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 $i$ and the outer shell carries an equal current in the opposite direction as shown in the figure. The magnetic induction at a distance $r$ from the axis, for $(b<r<c)$ is-
[Take permeability of the conductor as $\mu$]

• A. $\frac{\mu i}{2\pi r}$
• B. $\frac{\mu i}{2\pi r}\left(\frac{r^2-b^2}{c^2-b^2}\right)$
• C. $\frac{\mu i}{2\pi r}\left(\frac{c^2-r^2}{c^2-b^2}\right)$
• D. $\frac{\mu i}{2\pi r}\left(\frac{c^2-b^2}{r^2-b^2}\right)$

Answer 1

The correct option is C $\frac{\mu i}{2\pi r}\left(\frac{c^2-r^2}{c^2-b^2}\right)$

Applying Ampere's circuital law,

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

As given, permeability of the outer conductor is $\mu$

$\therefore \oint \overrightarrow{B}\cdot \overrightarrow{dl}=\mu i_{enc}.......\left(1\right)$

Where, $i_{enc}=$ Net current enclosed by the amperian loop


Let us consider an Amperian loop of radius $r$ as shown in figure, the net current enclosed by the Amperian loop in the outer conductor is,

$i_{out}=JdA=\frac{i}{\pi (c^2-b^2)}\times \pi (r^2-b^2)=\left(\frac{r^2-b^2}{c^2-b^2}\right)i$

So the net current enclosed by the Amperian loop is,

$i_{net}=i_{in}-i_{out}=i-i\left(\frac{r^2-b^2}{c^2-b^2}\right)$

$\therefore i_{net}=i[1-\frac{r^2-b^2}{c^2-b^2}]$

Putting the value of $i_{enc}$ in $\left(1\right)$ we get,

$B\times 2\pi r=\mu i[1-\frac{r^2-b^2}{c^2-b^2}]$

$B=\frac{\mu i}{2\pi r}\left(\frac{c^2-r^2}{c^2-b^2}\right)$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{C}\right)$ is the correct answer.