1. 1/4 2. 1/2 3. 1 4. 4

Answer: 3. 1 Key idea: use Ampere's law. Inside the wire, B=μ₀Ir/2π a² ; outside, B=μ₀I/2π r . At r=a/2 : B=μ₀I/4π a . At r=2a : B'=μ₀I/4π a . Ratio B/B'=1 .

Topic: Ampere's law: straight wire and solenoid· Magnetic Effects of Current and Magnetism

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 :

1. \(\frac{1}{4}\)

2. \(\frac{1}{2}\)

3. 1

4. 4

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Answer: 3. 1

Key idea: use Ampere's law. Inside the wire, \(B=\dfrac{\mu_0Ir}{2\pi a^2}\); outside, \(B=\dfrac{\mu_0I}{2\pi r}\).

At \(r=a/2\): \(B=\dfrac{\mu_0I}{4\pi a}\).
At \(r=2a\): \(B'=\dfrac{\mu_0I}{4\pi a}\).
Ratio \(B/B'=1\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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 :

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