Faraday's law worksheet answers for AP Physics 2 (free PDF download)

The problems are on the faraday's law worksheet. Try them on your own first, then compare with the answers below.

Quick answer: Electromagnetic induction produces an emf whenever the magnetic flux Φ = BA cos θ through a loop changes. These worked answers correct the 20 problems of the AP Physics 2 worksheet on faraday's law, from flux through a loop to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.

Faraday’s law worksheet answers for AP Physics 2 (free PDF download)

Each answer gives the relationship, the substitution and the final result in bold. For direction questions, check that you identified the change in flux before choosing the induced field.

Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 12

Problem 1 Flux through a loop

1. Φ = BA cos 0° = 0.50 × 0.020 = 1.0 × 10⁻² Wb.

2. A perpendicular field is parallel to the normal, so θ = 0°.

Problem 2 Flux at different angles

BA = 0.30 × 0.040 = 0.012 Wb.

1. 1.2 × 10⁻² Wb.

2. 0.012 × cos 60° = 6.0 × 10⁻³ Wb.

3. 0.012 × cos 90° = 0 Wb.

Problem 3 Ways to change flux

Change the field strength B, change the area A of the loop in the field, or change the angle θ by rotating the loop (or the field).

Problem 4 Average emf

1. ΔΦ = 0.060 − 0.020 = 0.040 Wb.

2. ε = ΔΦ/Δt = 0.040 / 0.40 = 0.10 V.

Problem 5 Approaching magnet

1. The flux through the loop increases as the field gets stronger.

2. The induced current is counterclockwise as seen from the magnet.

3. The loop must oppose the increase, so its face towards the magnet becomes a north pole and repels the incoming north pole.

Problem 6 Rod emf

ε = BLv = 0.50 × 0.40 × 3.0 = 0.60 V.

Problem 7 Units of flux

1. Φ = BA, so the unit is T·m². By definition this is the weber: 1 Wb = 1 T·m².

2. 25 cm² = 25 × 10⁻⁴ m² = 2.5 × 10⁻³ m².

3. From ε = ΔΦ/Δt, Wb/s is the volt.

Problem 8 Coil in a rising field

ΔΦ = 0.60 × 4.0 × 10⁻³ = 2.4 × 10⁻³ Wb. ε = NΔΦ/Δt = 150 × 2.4 × 10⁻³ / 0.20 = 1.8 V.

Problem 9 Flipping a coil

1. The flux changes from +BA to −BA: ΔΦ = 2BA = 2 × 0.20 × 0.010 = 4.0 × 10⁻³ Wb.

2. ε = NΔΦ/Δt = 50 × 4.0 × 10⁻³ / 0.10 = 2.0 V.

Problem 10 Rod on rails

1. ε = BLv = 0.40 × 0.30 × 5.0 = 0.60 V.

2. I = ε/R = 0.60 / 3.0 = 0.20 A.

3. F = BIL = 0.40 × 0.20 × 0.30 = 0.024 N.

4. P = Fv = 0.024 × 5.0 = 0.12 W, which equals εI = 0.12 W.

Problem 11 Reading a flux graph

1. The slope is 12 mWb / 2 s = 6.0 mWb/s. ε = NΔΦ/Δt = 20 × 6.0 × 10⁻³ = 0.12 V in magnitude.

2. The flux is constant, so ε = 0.

3. The slope is −12 mWb / 2 s = −6.0 mWb/s, so the magnitude is again 0.12 V, but with the opposite sign (opposite current direction).

Problem 12 Direction of the current

1. Increasing flux into the page: the loop opposes it with a field out of the page, so the current is counterclockwise.

2. Decreasing flux into the page: the loop supports it with a field into the page, so the current is clockwise.

3. Lenz’s law: the induced field always opposes the change in flux, not the field itself.

Problem 13 Falling magnet

1. ΔE = mgh = 0.050 × 9.81 × 0.80 = 0.39 J.

2. The moving magnet induces eddy currents in the copper. By Lenz’s law these currents create a field that opposes the motion, so there is an upward force on the magnet. The lost gravitational energy becomes thermal energy in the tube.