Question

A long cylindrical conductor of radius 'a' has two cylindrical cavities each of diameter 'a' through its entire length as shown in the figure. A current I is directed out of the page and is uniform throughout the cross-section of the conductor. The magnetic field at point $P_1$ is

• A. $\frac{{\mu }_{0}I}{2\pi r}\left(\frac{2r^2+a^2}{4r^2-a^2}\right)$ to the right
• B. $\frac{{\mu }_{0}I}{2\pi }\left(\frac{r^2+a^2}{r^2-a^2}\right)$ to the right
• C. $\frac{{\mu }_{0}I}{\pi r}\left(\frac{2r^2+a^2}{2r^2-a^2}\right)$ to the left
• D. $\frac{{\mu }_{0}I}{2\pi r}\left(\frac{2r^2-a^2}{4r^2-a^2}\right)$ to the left

Answer 1

The correct option is D

$\frac{{\mu }_{0}I}{2\pi r}\left(\frac{2r^2-a^2}{4r^2-a^2}\right)$ to the left

Let $I$ be the current through the complete conductor.
Since $I\propto A$, current through cavity will be $\frac{I}{4}$.
For the complete conductor, $B=\frac{{\mu }_{0}I}{2\pi r}$
For cavity 1, $B_1=\frac{{\mu }_0\frac{I}{4}}{2\pi (r-\frac{a}{2})}$
For cavity 2, $B_2=\frac{{\mu }_0\frac{I}{4}}{2\pi (r+\frac{a}{2})}$
Now, $B_{net}=B-(B_1+B_2)$
Solving, we get $B_{net}=\frac{{\mu }_{0}I}{2\pi r}\left(\frac{2r^2-a^2}{4r^2-a^2}\right)$
The direction of the field will be towards left.