Question

A long straight metal rod has a cavity of radius $a$ drilled parallel to the axis of the rod 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. $\frac{{\mu }_{0}ic}{\pi (b^2-a^2)}$
• B. $\frac{{\mu }_{0}ic}{2\pi (b^2-a^2)}$
• C. $\frac{{\mu }_{0}i(b^2-a^2)}{2\pi c}$
• D. $\frac{{\mu }_{0}ic}{2\pi (a^2-b^2)}$

Answer 1

The correct option is B $\frac{{\mu }_{0}ic}{2\pi (b^2-a^2)}$

As we know, field inside the cylindrical cavity is,

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

Here, $\overrightarrow{r}=\text{OC}=c$ and $\overrightarrow{J}=\frac{i}{(\pi b^2-\pi a^2)}$

Putting this values in (1) we get,

$\Rightarrow B=\frac{{\mu }_{0}ic}{2\pi (b^2-a^2)}$

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