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-secton 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 the circle is , therefore,
`ointvecB.dvecl=B2pix`
in each of the four parts of the figure.
(a) The current enclosed within the circle in part `b` is `i_(0)` so that `i_(0)/(pia^(2)).pix^(2)=(i_(0))/(a^(2))x^(2)`
`oint vecB.dvecl=mu_(0)i` gives
`B.2pix=(mu_(0)l_(0)x^(2))/a^(2)` or `B=(mu_(0)l_(0)x^(2))/(2pix)`
The direction will be along the tangent of the circle.
(b) The current enclosed within the circle in part `b` is so that
(c) The area of cross-section of the outer shells is `pic^(2)-pib^(2)` . The area of cross-section of the outer shell within the circle in part of the figure is `pix^(2)-pib^(2)`
Thus, the current through this pair is `l_(0)((x^(2)-b^(2)))/((c^(2)-b^(2))).` This is in the same direction to the current `i_(0)` in the inner wire.Thus, the net current enclosed by the circle is
`l_("net")=l_(0)(l_(0)((x^(2)-b^(2))))/((c^(2)-b^(2)))=(l_(0)(c^(2)-x^(2)-2b^(2)))/((c^(2)-b^(2))).`
From Ampere's law
`B 2pix=(l_(0)(c^(2)+x^(2)-2b^(2)))/((c^(2)-b^(2)))` or `B=(mu_(0)(l_(0)(c^(2)+x^(2)-2b^(2))))/(2pix(c^(2)-b^(2)))`
(d) The net current enclosed by the circle in part `d` of the figure is `2i_(0)` and hence
`B 2pix=mu_(0) 2i_(0)` or `B=(u_(0)l_(0))/(pix)`.