JEE Main 2023 · 6 April · morning · Single correct · +4 / −1
A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is
- (a) uniform in the region r < a and inversely proportional to distance r from the axis, in the region r > a
- (b) zero in the region r < a and inversely proportional to r in the region r > a
- (c) directly proportional to r in the region r < a and inversely proportional to r in the region r > a
- (d) inversely proportional to r in the region r < a and uniform throughout in the region r > a
Answer and solution
Answer (c) directly proportional to r in the region r < a and inversely proportional to r…
Solution
Physics sir · 4 steps
Draw an Amperian circle of radius r and ask one thing: how much current does it enclose?
- Set up the loop. Take a circle of radius r centred on the axis. By symmetry B has the same size all around it and points along it, so ∮ B⃗· dl⃗ = B(2π r), which must equal μ₀ I_(enc).
- Inside the wire. For r < a the current is spread uniformly, so the circle encloses only the area fraction (π r²)/(π a²) of it. That gives I_(enc) = Ir²/a².
B(2π r) = μ₀ I r²/a² ⇒ B = (μ₀ I r)/(2π a²) ∝ r
- Outside the wire. For r > a the circle encloses all of I, so B(2π r) = μ₀ I and B = (μ₀ I)/(2π r) ∝ 1/r. Both formulas give (μ₀ I)/(2π a) at r = a, so the field rises linearly from zero on the axis, peaks at the surface, then falls off.
- So. The field is directly proportional to r inside and inversely proportional to r outside, which is option (c).
The idea to keep. With Ampere's law, the field at radius r depends only on the current inside that radius. Uniform current through the volume gives B proportional to r inside and 1/r outside.
Answer from the NTA final answer key. Solution by PrepWiser.