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NEET 2016 (Phase 1) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from NEET 2016 (Phase 1)

Q53NEET 2016 · PHASE 1Easy

Ampere's Law, Cylinders, Solenoid & Toroid

A long straight wire of radius aa carries a steady current II. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B', at radial distance a2\frac{a}{2} and 2aa respectively, from the axis of the wire is

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Answer: (a)

  1. Inside at r=a2r = \frac{a}{2}: B=μ0I(a/2)2πa2=μ0I4πaB = \frac{\mu_0 I (a/2)}{2\pi a^2} = \frac{\mu_0 I}{4\pi a}.
  2. Outside at r=2ar = 2a: B′=μ0I2π(2a)=μ0I4πaB' = \frac{\mu_0 I}{2\pi (2a)} = \frac{\mu_0 I}{4\pi a}.
  3. BB′=1\frac{B}{B'} = 1
  4. So the answer is (a) 1.
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