Question

A long straight metal rod has a very long hole of radius $a$ drilled parallel to its axis, as shown in the figure. If the rod carries a current $i$, find the value of magnetic induction on the axis of the hole, where $\text{OC}=c.$

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

Answer 1

The correct option is B $\frac{{\mu }_{o}ic}{\pi (b^2-a^2)}$


As we know, magnetic field inside the cavity is,

$B=\frac{{\mu }_0}{2}(\overrightarrow{J}\times \overrightarrow{r}).......\left(1\right)$

Where, $J=\text{Current density};r=\text{Separation between the axis}$

The given system is the combination of a solid cylinder of radius $b$ and a cylindrical cavity of radius $a.$ The separation between their axis is $\text{OC}=c$ as shown in figure.

According to the given statement, current $i$ is the flowing in the remaining part.

$\therefore$ The current density $J$ is,

$J=\frac{i}{\pi b^2-\pi a^2}=\frac{i}{\pi (b^2-a^2)}$

Putting the value of $J$ in eq.$\left(1\right)$ we get,

$B=\frac{{\mu }_{o}ic}{2\pi (b^2-a^2)}$

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

Why this question ? Key concept : Field inside the cavity is uniform and independent of distance from the axis of cavity