Figure A and B shown to long straight wires of circular cross-section (a and b with a < b), carrying current I which is uniformly distributed across the cross-section. The magnitude of magnetic field B varies with radius r and can be represented as :

Solution

Physics sir · 3 steps

Both wires carry the same I, so the thinner wire packs it tighter: steeper rise, taller peak, earlier peak.

1. Inside the wire. Take an Amperian circle of radius r < R. It encloses only the fraction r²/R² of the current, so B · 2π r = μ₀ I r²/R². That gives B = (μ₀ I r)/(2π R²), a straight line from the origin whose slope is larger for the smaller radius.
2. Outside the wire. For r > R the circle encloses all of I, so B = (μ₀ I)/(2π r). This does not depend on R, so beyond r = b the two curves fall off as 1/r and coincide.

   Bₘₐₓ = (μ₀ I)/(2π R) at r = R

   Feature	Wire a (radius a)	Wire b (radius b)
   Slope inside	(μ₀ I)/(2π a²), steeper	(μ₀ I)/(2π b²), gentler
   Peak position	r = a, earlier	r = b, later
   Peak height	(μ₀ I)/(2π a), taller	(μ₀ I)/(2π b), shorter
   For r > b	(μ₀ I)/(2π r)	(μ₀ I)/(2π r), same

3. So. The correct graph has curve a peaking first and higher, curve b peaking later and lower, both decaying as 1/r outside. That is option (c).

The idea to keep. For a solid wire with uniform current, B rises linearly inside and peaks at the surface at (μ₀ I)/(2π R). Same current, smaller radius means a steeper line and a taller, earlier peak.

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