A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r < R) from its centre will be :

Solution

Physics sir · 3 steps

Inside the wire, the Amperian loop only encloses part of the current, and that part grows as r².

1. Choose the loop. Draw a circle of radius r < R centred on the axis. By symmetry, B has the same size everywhere on it and points along it, so ∮ B⃗· dl⃗ = B × 2π r.
2. Current enclosed. The current is spread uniformly over the area π R². The loop encloses only the area π r², so I_(enc) = I × (π r²)/(π R²) = (I r²)/R².

   B × 2π r = μ₀ (I r²)/R² ⇒ B = (μ₀ I)/(2π R²) r

3. So. Everything except r is constant, so inside the wire B ∝ r. The field is zero at the centre and rises linearly up to the surface, which is option (b).

The idea to keep. Apply Ampere's law with only the current inside your loop. For a uniform solid wire, B ∝ r inside and B ∝ 1/r outside, with the largest field at the surface.

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