Consider a coaxial cable which consists of an inner wire of radius a surrounded by an outer shell of inner and outer radii b and c respectively. The inner wire carries an electric current `i_0` and the outer shell carries an equal current in opposite direction. Find the magnetic field at a distance x from the axis where (a) `xlta,` (b) `altxltb`, (c) `bltxltc` and (d)`xgtc`. Assume that the current density is uniform in the inner wire and also uniform in the outer shell.

Explanation

A cross-section of the cable is shown in figure. Draw a circle of radius x with the centre at the axis of the cable. The parts a, b, c and d of the figure correspond to the four parts of the problem. By symmetry, the magnetic field at each point of a circle will have the same magnitude and will be tangential to it. The circulation of B along this circle is, therefore,
`ointvecB.dvecl=B2pix`
in each of the four parts of the figure.
The area of cross-section of the outer shell is `pic^(2) – pib^(2)`. The area of cross-section of the outer shell with in the circle in part c of the figure is `pix^(2) – pib^(2)`.
Thus, the current through this part is `(i_(0)(x^(2)-b^(2)))/((c^(2)-b^(2)))`
This is in the opposite direction to the current i0 in the inner wire. Thus, the net current enclosed by the circle is
`i_("net")=i_(0)-(i_(0)(x^(2)-b^(2)))/(c^(2)-b^(2))=(i_(0)(c^(2)-x^(2)))/(c^(2)-b^(2))`
From Ampere’s law,
`B2pix=(mu_(0)i_(0)(c^(2)-x^(2)))/(c^(2)-b^(2))or,B=(mu_(0)i_(0)(c^(2)-x^(2)))/(2pix(c^(2)-b^(2)))`