Consider a co axial cable which consists of an inner wilre of radius `a` surrounded by an outer shell of inner and outer radii `b` and `c` respectively. The inner wire caries a current `I` and outer shell carries an equal and opposite current. The magnetic field at a distance `x` from the axis where `bltxltc` is

Explanation

A cross section of the cable is shown in Fig. 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, `oint vecB.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=(i_0)/(pia^2) pix^2=(i_0)/(a^2)x^2`.
Ampere's law `oint vecB.dvecl=mu_0I` gives
`B2pix=(mu_0i_0x^2)/(a^2) or , B=(mu_0i_0x)/(2pia^2)`
The direction will be along the tangent to the circle.
(b) The current enclosed within the circle in part (b) is `i_0` so
that `B2pix=mu_0i_0 or, B=(mu_0i_0)/(2pix)`
(c) current density of outer shell : `J=(i_0)/(pic^2-pib^2)`
So current from `x=0 to x=x`:
`I=i_0-J(pix^2-pib^2)`
`=i_0-i_0((x^2-b^2)/(c^2-b^2))=(i_0(c^2-x^2))/(c^2-b^2)`
`B=(mu_0I)/(2pix)=(mu_0i_0(c^2-x^2))/(2pix(c^2-b^2))`
(d) For `xgtc`, magnetic field will zero, because net current
is zero.