Question

The figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius $R$ and outer radius $2R$, cylinder carrying uniformly distributed current $I$ along its axis. The magnetic induction at a point $\text{P}$ at distance $\frac{3R}{2}$ from the axis of the cylinder will be:

• A. Zero
• B. $\frac{5{\mu }_{0}I}{72\pi R}$
• C. $\frac{7{\mu }_{0}I}{18\pi R}$
• D. $\frac{5{\mu }_{0}I}{36\pi R}$

Answer 1

The correct option is D $\frac{5{\mu }_{0}I}{36\pi R}$


Let that current is $I$ flowing along $+vez$- axis (perpendicularly outward).

Current density of the conductor is given by

$J=\frac{I}{\pi \left[\right(2R{)}^2-R^2]}=\frac{I}{3\pi R^2}$

Now constructing the Amperian loop at a distance $\frac{3R}{2}$ from axis

So the urrent enclosed by amperian loop is,

$I^{\prime }=\left(J\right)\times A=J(\pi {\left(\frac{3R}{2}\right)}^2-\pi R^2)$

$I^{\prime }=\frac{I}{3\pi R^2}\left[\frac{5\pi R^2}{4}\right]=\frac{5I}{12}$

Applying Ampere's circuital law,

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{I}_{en}$

Here the direction of $\overrightarrow{B}$ is along the circular amperian loop.

$\therefore B\left[2\pi \left(\frac{3R}{2}\right)\right]={\mu }_0{I}^{\prime }=\frac{5I{\mu }_0}{12}$

$\left(3\pi R\right)B=\frac{5{\mu }_{0}I}{12}$

$\therefore B=\frac{5{\mu }_{0}I}{36\pi R}$

Hence, option $\left(d\right)$ is correct.