NEET 2016 Phase 1 · PhysicsSelf-inductionPrevious Year Question
A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4 × 10⁻³ Wb. The self-inductance of the solenoid is:
Answer: (D) 1 H. Answer: (d) Solution: Self-inductance links total flux to current: L = NΦ/I, where NΦ is the total flux linkage.
- A.4 H
- B.3 H
- C.2 H
- D.1 H✓
Correct Answer
(D) 1 H
Solution & Explanation
Answer: (d) Solution: Self-inductance links total flux to current: L = NΦ/I, where NΦ is the total flux linkage. Here N = 1000, flux per turn Φ = 4 × 10⁻³ Wb, current I = 4 A. L = (1000 × 4 × 10⁻³) / 4 = 4 / 4 = 1 H. So the answer is D (1 H).
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