A-level Physics

9.4 Electromagnetic induction — Learn

Prerequisites: magnetic flux and flux linkage from Section 9.3; the force on a wire, F=BILF = BIL, from Section 9.3; work, energy and power from Section 3.3; emf and current from Section 5.1; graphs and gradients from Section 3.1.

By the end of this section you can: explain electromagnetic induction; use Faraday's law with flux linkage; state and apply Lenz's law; find the emf induced in a moving conductor; describe an alternating-current generator and use ε0=NABω\varepsilon_0 = NAB\omega; and explain how a search coil, a transformer and eddy currents use induction.

1. Induction

Move a magnet towards a coil connected to a galvanometer and the needle moves. Move the magnet away and the needle moves the other way. Hold the magnet still and the needle returns to zero. A changing magnetic field near a conductor causes a current. This is electromagnetic induction, and the current is an induced current. The emf that drives it is an induced emf.

Three things control the size of the induced emf:

Nothing is induced when the flux is steady, however large it is. Only a change in flux induces an emf.

2. Faraday's law

The magnetic flux linkage of a coil is NΦN\Phi. Faraday's law says that the induced emf is equal to the rate of change of the flux linkage:

ε=−NΔΦΔt\varepsilon = -N\frac{\Delta\Phi}{\Delta t}

The unit of emf is the volt. For a coil of NN turns and area AA in a field that changes at a steady rate, this becomes

ε=−NAΔBΔt\varepsilon = -NA\frac{\Delta B}{\Delta t}

The minus sign is the direction, and it comes from Lenz's law in the next section.

Worked example 1. A coil of 50 turns has an area of 0.020 m20.020\,\text{m}^{2}. The flux density through it rises from 0.10 T to 0.50 T at a steady rate in 0.20 s. Find the induced emf.

First find the change in flux density and then the change in flux linkage.

ΔB=0.50−0.10=0.40 TNΔΦ=NA ΔB=50×0.020×0.40=0.40 Wb\begin{aligned} \Delta B &= 0.50 - 0.10 = 0.40\,\text{T} \\ N\Delta\Phi &= NA\,\Delta B = 50 \times 0.020 \times 0.40 \\ &= 0.40\,\text{Wb} \end{aligned}

Then divide by the time.

ε=NΔΦΔt=0.400.20=2.0 V\varepsilon = \frac{N\Delta\Phi}{\Delta t} = \frac{0.40}{0.20} = 2.0\,\text{V}

The induced emf is 2.0 V.

3. Lenz's law

Lenz's law says that the induced current flows in the direction that opposes the change that produced it. If the flux through a coil increases, the induced current creates a field that opposes the increase. If the flux decreases, the induced current creates a field that tries to keep it.

Lenz's law is a statement of the conservation of energy. If the induced current helped the change, the change would grow, the current would grow with it, and we would get energy from nothing. The induced current must instead oppose the change, so work must be done to keep it going.

A magnet dropped through a copper tube falls slowly. The changing flux induces eddy currents in the tube, and those currents oppose the motion of the magnet. Energy is transferred from the falling magnet to the tube, which warms up. The same effect is used in an electromagnetic brake on a train.

4. The emf induced in a moving conductor

A conductor moving through a magnetic field also has a changing flux through it, so an emf is induced. Consider a straight rod of length LL moving at speed vv at right angles to a field of flux density BB. In a time Δt\Delta t the rod sweeps out an area LvΔtLv\Delta t, so the flux it cuts is BLvΔtBLv\Delta t. Faraday's law gives

ε=ΔΦΔt=BLvΔtΔt=BLv\varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{BLv\Delta t}{\Delta t} = BLv

The rod acts like a battery of emf BLvBLv. If the rod is part of a complete circuit, the emf drives a current. That current feels a magnetic force F=BILF = BIL which, by Lenz's law, opposes the motion. To keep the rod moving you must pull with a force BILBIL, doing work at a rate Fv=BILvFv = BILv. The electrical power is εI=BLvI\varepsilon I = BLvI, which is the same. Mechanical energy becomes electrical energy, and no energy is created.

Worked example 2. The spacecraft's tether is 20 m long and moves at 7.67×103 m s−17.67 \times 10^{3}\,\text{m}\,\text{s}^{-1} at right angles to the Earth's magnetic field, whose flux density is 3.0×10−5 T3.0 \times 10^{-5}\,\text{T}. Find the emf induced across the tether.

ε=BLv=3.0×10−5×20×7.67×103=4.6 V\begin{aligned} \varepsilon &= BLv \\ &= 3.0 \times 10^{-5} \times 20 \times 7.67 \times 10^{3} \\ &= 4.6\,\text{V} \end{aligned}

The emf across the tether is 4.6 V. The tether drives a current of 0.80 A, so it delivers a power of εI=4.6×0.80=3.7 W\varepsilon I = 4.6 \times 0.80 = 3.7\,\text{W}. The current feels a magnetic force

F=BIL=3.0×10−5×0.80×20=4.8×10−4 NF = BIL = 3.0 \times 10^{-5} \times 0.80 \times 20 = 4.8 \times 10^{-4}\,\text{N}

which opposes the spacecraft's motion. The mechanical power removed is Fv=4.8×10−4×7.67×103=3.7 WFv = 4.8 \times 10^{-4} \times 7.67 \times 10^{3} = 3.7\,\text{W}, the same as the electrical power. The tether is a generator that takes energy from the orbit.

5. The alternating-current generator

A coil rotating in a magnetic field is a generator. As the coil turns, the flux linkage through it changes continuously. When the plane of the coil is at right angles to the field, the flux linkage is largest but its rate of change is zero, so the emf is zero. When the plane of the coil is parallel to the field, the flux linkage is zero but its rate of change is largest, so the emf is largest.

The emf follows a sine curve. Its peak value is

ε0=NABω\varepsilon_0 = NAB\omega

where ω\omega is the angular speed of the coil in radians per second. The emf at any instant is ε=ε0sin⁡ωt\varepsilon = \varepsilon_0 \sin\omega t. This is how a power station generator produces alternating current.

Worked example 3. A generator coil has 200 turns and an area of 4.0×10−3 m24.0 \times 10^{-3}\,\text{m}^{2}. It turns at 50 rad s⁻¹ in a field of flux density 0.25 T. Find the peak emf.

ε0=NABω=200×4.0×10−3×0.25×50=10 V\begin{aligned} \varepsilon_0 &= NAB\omega \\ &= 200 \times 4.0 \times 10^{-3} \times 0.25 \times 50 \\ &= 10\,\text{V} \end{aligned}

The peak emf is 10 V, and the emf at any instant is 10sin⁡50t10\sin 50t volts.

6. Other applications of induction

Watch out. The induced emf depends on the rate of change of the flux linkage, not on the flux linkage itself. A coil sitting in a strong steady field has zero induced emf, because nothing is changing. A second trap is to drop the number of turns: the flux linkage is NΦN\Phi, so a 200-turn coil induces 200 times the emf of a single turn. A third trap is a sign error with Lenz's law. The induced current opposes the change, not the field. Check the direction by asking what the change is and which way a current would have to flow to oppose it.

7. Try it: the moving rod

The figure shows a conducting rod sliding along two rails through a magnetic field. Four sliders set the flux density, the length of the rod, its speed and the resistance of the circuit. The readout gives the induced emf, the current, the force that opposes the motion, and the mechanical and electrical power.

Predict first. Do not move the sliders yet. Write down your answers.

  1. If the rod stops moving, what happens to the induced emf?
  2. Compare the mechanical power you supply with the electrical power in the circuit. Which is larger?
  3. If you double the speed, what happens to the emf and to the opposing force?

Test. Set a flux density, a length, a speed and a resistance, and read all five values. Then double the speed and check prediction 3. Compare the two powers at every setting.

Explain. The two powers are always equal. Explain why, using Lenz's law and the conservation of energy. Then explain why you must keep pulling the rod to keep it moving at a steady speed. Write three or four sentences.

Progress check

Return to the spacecraft. Its tether is 20 m long and moves at 7.67×103 m s−17.67 \times 10^{3}\,\text{m}\,\text{s}^{-1} at right angles to the Earth's field of flux density 3.0×10−5 T3.0 \times 10^{-5}\,\text{T}. It drives a current of 0.80 A.

  1. Find the emf induced across the tether.
  2. Find the electrical power delivered.
  3. Find the magnetic force that opposes the motion, and show that the mechanical power removed from the orbit equals the electrical power.
  4. Explain why the tether slows the spacecraft down instead of speeding it up.
  5. A search coil of 2000 turns and area 1.5×10−4 m21.5 \times 10^{-4}\,\text{m}^{2} is in a field that changes at 0.20 T s⁻¹. Find the emf induced in it.

If question 1 goes wrong, use ε=BLv\varepsilon = BLv. If question 2 goes wrong, use P=εIP = \varepsilon I. If question 3 goes wrong, use F=BILF = BIL and P=FvP = Fv. If question 4 goes wrong, apply Lenz's law to the current in the tether. If question 5 goes wrong, use ε=NAΔBΔt\varepsilon = NA\dfrac{\Delta B}{\Delta t}.

You should now be able to

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