9.4 Electromagnetic induction — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Find the change in the flux linkage before you divide by the time, and use Lenz's law to check every direction.

Warm-up: remember

A current of 1.5 A flows through a resistor of resistance $12\,\Omega$ . Find the power dissipated.
A coil of area $0.010\,\text{m}^{2}$ is at right angles to a magnetic field of flux density 0.50 T. Find the magnetic flux through the coil.
A motor lifts a 2.0 kg mass through 3.0 m in 5.0 s. Find the useful power output.

Answers

$P = I^{2}R = 1.5^{2} \times 12 = 27\,\text{W}$ .
$\Phi = BA = 0.50 \times 0.010 = 5.0 \times 10^{-3}\,\text{Wb}$ .
$P = \dfrac{mgh}{t} = \dfrac{2.0 \times 9.81 \times 3.0}{5.0} = 12\,\text{W}$ .

If question 1 goes wrong, revisit electrical power in Section 5.2. If question 2 goes wrong, revisit magnetic flux in Section 9.3. If question 3 goes wrong, revisit work, energy and power in Section 3.3.

Practise with help

1. A coil of 100 turns has an area of $0.015\,\text{m}^{2}$ . The flux density through it rises steadily from 0.20 T to 0.60 T in 0.40 s. Find the induced emf.

Hint

Faraday's law uses the change in the flux linkage, not the final flux.

Next step

Find $\Delta B$ first, then multiply by $N$ and $A$ to get the change in flux linkage. Divide by the time last.

Full working

\Delta B = 0.60 - 0.20 = 0.40\,\text{T}

N\Delta\Phi = NA\,\Delta B = 100 \times 0.015 \times 0.40 = 0.60\,\text{Wb}

\varepsilon = \frac{N\Delta\Phi}{\Delta t} = \frac{0.60}{0.40} = 1.5\,\text{V}

The induced emf is 1.5 V.

2. A rod of length 0.25 m moves at $6.0\,\text{m}\,\text{s}^{-1}$ at right angles to a magnetic field of flux density 0.40 T. Find the emf induced across the rod.

Hint

A moving conductor is a generator. The emf depends on the length, the speed and the field.

Next step

Use $\varepsilon = BLv$ .

Full working

\varepsilon = BLv = 0.40 \times 0.25 \times 6.0 = 0.60\,\text{V}

The emf across the rod is 0.60 V.

3. The rod in question 2 is part of a circuit of total resistance $2.0\,\Omega$ . Find the current and the magnetic force that opposes the motion.

Hint

Use Ohm's law for the current, then the force on a current-carrying wire.

Next step

Find $I = \dfrac{\varepsilon}{R}$ first, then use $F = BIL$ .

Full working

I = \frac{\varepsilon}{R} = \frac{0.60}{2.0} = 0.30\,\text{A}

F = BIL = 0.40 \times 0.30 \times 0.25 = 0.030\,\text{N}

The current is 0.30 A, and the opposing force on the rod is 0.030 N. By Lenz's law this force points opposite to the velocity.

4. A search coil of 500 turns and area $2.0 \times 10^{-4}\,\text{m}^{2}$ is in a magnetic field that changes at a rate of $0.30\,\text{T}\,\text{s}^{-1}$ . Find the emf induced in the coil.

Hint

The rate of change of the field is already given, so you do not need a time interval.

Next step

Use $\varepsilon = NA\dfrac{\Delta B}{\Delta t}$ .

Full working

\begin{aligned} \varepsilon &= NA\frac{\Delta B}{\Delta t} \\ &= 500 \times 2.0 \times 10^{-4} \times 0.30 \\ &= 0.030\,\text{V} \end{aligned}

The induced emf is 0.030 V, or 30 mV.

5. A generator coil of 150 turns and area $5.0 \times 10^{-3}\,\text{m}^{2}$ turns at $60\,\text{rad}\,\text{s}^{-1}$ in a field of flux density 0.20 T. Find the peak emf.

Hint

Use the peak value of the alternating emf, not the value at an instant.

Next step

Use $\varepsilon_0 = NAB\omega$ , and check that the angular speed is in radians per second.

Full working

\begin{aligned} \varepsilon_0 &= NAB\omega \\ &= 150 \times 5.0 \times 10^{-3} \times 0.20 \times 60 \\ &= 9.0\,\text{V} \end{aligned}

The peak emf is 9.0 V.

6. Explain why the induced current in a conductor always opposes the change that produced it.

Hint

Think about what would happen if the induced current helped the change.

Next step

Use the conservation of energy, and say what would happen to the current if it reinforced the change.

Full working

If the induced current helped the change that produced it, it would make the flux change faster, which would make the current larger, which would make the flux change faster still. The current would grow without any input of energy, which would break the conservation of energy. Lenz's law says the induced current must instead oppose the change. Work must then be done to keep the change going, and that work becomes electrical energy.

7. A transformer has 1200 turns on its primary coil and 100 turns on its secondary coil. The primary coil is connected to a 240 V alternating supply. Find the secondary voltage.

Hint

The turns ratio sets the voltage ratio.

Next step

Use $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ and rearrange for $V_s$ .

Full working

V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{1200} = 20\,\text{V}

The secondary voltage is 20 V. This is a step-down transformer.

8. A magnet is dropped down a vertical copper tube and falls much more slowly than the same magnet falls through the air. Explain why.

Hint

The moving magnet changes the flux through the copper.

Next step

Say what is induced in the copper, and then what Lenz's law says about the force on the magnet.

Full working

As the magnet moves, the magnetic flux through the copper changes. By Faraday's law this induces eddy currents in the tube. By Lenz's law the induced currents oppose the change that produced them, so they exert an upward force on the magnet, opposing its fall. The magnet still falls, because it is not held, but it reaches a much lower terminal speed. The kinetic energy lost by the magnet becomes internal energy in the copper.