Answer:
Option A
Explanation:
$B_{R}=B_{T}-B_{C}$
R= Remaining portion
T= Total portion and
C= cavity
$B_{R}= \frac{\mu_{0}l_{T}}{2a \pi}- \frac{\mu_{0}l_{C}}{2(3a/2)\pi}$.......(i)
$l_{T}=J(\pi a^{2})$
$l_{C}=J\left(\frac{\pi a^{2}}{4}\right)$
Substituting the values in Eq. (i), we
have
$B_{R}=\frac{\mu_{0}}{a \pi}\left[\frac{l_{T}}{2}-\frac{l_{C}}{3}\right]$
$=\frac{\mu_{0}}{a \pi}\left[\frac{\pi a^{2}J}{2}-\frac{\pi a^{2} J}{12}\right]$
$=\frac{5\mu_{0}aJ}{12}$
$\therefore$ N=5