In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero :

Solution

Physics sir · 2 steps

Draw an Amperian circle in each region and ask one thing: how much net current does it enclose?

1. Use Ampere's law. By symmetry, B has the same size all along a circle of radius r centred on the axis. So the line integral is B times the circumference, and B depends only on the net current inside that circle.

   ∮ B⃗· dl⃗ = B (2π r) = μ₀ I_(enc)

   Region	Net current enclosed	Field B
   Inside inner conductor (r < a)	Ir²/a²	(μ₀ I r)/(2π a²), not zero
   Between the conductors (a < r < b)	I	(μ₀ I)/(2π r), not zero
   Within outer conductor (b < r < c)	I minus part of the return current	not zero, falls to 0 only at r = c
   Outside the cable (r > c)	I - I = 0	zero

2. So. Only a circle drawn outside the whole cable encloses both currents, which cancel exactly. The field is zero outside the cable, option (b).

The idea to keep. With Ampere's law, the field on a symmetric loop depends only on the net current it encloses. Equal and opposite currents both inside the loop mean zero field there.

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