Questions on electromagnetic induction: Faraday and Lenz

Question 1

State Faraday's law of electromagnetic induction.

Accepted Answer

The magnitude of the induced emf equals the rate of change of flux linkage (1); emf = N ΔΦ/Δt (1).

Question 2

The flux linkage through a coil changes by 0.60 Wb-turns in 0.20 s. Calculate the magnitude of the average induced emf.

Accepted Answer

emf = Δ(NΦ)/Δt = 0.60/0.20 (1) emf = 3.0 V (1)

Question 3

State Lenz's law.

Accepted Answer

The direction of the induced current (or emf) is such as to oppose the change in flux that produces it (1).

Question 4

State the SI unit of the rate of change of flux linkage.

Accepted Answer

The volt: a rate of change of flux linkage of 1 Wb-turn per second induces an emf of 1 V (1).

Question 5

A coil sits at rest in a steady magnetic field, so its flux linkage is constant. State the emf induced in the coil and explain your answer.

Accepted Answer

The emf is zero (1); the induced emf equals the rate of change of flux linkage, and here the flux linkage is not changing (1).

Question 6

A straight conducting rod of length 0.30 m moves at 4.0 m s⁻¹ perpendicular to a magnetic field of flux density 0.25 T. Calculate the emf induced across the ends of the rod.

Accepted Answer

emf = Blv = 0.25 × 0.30 × 4.0 (1) emf = 0.30 V (1)

Question 7

A coil of 200 turns and area 0.01 m² lies with its plane perpendicular to a magnetic field. The flux density falls steadily from 0.50 T to 0.10 T in 0.40 s. Calculate the average emf induced in the coil.

Accepted Answer

emf = NAΔB/Δt (1) = 200 × 0.01 × (0.50 − 0.10)/0.40 (1) emf = 2.00 V (1)

Question 8

The flux linkage of a coil collapses from 0.80 Wb-turns to zero in 0.010 s. Calculate the magnitude of the average induced emf.

Accepted Answer

emf = Δ(NΦ)/Δt = 0.80/0.010 (1) emf = 80 V (1)

Question 9

Explain how Lenz's law is a consequence of the conservation of energy.

Accepted Answer

If the induced current acted to aid the change in flux, the change would reinforce itself and electrical energy would be produced from nothing (1). The induced current instead opposes the change, so work must be done against the opposing force to keep the change going (1); this work input is the source of the electrical energy generated, consistent with energy conservation (1).

Question 10

The flux through each turn of a 250-turn coil increases steadily by 3.2 × 10⁻⁴ Wb in 40 ms. The coil is connected in a circuit of total resistance 4.0 Ω. Calculate the average induced emf and the average current in the circuit.

Accepted Answer

emf = NΔΦ/Δt = 250 × 3.2 × 10⁻⁴/0.040 (1) emf = 2.0 V (1) I = emf/R = 2.0/4.0 = 0.50 A (1)

Question 11

An aeroplane with a wingspan of 60 m flies horizontally at 250 m s⁻¹ at a location where the vertical component of the Earth's magnetic field is 45 μT. Calculate the emf induced between its wingtips, and state why only the vertical component of the field contributes.

Accepted Answer

emf = Blv = 45 × 10⁻⁶ × 60 × 250 (1)
emf = 0.68 V (1)
The wings sweep out a horizontal area, so only the flux of the vertical component passes through the area swept; the horizontal component lies in the plane of sweep and contributes no flux (1)

Question 12

Two flat coils lie side by side. When a switch in coil A's circuit is closed, a meter in coil B's circuit deflects briefly and then returns to zero, even though the current in coil A remains switched on. Explain these observations.

Accepted Answer

While the current in A grows, the flux it produces through B is changing, so an emf is induced in B and the meter deflects (1); once the current in A is steady, the flux through B is constant, so there is no rate of change of flux linkage and no emf (1).

Question 13

A rectangular coil moves at constant velocity into, through and out of a region of uniform magnetic field perpendicular to its plane. Explain why an emf is induced only while the coil enters and leaves the region, and how the two emfs compare.

Accepted Answer

Entering, the flux linkage increases at a steady rate, so a constant emf is induced (1); fully inside, the flux linkage is constant, so the emf is zero (1); leaving, the flux linkage falls at the same rate, so the emf has the same magnitude but the opposite sense (1).

Question 14

A coil of 100 turns and area 0.02 m² rotates at a frequency of 50 Hz in a uniform magnetic field of flux density 0.30 T. Calculate the peak emf induced in the coil.

Accepted Answer

ω = 2πf = 2π × 50 (1) peak emf = BANω = 0.30 × 0.02 × 100 × (2π × 50) (1) peak emf = 188 V (1)

Question 15

A conducting rod of length 0.20 m slides at a constant 3.0 m s⁻¹ along horizontal rails in a magnetic field of flux density 0.40 T. The circuit is completed by a 2.0 Ω resistor. Calculate the induced emf, the current in the circuit, and the force needed to keep the rod moving at constant speed.

Accepted Answer

emf = Blv = 0.40 × 0.20 × 3.0 = 0.24 V (1) I = emf/R = 0.24/2.0 = 0.12 A (1) Applied force balances the opposing magnetic force: F = BIl (1) F = 0.40 × 0.12 × 0.20 = 0.0096 N (1)

Question 16

A bar magnet is dropped north pole first through a coil connected to a meter. Use Lenz's law to explain the direction of the induced current as the magnet enters the coil, and state what happens to the current direction as the magnet leaves.

Accepted Answer

As the magnet enters, the flux linkage of the coil increases (1); the induced current flows so that the top face of the coil becomes a north pole, opposing the approaching north pole (1). As the magnet leaves, the flux falls, so the current reverses to attract the receding magnet (1).

Question 17

A generator must produce a peak emf of at least 50 V when its coil rotates at 50 Hz in a magnetic field of flux density 0.35 T. Two coils are available.
Coil P: 200 turns, area 2.5 × 10⁻³ m²
Coil Q: 350 turns, area 1.2 × 10⁻³ m²
Deduce which coil, if either, meets the requirement.

Accepted Answer

ω = 2πf = 2π × 50 = 314 rad s⁻¹ (1)
Peak emf of P = BANω = 0.35 × 2.5 × 10⁻³ × 200 × 314 = 55.0 V (1)
Peak emf of Q = 0.35 × 1.2 × 10⁻³ × 350 × 314 = 46.2 V (1)
Only P reaches 50 V, so coil P should be chosen (1)

Question 18

A coil rotates at a steady rate in a uniform magnetic field. Explain why the induced emf is zero at the instant the coil is face-on to the field, and a maximum when the coil is edge-on.

Accepted Answer

The emf equals the rate of change of flux linkage, not the flux linkage itself (1); face-on, the flux linkage is at its maximum but momentarily not changing, so the emf is zero (1); edge-on, the flux linkage is passing through zero at its greatest rate, so the emf is a maximum (1).

Question 19

A thin aluminium plate hangs from a pivot and swings between the poles of a strong magnet. Aluminium is not ferromagnetic, yet the plate comes to rest after very few swings. Explain fully why the plate stops so quickly and what happens to its energy.

Accepted Answer

As the plate swings, the flux through each region of the plate changes as it enters and leaves the field (1). By Faraday's law, emfs are induced in the plate (1), driving circulating (eddy) currents within the metal (1). By Lenz's law these currents flow so as to oppose the change producing them, so the forces on them oppose the plate's motion whichever way it swings (1). The opposing force decelerates the plate, so the oscillation dies away quickly (1). The plate's kinetic energy is transferred to internal energy in the aluminium by the eddy currents flowing through its resistance (1).