Question

A coaxial cable 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 $i_0$, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance $x$ from the axis when (i) $x<a$ and (ii) $a<x<b$ ?

• A. $\frac{b^2-a^2}{x^2}$
• B. $\frac{x^2}{b^2-a^2}$
• C. $\frac{a^2}{x^2}$
• D. $\frac{x^2}{a^2}$

Answer 1

The correct option is D $\frac{x^2}{a^2}$


Using Amperes circuital law,

For $x<a,$

$B\left(2\pi x\right)={\mu }_0\frac{i_0}{\pi a^2}\pi x^2$

$B=\frac{{\mu }_0{i}_{0}x}{2\pi a^2}\dots \left(i\right)$

For $a<x<b$

$B\left(2\pi x\right)={\mu }_0{i}_0$
$B=\frac{{\mu }_0{i}_0}{2\pi x}\dots \left(ii\right)$

Ratio of magnetic fields is
$\frac{\frac{{\mu }_0{i}_{0}x}{2\pi a^2}}{\frac{{\mu }_0{i}_0}{2\pi x}}=\frac{x^2}{a^2}$