A long solenoid of diameter 0.1 m has 2 × 10⁴ turns per metre. A 100-turn coil of radius 0.01 m and resistance 10π² Ω is placed coaxially inside it. The solenoid current falls uniformly from 4 A to 0 A in 0.05 s. Find the total induced charge passing through the coil.
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Question

A long solenoid of diameter 0.1 m has 2 × 10⁴ turns per metre. A 100-turn coil of radius 0.01 m and resistance 10π² Ω is placed coaxially inside it. The solenoid current falls uniformly from 4 A to 0 A in 0.05 s. Find the total induced charge passing through the coil.

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G1 Physics · Accepted Answer

Step 1: The field inside a solenoid is B = μ₀ nI. Current change ΔI = 0−4 = −4 A. Step 2: Coil flux linkage change is |Δ(NΦ)| = N A μ₀n|ΔI|, where N=100, A=π(0.01)² m², n=2×10⁴ m⁻¹. Step 3: Total induced charge Q=∫i dt = |Δ(NΦ)|/R, independent of the duration for constant resistance. Step 4: Q = [100·π(0.01)²·(4π×10⁻⁷)·(2×10⁴)·4]/(10π²) = 3.2×10⁻⁵ C = 32 μC. Final answer: (C) 32 μC.