not always. The current opposes the CHANGE in flux. When a magnet is pulled away, the current makes a field in the SAME direction as the magnet's field inside the coil, trying to keep the flux from falling (Figure 2.1b).

Key points

Lenz's law: the induced current opposes the change that causes it; this is the minus sign in $\varepsilon = -N\frac{d\Phi}{dt}$ and a consequence of energy conservation.
Motional emf: $\varepsilon = BLv$ ; direction by Fleming's right-hand rule; the force $B I L$ on the moving conductor opposes its motion.
A rod rotating about one end: $\varepsilon = \frac{1}{2}B\omega L^2$ .

Sources and credits

13.3: Lenz's Law and 13.4: Motional Emf, from University Physics Volume 2 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
23.6: Motional Emf, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Video: Lenz's Law with Copper Pipe, MIT Physics Instructional Resources Lab, embedded from YouTube.

The experiment, the two diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Why a generator is hard to turn

2Lenz's law

3Experiment 2.1: finding the direction of an induced current

4Motional emf: a conductor moving through a field