The problems are on the electromagnetic induction worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Electromagnetic induction is the creation of an emf in a circuit when the magnetic flux linked with it changes. These worked answers correct the 20 problems of the Physics 11-12 worksheet on electromagnetic induction, from flux through a coil to spot the error: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit. For direction questions, always say what is changing, then state how the induced current must oppose it. Small rounding differences in the last digit are acceptable.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB D.4
Problem 2 Tilted coil
Φ = BA cos θ = 0.50 × 0.040 × cos 60° = 0.020 × 0.50 = 0.010 Wb.Problem 3 Single loop emf
ε = ΔΦ/Δt = (8.0 − 2.0) × 10⁻³ / 0.30 = 20 mV (0.020 V).Problem 4 More turns
1. ε = NΔΦ/Δt = 150 × 6.0 × 10⁻³ / 0.30 = 3.0 V.2. The emf is proportional to N, because each turn contributes the same emf in series.
Problem 5 Aircraft wing
ε = BLv = 4.0 × 10⁻⁵ × 35 × 250 = 0.35 V. The emf is small and cannot be used to light a bulb.Problem 6 Which pole forms?
1. A north pole forms on the near end.2. The flux through the coil from the magnet is increasing. The induced current produces a field that opposes this increase, so the near end of the coil repels the approaching north pole.
Problem 7 Unit conversion
1. A = 0.12 × 0.080 = 0.0096 m² (9.6 × 10⁻³ m²).2. 35 mT = 0.035 T.
3. Φ = BA = 0.035 × 9.6 × 10⁻³ = 3.36 × 10⁻⁴ Wb = 336 µWb.
Problem 8 Coil in a fading field
1. ε = N A ΔB/Δt = 500 × 4.0 × 10⁻³ × 0.60 / 0.20 = 6.0 V.2. I = ε/R = 6.0 / 25 = 0.24 A.
Problem 9 Sliding bar and force
1. ε = BLv = 0.50 × 0.40 × 2.5 = 0.50 V.2. I = ε/R = 0.50 / 0.50 = 1.0 A.
3. F = BIL = 0.50 × 1.0 × 0.40 = 0.20 N.
Problem 10 Reading a flux graph
1. ε = −ΔΦ/Δt = −4.0 × 10⁻³ / 0.2 = −20 mV (magnitude 20 mV).2. The flux is constant, so ε = 0.
3. ε = −(0 − 4.0 × 10⁻³) / 0.1 = +40 mV. It is twice as large and has the opposite sign, because the flux falls instead of rises.
Problem 11 Peak emf of a generator
1. ω = 2πf = 2π × 50 = 314.2 rad/s.2. ε₀ = NBAω = 100 × 0.15 × 0.020 × 314.2 = 94.2 V.
3. ε = ε₀ sin ωt = 94.2 × sin(314.2 × 0.0020) = 94.2 × sin(0.628 rad) = 55.4 V.
Problem 12 Bicycle dynamo
1. ε₀ = NBAω is proportional to f, so the peak emf doubles to 24 V.2. The emf frequency equals the rotation frequency, 16 Hz.
3. The flux changes faster at higher speed, so the emf and the power in the lamp are larger.
Problem 13 Magnet in a metal tube
1. As the magnet falls, the flux through each ring of the tube changes, inducing eddy currents. By Lenz’s law these currents produce fields that oppose the motion, giving an upward force that nearly balances the weight.2. The energy is transferred to the tube as thermal energy by the eddy currents.
Problem 14 Pulling a coil faster
1. ε = BLv, so the emf is doubled.2. The current doubles, so F = BIL doubles.
3. P = Fv: force doubles and speed doubles, so the power is four times greater.
Problem 15 Direction of induced current
1. The area of the loop and the flux out of the page both increase. The induced current must create a field into the page inside the loop, which means a clockwise current.2. Clockwise means the current flows down the rod; with B out of the page, F = IL × B points to the left, opposing the motion.
Problem 16 Terminal speed of a falling rod
1. ε = BLv, I = BLv/R, F = BIL = B²L²v/R.2. At terminal speed B²L²v/R = mg, so v = mgR/(B²L²) = 0.020 × 9.81 × 0.50 / (0.16 × 0.090) = 6.8 m/s.
3. As v rises, the upward magnetic force grows until it equals the weight; the resultant force is then zero.
Problem 17 Heating by induction
1. ε = A ΔB/Δt = 0.15 × 0.20 × 0.50 = 0.015 V.2. I = ε/R = 0.015 / 0.30 = 0.05 A.
3. E = I²Rt = 0.05² × 0.30 × 4.0 = 3.0 mJ (3.0 × 10⁻³ J).
Problem 18 Spinning metal disc
1. Area per second = πr² × f = π × 0.15² × 10 = 0.707 m²/s.2. ε = B × (area swept per second) = 0.20 × 0.707 = 0.141 V, about 0.14 V.
Problem 19 Search coil measurement
1. The flux changes from +BA to −BA, so ΔΦ = 2BA per turn.2. ε = N(2BA)/Δt, so B = εΔt/(2NA) = 0.45 × 0.040 / (2 × 60 × 2.0 × 10⁻⁴) = 0.018 / 0.024 = 0.75 T.
Problem 20 Spot the error
1. The student used the final field instead of the change in field: ΔB = 0.60 − 0.20 = 0.40 T.2. ε = NAΔB/Δt = 30 × 0.010 × 0.40 / 0.10 = 1.2 V.
3. 1.8 V is correct for a field rising from 0 to 0.60 T in 0.10 s.
