Question

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 the same direction. Find the magnetic field at a distance $x$ from the axis where $b<x<c$.
(Assume current density to be uniform in the inner wire and outer shell.)

• A. $\frac{{\mu }_0{i}_{0}x}{2\pi a^2}$
• B. $\frac{{\mu }_0{i}_0(2c^2+x^2)}{\pi x(c^2-x^2)}$
• C. $\frac{{\mu }_0{i}_0}{2\pi x}$
• D. $\frac{{\mu }_0{i}_0(c^2+x^2-2b^2)}{2\pi x(c^2-b^2)}$

Answer 1

The correct option is D $\frac{{\mu }_0{i}_0(c^2+x^2-2b^2)}{2\pi x(c^2-b^2)}$

Let us draw an Amperian loop as a circle of radius $x$ from axis of cable. By symmetry, the value of magnetic field will be same at each point of circle and direction will be tangential along the path.


$\therefore \oint \overrightarrow{B}\cdot \overrightarrow{dl}=B\left(2\pi x\right)=2B\pi x$

From Ampere's circuital law

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0{i}_{en}$

$2B\pi x={\mu }_0{i}_{en}.......\left(1\right)$

Since current $i_0$ is flowing through outer cable, so current density of outer shell will be

$J=\frac{i_0}{\pi (c^2-b^2)}$

And the area of cross- section of outer shell with $x$ as outer radius is

$A=\pi (x^2-b^2)$

So, the current enclosed by circular amperian loop of radius $x$ will be

$i_{en}=i_0+(\frac{i_0}{\pi (c^2-b^2)}\times \pi (x^2-b^2))$

$(\because$ the current in inner wire $\left(i_0\right)$ gets completely enclosed by loop)

$\therefore i_{en}=i_0+\frac{i_0(x^2-b^2)}{(c^2-b^2)}$

Using eq $\left(1\right)$;

$2B\pi x={\mu }_0\frac{i_0(c^2+x^2-2b^2)}{c^2-b^2}$

$\therefore B=\frac{{\mu }_0{i}_0(c^2+x^2-2b^2)}{2\pi x(c^2-b^2)}$

Therefore, option $\left(d\right)$ is the correct answer.