Specification: AQA A-level Physics (7408) 3.7.5.4 | Paper: 2 | Data and formulae booklet: given in the exam
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Magnetic flux and flux linkage: magnetic flux Φ = BA, where B is the flux density perpendicular to the area A; it is measured in webers (Wb). Flux linkage is NΦ, the flux through one turn multiplied by the number of turns, measured in Wb turns.
Faraday’s law: the induced emf is directly proportional to the rate of change of flux linkage, ε = N ΔΦ ÷ Δt.
Lenz’s law: the direction of the induced current (or emf) is always such that it opposes the change that produces it.
The equation
induced emf = number of turns × rate of change of flux ε = N ΔΦ ÷ Δt and, for a coil rotating in a uniform field, ε = BANω sin ωt
| Quantity | Symbol | Unit |
|---|---|---|
| induced emf | ε | volts (V) |
| number of turns on the coil | N | no unit |
| change in magnetic flux through one turn | ΔΦ | webers (Wb), equivalent to T m² |
| change in flux linkage | N ΔΦ | weber turns (Wb turns) |
| time taken for the change | Δt | seconds (s) |
| magnetic flux density | B | teslas (T) |
| area of the coil | A | square metres (m²) |
| angular speed of the coil | ω | radians per second (rad s⁻¹) |
Rearranged: N ΔΦ = εΔt | Δt = N ΔΦ ÷ ε | ΔΦ = ΔB × A when the field is perpendicular to the coil | peak emf ε0 = BANω, with ω = 2πf | Convert cm² to m² (× 10−4) and ms to s before you substitute. The area under an emf–time graph equals the change in flux linkage.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) The magnetic flux through each turn of a 200-turn coil increases steadily from zero to 3.0 × 10−4 Wb in 0.050 s. Calculate the emf induced in the coil.
Show solution
F: ε = N ΔΦ ÷ Δt
I: ε = 200 × 3.0 × 10−4 ÷ 0.050
F: ε is already the subject
A: 1.2 V
Q2 (Core) The flux linkage of a coil changes by 0.60 Wb turns in 0.15 s. Calculate the average emf induced in the coil.
Show solution
F: ε = N ΔΦ ÷ Δt
I: ε = 0.60 ÷ 0.15 (N ΔΦ is the change in flux linkage)
F: ε is already the subject
A: 4.0 V
Q3 (Core) A coil of 500 turns has an area of 4.0 cm². It is placed with its plane perpendicular to a uniform magnetic field of flux density 0.30 T. The field is switched off and falls steadily to zero in 20 ms. Calculate the average induced emf.
Show solution
Convert first: 4.0 cm² = 4.0 × 10−4 m²; 20 ms = 0.020 s
Step 1 – change in flux linkage
N ΔΦ = N × ΔB × A = 500 × 0.30 × 4.0 × 10−4 = 0.060 Wb turns
Step 2 – emf
F: ε = N ΔΦ ÷ Δt
I: ε = 0.060 ÷ 0.020
F: ε is already the subject
A: 3.0 V
Watch out: 1 cm² = 10−4 m², not 10−2 m². Square the conversion factor as well as the unit.
Q4 (Core) A coil of 150 turns and area 2.5 cm² is placed perpendicular to a magnetic field. The flux density increases steadily from zero to 0.80 T and the average induced emf is 0.12 V. Calculate the time taken for the field to increase.
Show solution
Convert first: 2.5 cm² = 2.5 × 10−4 m²
F: ε = N ΔΦ ÷ Δt
I: 0.12 = 150 × 0.80 × 2.5 × 10−4 ÷ Δt = 0.030 ÷ Δt
F: Δt = 0.030 ÷ 0.12
A: 0.25 s
Q5 (Exam) The north pole of a bar magnet is pushed into one end of a 400-turn coil connected to a sensitive voltmeter. The flux through each turn rises steadily from zero to 6.0 × 10−5 Wb in 0.30 s. (a) Calculate the average induced emf. (b) Use Lenz’s law to state the magnetic polarity of the end of the coil nearest the magnet while the magnet is moving in. (c) The magnet is then pulled out in 0.12 s. Calculate the new average emf and describe how the voltmeter reading differs.
Show solution
F: ε = N ΔΦ ÷ Δt
I: ε = 400 × 6.0 × 10−5 ÷ 0.30 = 0.024 ÷ 0.30
F: ε is already the subject
A: (a) 0.080 V
(b) Lenz’s law: the induced current must oppose the change that causes it, which is the north pole approaching. So the end nearest the magnet becomes a north pole, which repels the incoming magnet. Work must be done to push the magnet in, and this work becomes the electrical energy.
(c) ε = 0.024 ÷ 0.12 = 0.20 V. The reading is larger (the flux changes faster) and in the opposite direction, because the flux is now decreasing; the near end becomes a south pole, attracting the magnet back.
Watch out: “opposes the change” does not mean “opposes the magnet” in a fixed way. When the magnet leaves, the coil attracts it, opposing the decrease in flux.
Q6 (Exam) A simple ac generator has a rectangular coil of 150 turns and area 4.0 × 10−3 m². It rotates at 50 revolutions per second in a uniform magnetic field of flux density 0.12 T. (a) Calculate the peak emf, to 3 significant figures. (b) At t = 0 the plane of the coil is perpendicular to the field. Calculate the emf at t = 2.0 ms, to 3 significant figures.
Show solution
Convert first: ω = 2πf = 2π × 50 = 314.1… rad s⁻¹; 2.0 ms = 2.0 × 10−3 s
F: ε0 = BANω
I: ε0 = 0.12 × 4.0 × 10−3 × 150 × 314.1…
F: ε0 is already the subject
A: (a) 22.61… V = 22.6 V (3 s.f.)
(b) ε = BANω sin ωt; ωt = 314.1… × 2.0 × 10−3 = 0.6283… rad; sin 0.6283… = 0.5877…
ε = 22.61… × 0.5877… = 13.29… V = 13.3 V (3 s.f.)
Watch out: ωt is in radians. With your calculator in degree mode you get sin 0.628° and an emf of about 0.25 V.
Q7 (Exam) A graph shows the emf induced in a 250-turn coil of area 5.0 cm² as the magnetic field through it changes. The emf rises steadily from 0 to 0.40 V in the first 0.20 s, then falls steadily back to 0 at t = 0.50 s. The field is perpendicular to the coil and starts at zero. Calculate (a) the change in flux linkage, (b) the change in flux through each turn and (c) the final flux density.
Show solution
Convert first: 5.0 cm² = 5.0 × 10−4 m²
Step 1 – (a) area under the emf–time graph
the graph is a triangle, so area = ½ × base × height = ½ × 0.50 × 0.40 = 0.10 V s
A: change in flux linkage N ΔΦ = 0.10 Wb turns
Step 2 – (b) flux per turn
ΔΦ = 0.10 ÷ 250 = 4.0 × 10−4 Wb
Step 3 – (c) flux density
F: Φ = BA
I: 4.0 × 10−4 = B × 5.0 × 10−4
F: B = 4.0 × 10−4 ÷ 5.0 × 10−4
A: 0.80 T
Watch out: the area under the graph is the change in flux linkage (NΔΦ), not the flux. Divide by N to get the flux through one turn.
Q8 (Exam) A generator coil of 200 turns and area 2.5 × 10−3 m² rotates at 25 revolutions per second in a uniform magnetic field. The peak emf is 6.0 V. (a) Calculate the magnetic flux density, to 2 significant figures. (b) State the peak emf and the frequency of the output if the rotation rate is doubled.
Show solution
Convert first: ω = 2πf = 2π × 25 = 157.0… rad s⁻¹
F: ε0 = BANω
I: 6.0 = B × 2.5 × 10−3 × 200 × 157.0…
F: B = 6.0 ÷ (2.5 × 10−3 × 200 × 157.0…) = 6.0 ÷ 78.53…
A: (a) 0.07639… T = 0.076 T (2 s.f.)
(b) ε0 ∝ ω, so doubling the rotation rate doubles the peak emf to 12 V, and the frequency doubles to 50 Hz.
Watch out: use ω = 2πf, not f. Putting 25 in place of ω makes B 2π times too big.
Q9 (Stretch) A metal rod of length 0.25 m slides at a constant 3.0 m s⁻¹ along two horizontal rails, at right angles to a vertical uniform magnetic field of flux density 0.40 T. The rod and rails form a circuit of total resistance 1.5 Ω. (a) Show that the induced emf is ε = Blv and calculate it. (b) Calculate the current and the magnetic force on the rod. (c) Show that the power needed to keep the rod moving equals the electrical power produced.
Show solution
Step 1 – (a) emf
in time Δt the rod sweeps out an area ΔA = lvΔt, so the flux cut is ΔΦ = BlvΔt. With N = 1, ε = ΔΦ ÷ Δt = Blv
I: ε = 0.40 × 0.25 × 3.0
A: 0.30 V
Step 2 – (b) current and force
I = ε ÷ R = 0.30 ÷ 1.5 = 0.20 A
F = BIl = 0.40 × 0.20 × 0.25 = 0.020 N, opposing the motion (Lenz’s law)
Step 3 – (c) power
mechanical power = Fv = 0.020 × 3.0 = 0.060 W
electrical power = I²R = 0.20² × 1.5 = 0.060 W
A: both are 0.060 W, so the work done pushing the rod against the magnetic force becomes electrical energy
Watch out: if the induced current helped the motion, the rod would speed up with no energy input. Lenz’s law is conservation of energy.
Q10 (Stretch) A coil of 120 turns and area 3.0 × 10−3 m² rotates at 25 revolutions per second in a uniform field of flux density 0.20 T. (a) Calculate the maximum flux linkage. (b) Calculate the peak emf, to 3 significant figures. (c) The coil turns through a quarter of a revolution from the position of maximum flux linkage. Use the area under the emf–time graph to find the average emf during this quarter turn, and express it as a fraction of the peak emf, to 3 significant figures.
Show solution
Step 1 – (a) maximum flux linkage
NΦ = BAN = 0.20 × 3.0 × 10−3 × 120
A: 0.072 Wb turns
Step 2 – (b) peak emf
ω = 2π × 25 = 157.0… rad s⁻¹; ε0 = BANω = 0.072 × 157.0… = 11.30… V
A: 11.3 V (3 s.f.)
Step 3 – (c) average emf over a quarter turn
the flux linkage falls from 0.072 Wb turns to zero, so the area under the emf–time graph is 0.072 Wb turns
time for a quarter turn = 1 ÷ (4 × 25) = 0.010 s; average emf = 0.072 ÷ 0.010 = 7.2 V
7.2 ÷ 11.30… = 0.637 (3 s.f.), which is 2 ÷ π
Watch out: the average emf over a quarter turn is not ½ε0. The emf follows a sine curve, not a straight line.
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