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The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 30 — Inductance

Problem sheet · 13 points · 3 extra credit

Chapter 30 — Inductance

Core: 13 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

  1. 1. (1 pt •) Two coils are wound on a common core. When the current in coil 1 changes at the rate di1/dt=850 A/sdi_1/dt = 850\,\text{A/s}, an emf of magnitude 2.72 V2.72\,\text{V} is induced in coil 2.

    (a) Find the mutual inductance MM of the pair.

    (b) When a steady current i1=6.0 Ai_1 = 6.0\,\text{A} flows in coil 1, what is the total flux linkage N2ΦB2N_2\Phi_{B2} through coil 2?

  2. 2. (2 pts ••) A long air-core solenoid has N=400N = 400 turns wound over a length l=0.25 ml = 0.25\,\text{m}, with a circular cross-section of radius r=2.0 cmr = 2.0\,\text{cm}.

    (a) Compute its self-inductance LL.

    (b) The current is i=5.0 Ai = 5.0\,\text{A} and steady. Find the magnetic field BB inside the solenoid.

    (c) Using the magnetic energy density u=B2/2μ0u = B^2/2\mu_0 and the solenoid's interior volume, find the total stored energy. Check that it equals 12Li2\tfrac12 L i^2.

  3. 3. (2 pts ••) A toroidal coil has N=500N = 500 turns, a rectangular cross-section of height h=1.5 cmh = 1.5\,\text{cm}, inner radius a=3.0 cma = 3.0\,\text{cm}, and outer radius b=5.0 cmb = 5.0\,\text{cm}. Its self-inductance is

    L=μ0N2h2πln⁡ ⁣ba.L = \frac{\mu_0 N^2 h}{2\pi}\ln\!\frac{b}{a}.

    (a) Evaluate LL.

    (b) An inductor is useful because it opposes changes in current. If the current in this toroid is ramped from 00 to 3.0 A3.0\,\text{A} in 4.0 ms4.0\,\text{ms}, what is the magnitude of the self-induced emf?

  4. 4. (2 pts ••) An inductor of L=0.15 HL = 0.15\,\text{H} carries a current that increases uniformly from 00 to 4.0 A4.0\,\text{A} in 20 ms20\,\text{ms}.

    (a) Find the magnitude of the self-induced emf during the ramp, using E=−L di/dt\mathcal{E} = -L\,di/dt.

    (b) Find the energy stored in the inductor once the current reaches 4.0 A4.0\,\text{A}.

  5. 5. (2 pts ••) A series R–L circuit has L=0.30 HL = 0.30\,\text{H}, R=25 ΩR = 25\,\Omega, and a 12 V12\,\text{V} battery. At t=0t = 0 the switch is closed with zero initial current.

    (a) Find the time constant τ\tau and the final (steady-state) current.

    (b) Find the current after one time constant has elapsed.

    (c) How long after closing does the current reach 90%90\% of its final value?

  6. 6. (2 pts ••) In the circuit of Problem 5, the current has reached its steady value I0=0.48 AI_0 = 0.48\,\text{A} when, at a new t=0t = 0, the battery is removed and the inductor discharges through the same resistor.

    (a) Write the current i(t)i(t) during the decay.

    (b) Find the time for the current to fall to half of I0I_0.

    (c) A student claims "the energy just disappears when the battery is removed." In one or two sentences, say where the stored magnetic energy actually goes.

  7. 7. (2 pts ••) An L–C circuit has L=0.40 HL = 0.40\,\text{H} and C=8.0 μFC = 8.0\,\mu\text{F}. The capacitor is initially charged to V0=6.0 VV_0 = 6.0\,\text{V} and released at t=0t = 0 with no current.

    (a) Find the angular frequency ω\omega and period TT of the oscillation.

    (b) Find the maximum charge Q0Q_0 on the capacitor and the maximum current ImaxI_\text{max}.

    (c) At the instant the charge on the capacitor is q=Q0/2q = Q_0/2, what fraction of the total energy is stored in the capacitor? (In a real L–R–C series circuit the resistance would slowly drain this energy, damping the oscillation — but ignore resistance here.)

  8. Extra Credit (3 pts •••)

    Consider the series R–L circuit of Problem 5: a battery of emf E\mathcal{E}, resistor RR, and inductor LL in series, switch closed at t=0t = 0 with i(0)=0i(0) = 0.

    (a) Apply the loop rule to write the first-order differential equation governing i(t)i(t), being careful with the sign of the inductor's voltage.

    (b) Solve this differential equation from first principles (by separating variables and integrating) to obtain the closed-form current growth i(t)i(t). Show every step.

    (c) From your i(t)i(t), derive the power delivered by the battery, the power dissipated in the resistor, and the rate at which energy is stored in the inductor, and show explicitly that these three balance at every instant (conservation of energy).