9.4 Electromagnetic induction — Practise

Work through these on your own. Find the change in flux linkage before you divide by the time, and use Lenz's law to check the direction of every induced current. Write the unit of every answer. In the questions that ask for a reason, write your answer in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A coil of 80 turns has an area of 0.020 m^2. The flux density through it rises steadily from 0.15 T to 0.55 T in 0.25 s. Find the induced emf.
A rod of length 0.30 m moves at 5.0 m/s at right angles to a magnetic field of flux density 0.50 T. Find the emf induced across it.
The rod in question 2 is part of a circuit of total resistance 1.5 ohms. Find the current and the force that opposes the motion.
A search coil of 800 turns and area 1.2e-4 m^2 is in a magnetic field that changes at 0.25 T/s. Find the emf induced in the coil.
A generator coil of 250 turns and area 4.0e-3 m^2 turns at 40 rad/s in a field of flux density 0.30 T. Find the peak emf.
A coil of 400 turns and area 6.0e-3 m^2 is turned from parallel to perpendicular to a field of 0.20 T in 0.050 s. Find the average induced emf.
A transformer has 2000 turns on the primary coil and 50 turns on the secondary coil. The primary is connected to a 230 V alternating supply. Find the secondary voltage.
Explain, using the conservation of energy, why the force on a conductor that carries an induced current opposes the motion producing it.
A rod moving through a magnetic field induces an emf of 0.60 V. The flux density has an uncertainty of 2.0%, the length 0.5% and the speed 1.0%. Find the absolute uncertainty in the emf.
A search-coil calculation gives an emf of 0.02431 V. Write the answer to two significant figures.
An induced emf is 30 mV. Write it in volts.
A rod of length 0.20 m moves at 4.0 m/s at right angles to a field of 0.30 T. It completes a circuit of resistance 0.60 ohms. Find the power dissipated in the circuit.
A coil is rotated at a steady rate in a uniform magnetic field. Explain how to find the maximum emf, and why the emf is zero at two positions in each turn.
A conducting tether of length 25 m moves at 7.5e3 m/s at right angles to a field of flux density 4.0e-5 T. It drives a current of 0.60 A. Find the emf across the tether and the electrical power delivered.

Answers and concise solutions

1. The induced emf

Answer: The change in flux linkage is $N\Delta\Phi = NA\Delta B = 80 \times 0.020 \times 0.40 = 0.64\,\text{Wb}$ , so $\varepsilon = \dfrac{0.64}{0.25} = 2.6\,\text{V}$ .

2. The induced emf

Answer: $\varepsilon = BLv = 0.50 \times 0.30 \times 5.0 = 0.75\,\text{V}$ .

3. The current and the force

Answer: $I = \dfrac{\varepsilon}{R} = \dfrac{0.75}{1.5} = 0.50\,\text{A}$ , and $F = BIL = 0.50 \times 0.50 \times 0.30 = 0.075\,\text{N}$ , opposite to the velocity.

4. The induced emf

Answer: $\varepsilon = NA\dfrac{\Delta B}{\Delta t} = 800 \times 1.2 \times 10^{-4} \times 0.25 = 0.024\,\text{V}$ , or 24 mV.

5. The peak emf

Answer: $\varepsilon_0 = NAB\omega = 250 \times 4.0 \times 10^{-3} \times 0.30 \times 40 = 12\,\text{V}$ .

6. The average induced emf

Answer: The change in flux linkage is $NBA = 400 \times 0.20 \times 6.0 \times 10^{-3} = 0.48\,\text{Wb}$ , so the average emf is $\dfrac{0.48}{0.050} = 9.6\,\text{V}$ .

7. The secondary voltage

Answer: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ , so $V_s = 230 \times \dfrac{50}{2000} = 5.8\,\text{V}$ .

8. Explain, using the conservation of energy, why the force on a conductor that carries an induced current opposes the motion producing it.

A changing flux induces a current in the conductor.
By Lenz's law the induced current opposes the change in flux.
If the current helped the change, the change would grow and the current would grow with it.
That would create electrical energy from nothing, which breaks the conservation of energy.
So the force opposes the motion, and work must be done against it to keep the motion going. That work becomes electrical energy.

9. The absolute uncertainty

Answer: The emf is $\varepsilon = BLv$ . A product adds percentage uncertainties, so the percentage uncertainty is $2.0 + 0.5 + 1.0 = 3.5\%$ . The absolute uncertainty is $\dfrac{3.5}{100} \times 0.60 = 0.021\,\text{V}$ . So $\varepsilon = 0.60 \pm 0.02\,\text{V}$ .

10. The emf to two significant figures

Answer: $0.02431\,\text{V}$ is $0.024\,\text{V}$ to two significant figures.

11. The emf in volts

Answer: $30\,\text{mV} = 30 \times 10^{-3}\,\text{V} = 0.030\,\text{V}$ .

12. The power dissipated

Answer: The emf is $\varepsilon = BLv = 0.24\,\text{V}$ , so $I = \dfrac{0.24}{0.60} = 0.40\,\text{A}$ . The power is $P = \varepsilon I = 0.24 \times 0.40 = 0.096\,\text{W}$ , or 96 mW.

13. A coil is rotated at a steady rate in a uniform magnetic field. Explain how to find the maximum emf, and why the emf is zero at two positions in each turn.

The maximum emf is $\varepsilon_0 = NAB\omega$ .
The emf depends on the rate of change of the flux linkage, not on the flux linkage itself.
When the coil is at right angles to the field, the flux linkage is a maximum but its rate of change is zero, so the emf is zero.
When the coil is parallel to the field, the flux linkage is zero but it is changing fastest, so the emf is a maximum.

14. The emf and the power

Answer: $\varepsilon = BLv = 4.0 \times 10^{-5} \times 25 \times 7.5 \times 10^{3} = 7.5\,\text{V}$ , and $P = \varepsilon I = 7.5 \times 0.60 = 4.5\,\text{W}$ .