Consider a long straight wire of a circular cross-section (radius a ) carrying a steady current I . The current is uniformly distributed across this cross-section. The distances from the centre of the wire’s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be :

Solution

Physics sir · 4 steps

Find where B peaks first, at the surface, then halve it once inside and once outside the wire.

1. Field inside. Take an Amperian circle of radius r < a. It encloses current I r²/a², so B · 2π r = μ₀ I r²/a², which gives B = (μ₀ I r)/(2π a²). The field grows linearly with r.
2. Field outside. For r > a the circle encloses all of I, so B = (μ₀ I)/(2π r). The field falls off as 1/r.

   Bₘₐₓ = B(a) = (μ₀ I)/(2π a)

3. Halve it. Inside: (μ₀ I r)/(2π a²) = 1/2 · (μ₀ I)/(2π a) gives r = a/2. Outside: (μ₀ I)/(2π r) = 1/2 · (μ₀ I)/(2π a) gives r = 2a.
4. So. The field is half its maximum at [a/2, 2a], which is option (c).

The idea to keep. For a uniformly filled wire, B rises as r inside and falls as 1/r outside, peaking at the surface. Inside, a fraction of Bₘₐₓ scales r directly; outside, it scales r inversely.

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