A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a/2 and 2 a from axis of the wire is :

Solution

Physics sir · 3 steps

One point is inside the wire and one is outside, so use a different Ampere loop for each.

1. Inside the wire. A circular Amperian loop of radius r < a encloses only part of the current: I_(enc) = Ir²/a². Ampere's law gives B · 2π r = μ₀ I r²/a², so B = (μ₀ I r)/(2π a²).
2. Outside the wire. A loop of radius r > a encloses the full current I. So B · 2π r = μ₀ I, which gives B = (μ₀ I)/(2π r).

   B(a/2) = (μ₀ I (a/2))/(2π a²) = (μ₀ I)/(4π a), B(2a) = (μ₀ I)/(2π (2a)) = (μ₀ I)/(4π a)

3. So. Both fields equal (μ₀ I)/(4π a), so the ratio is 1 : 1, option (c).

The idea to keep. For a solid wire with uniform current, B rises as r inside, peaks at the surface and falls as 1/r outside. Always check which side of the surface each point lies on.

Answer from the NTA final answer key. Solution by PrepWiser.