The problems are on the faraday's law worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Faraday's law states that a changing magnetic flux through a coil induces an emf equal to −N dΦ/dt, with its direction set by Lenz's law. These worked answers correct the 20 problems of the AP Physics C worksheet on faraday's law, from flux through a loop to timing in an LR circuit: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result in bold. Check your method as well as your final value, because free-response marks reward the reasoning.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 13
Problem 1 Flux through a loop
1. Φ = BA = 0.30 × 0.020 = 6.0 × 10⁻³ Wb.2. Φ = BA cos 60° = 6.0 × 10⁻³ × 0.50 = 3.0 × 10⁻³ Wb.
3. 1 Wb = 1 T·m².
Problem 2 Emf from a changing flux
1. ΔΦ = 0.5 − 2.0 = −1.5 mWb = −1.5 × 10⁻³ Wb.2. |ε| = N|ΔΦ|/Δt = 50 × 1.5 × 10⁻³ / 0.10 = 0.75 V.
3. 1 Wb = 1 V·s, so 1 Wb/s = 1 V.
Problem 3 Direction of the induced current
1. The induced field opposes the growing inward field, so it points out of the page.2. An out-of-page field inside the loop needs an anticlockwise current (right-hand rule).
3. If the field decreased, the loop would try to keep the inward flux, so the induced field would point into the page and the current would be clockwise.
Problem 4 Motional emf of a rod
1. ε = Bℓv = 0.25 × 0.30 × 4.0 = 0.30 V.2. I = ε/R = 0.30/0.50 = 0.60 A.
Problem 5 Inductance of a coil
1. NΦ = 400 × 2.5 × 10⁻⁵ = 1.0 × 10⁻² Wb.2. L = NΦ/I = 1.0 × 10⁻²/0.50 = 0.020 H = 20 mH.
Problem 6 Self-induced emf
1. dI/dt = 2.0/0.050 = 40 A/s.2. |ε| = L dI/dt = 0.040 × 40 = 1.6 V.
Problem 7 Time constant
1. τ = L/R = 0.30/60 = 5.0 × 10⁻³ s = 5.0 ms.2. 5τ = 25 ms (the current is then about 99.3 % of its final value).
Problem 8 Energy in an inductor
1. U = ½LI² = ½ × 0.25 × 3.0² = 1.125 J ≈ 1.1 J.2. U depends on I², so doubling I gives a factor of 4 (4.5 J).
Problem 9 Inductance of a solenoid
1. A = πr² = π(0.015)² = 7.07 × 10⁻⁴ m².2. L = μ₀N²A/ℓ = (4π × 10⁻⁷)(1200)²(7.07 × 10⁻⁴)/0.60 = 2.13 × 10⁻³ H ≈ 2.1 mH.
3. L ∝ N², so doubling N multiplies L by 4.
Problem 10 Alternating emf from a changing field
1. Φ = BA = 0.40 × 0.010 sin(100πt) = 4.0 × 10⁻³ sin(100πt) Wb.2. ε = −N dΦ/dt = −100 × 4.0 × 10⁻³ × 100π cos(100πt) = −40π cos(100πt) V.
3. εmax = 40π = 125.7 V ≈ 1.3 × 10² V.
Problem 11 Reading a current graph
1. The curve levels off at 0.60 A.2. 63 % of 0.60 A is 0.38 A, reached at 12 ms, so τ = 12 ms.
3. R = ε/I = 12/0.60 = 20 Ω. L = τR = 0.012 × 20 = 0.24 H, so L = 0.24 H.
Problem 12 Growth of current
1. τ = L/R = 0.80/40 = 0.020 s.2. I∞ = 24/40 = 0.60 A. At 40 ms = 2τ, I = 0.60 (1 − e⁻²) = 0.60 × 0.8647 = 0.52 A.
Problem 13 Decay of current
1. I₀ = ε/R = 24/40 = 0.60 A.2. 30 ms = 1.5τ, so I = 0.60 e−1.5 = 0.60 × 0.2231 = 0.13 A.
Problem 14 Initial rate of change
1. The current is zero, so VR = 0.2. The inductor carries the whole 24 V: dI/dt = ε/L = 24/0.80 = 30 A/s.
3. At steady state dI/dt = 0, so the inductor has no emf and only R limits the current: I = ε/R, independent of L.
Problem 15 Frequency of an LC circuit
1. ω = 1/√(LC) = 1/√(2.0 × 10⁻³ × 50 × 10⁻⁶) = 1/√(1.0 × 10⁻⁷) = 3.16 × 10³ rad/s ≈ 3.2 × 10³ rad/s.2. f = ω/2π = 503 Hz ≈ 5.0 × 10² Hz.
3. ω ∝ 1/√C, so multiplying C by 4 halves the frequency.
Problem 16 Terminal speed of a falling rod
1. The emf is Bℓv, the current is Bℓv/R, so F = BIℓ = B²ℓ²v/R (upwards).2. Set B²ℓ²v/R = mg: v = mgR/(B²ℓ²) = (0.050 × 9.81 × 0.20)/(0.25 × 0.16) = 0.0981/0.040 = 2.45 m/s ≈ 2.5 m/s.
3. At terminal speed the kinetic energy is constant, so all the gravitational energy lost heats the resistor (mgv = I²R).
Problem 17 Decaying field through a loop
1. A = π(0.050)² = 7.85 × 10⁻³ m².2. dB/dt = −(0.80/0.40) e−t/0.40, so at t = 0 it is −2.0 T/s. |ε| = A|dB/dt| = 7.85 × 10⁻³ × 2.0 = 1.57 × 10⁻² V ≈ 16 mV.
3. At t = 0.40 s multiply by e⁻¹ = 0.368: |ε| = 15.7 × 0.368 = 5.8 mV, so 5.8 mV.
Problem 18 Energy dissipated while the current decays
1. U₀ = ½LI₀² = ½ × 0.80 × 0.60² = 0.144 J.2. U ∝ I², so U(τ) = U₀e⁻² = 0.144 × 0.1353 = 0.0195 J ≈ 0.019 J.
3. Dissipated = 0.144 − 0.0195 = 0.1245 J ≈ 0.12 J.
Problem 19 Charge and current in an LC circuit
1. U = ½CV² = ½ × 1.0 × 10⁻⁴ × 12² = 7.2 × 10⁻³ J.2. Energy conservation: ½LImax² = ½CV², so Imax = V√(C/L) = 12 × √(1.0 × 10⁻⁴/0.025) = 12 × 0.0632 = 0.76 A.
3. T = 2π√(LC) = 2π√(0.025 × 1.0 × 10⁻⁴) = 2π × 1.58 × 10⁻³ = 9.9 ms.
4. At I = Imax/2 the inductor holds ¼ of the energy, so the capacitor holds ¾: q²/2C = ¾ Q²/2C, q = (√3/2)Q. With Q = CV = 1.2 × 10⁻³ C, q = 0.866 × 1.2 × 10⁻³ = 1.04 × 10⁻³ C ≈ 1.0 × 10⁻³ C.
Problem 20 Timing in an LR circuit
1. I = I∞(1 − e−t/τ) = 0.9 I∞ gives t = τ ln 10. With τ = 0.50/25 = 0.020 s, t = 0.020 × 2.303 = 0.046 s = 46 ms.2. U ∝ I², so half the final energy means I = I∞/√2 = 0.7071 I∞. Then e−t/τ = 0.2929 and t = τ ln(1/0.2929) = 0.020 × 1.228 = 0.0246 s ≈ 25 ms.
3. Half the energy needs only 71 % of the final current, and the current rises fastest at the start, so that value is reached sooner than 90 %.
