A coaxial cable consists of a central conducting core wire of radius ‘a’ and a coaxial cylindrical outer conductor of radius ‘b’. The two conductors carry equal current in opposite directions, in and out of the plane of the paper. What will be the ma…

Answer: (i) \(B = \dfrac{\mu_0 I}{2\pi r}\), along circles concentric with the cable (tangential), sense given by the right-hand rule for the core current. (ii) \(B = 0\).

By symmetry, B has the same magnitude at all points at distance \(r\) from the axis and is tangential to a circle of radius \(r\) centred on the axis. Take such a circle as the Amperian loop. Ampere’s law: \(\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}\), so \(B(2\pi r) = \mu_0 I_{\text{enc}}\).

(i) \(a < r < b\): the loop encloses only the core current \(I\).

Direction: the field lines are circles around the axis. If the core current comes out of the plane of the paper, B is anticlockwise (as seen on the paper); if it goes into the paper, B is clockwise (right-hand thumb rule).

(ii) \(r > b\): the loop encloses the core current \(I\) and the equal, opposite current \(-I\) in the outer conductor, so \(I_{\text{enc}} = I - I = 0\).

So there is no magnetic field outside a coaxial cable; this is why such cables do not disturb nearby circuits.