NEET 2017 · PhysicsCharge flowPrevious Year Question
A long solenoid of diameter 0.1 m has 2 × 10⁴ turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10 π² ohm, the total charge flowing through the coil during this time is
Answer: (C) 32 µC. Answer: (c) Solution: Solenoid field: B = µ₀nI, so ΔB = µ₀n·ΔI = (4π × 10⁻⁷)(2 × 10⁴)(4) = 32π × 10⁻³ T.
- A.32π µC
- B.16 µC
- C.32 µC✓
- D.16π µC
Correct Answer
(C) 32 µC
Solution & Explanation
Answer: (c) Solution: Solenoid field: B = µ₀nI, so ΔB = µ₀n·ΔI = (4π × 10⁻⁷)(2 × 10⁴)(4) = 32π × 10⁻³ T. Flux linkage change in the coil: ΔΦ_link = N·ΔB·(πa²) = 100 × (32π × 10⁻³) × π(0.01)² = 3.2π² × 10⁻⁴ Wb. Charge q = ΔΦ_link / R = (3.2π² × 10⁻⁴) / (10π²) = 3.2 × 10⁻⁵ C = 32 µC. So the answer is C (32 µC).
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