NEET 2016 Phase 1 · PhysicsAmpere thick wirePrevious Year Question
A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B′ at radial distances a/2 and 2a respectively, from the axis of the wire, is:
Answer: (C) 1. Answer: (C) Solution: Inside (r = a/2 < a): by Ampère's law B = μ₀Ir/2πa² = μ₀I(a/2)/(2πa²) = μ₀I/4πa.
- A.1/4
- B.1/2
- C.1✓
- D.4
Correct Answer
(C) 1
Solution & Explanation
Answer: (C) Solution: Inside (r = a/2 < a): by Ampère's law B = μ₀Ir/2πa² = μ₀I(a/2)/(2πa²) = μ₀I/4πa. Outside (r = 2a > a): B′ = μ₀I/2πr = μ₀I/2π(2a) = μ₀I/4πa. Ratio B : B′ = 1. → option C.
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