In the figure a cross-section has been shown of a long conductor of a type called a coaxial cable of radii a, b, and c. Equal but antiparallel uniformly distributed currents i exist in the two conductors. We have to derive expressions for in the ranges (a) , (b) , (c) , and (d) .

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In the figure a cross-section has been shown of a long conductor of a type called a coaxial cable of radii a, b, and c. Equal but antiparallel uniformly distributed currents i exist in the two conductors.

We assume that the current in the conductor of inner radius and outer radius is and it is in the direction of the outward perpendicular to the page. The current in the conductor of radius will also be , but its direction will that be of the inward perpendicular to the page.

We will take appropriate Amperian circular loops for finding the magnetic field using the Ampere’s law.

(a)

For ,

(b)

For ,

(c)

For ,

(d)

For ,

The Amperian loop encloses two currents of equal magnitude but opposite in signs as the currents flow in opposite directions. Therefore, contributions to the right-hand side of the Ampere’s law cancel each other. Therefore,

.