A long straight wire of a circular cross-section with radius ' a ' carries a steady current I . The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by

Solution

Physics sir · 4 steps

Draw an Amperian circle inside and outside: inside, B grows with r; outside, it falls as 1/r.

1. Inside the wire. Take a circle of radius r < a. Since the current is uniform, it encloses only the fraction r²/a² of I. Ampere's law gives B · 2π r = μ₀ I r²/a², so B rises linearly from zero at the centre.
2. Outside the wire. For r > a the circle encloses all of I. Then B · 2π r = μ₀ I, so B falls off as 1/r.

   B = (μ₀ I r)/(2π a²), r ≤ a ; (μ₀ I)/(2π r), r ≥ a

3. At the surface. Both expressions give (μ₀ I)/(2π a) at r = a. So the graph is continuous and peaks at the surface. There is no jump.
4. So. The plot is a straight line from the origin up to a peak at r = a, then a 1/r fall. That is option (b).

The idea to keep. For a solid wire with uniform current, B is zero at the axis, grows as r inside, peaks at the surface and decays as 1/r outside. Only the current enclosed by your loop counts.

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