The figure shows a long coaxial cable in which a current I flows down through the inner cylinder of radius a and the same current flows back up through the outer cylinder of radius 2 a . The cylinders are insulated from each other and the current is uniformly distributed over the area of the cross-section in each cylinder. The strength of the magnetic field at a distance 3 a 2 from the axis of the cable is

Step-by-step solution

Using ampere's law for a closed loop of radius r = 3 a 2 around the axis of the cable, ∮ B → . d l → = μ 0 I - I π r 2 - π a 2 4 π a 2 - π a 2 B × 2 π r = μ 0 I - I π r 2 - π a 2 4 π a 2 - π a 2 B × 2 π 3 a 2 = μ 0 I - I π 9 a 2 4 - π a 2 3 π a 2 B × 3 π a = μ 0 I - 5 I 12 B = 7 μ 0 I 36 π a