Question

A wire loop $P Q R S P$ is constructed by joining two semi-circular coils of radii $r_{1}$ and $r_{2},$ respectively as shown in the figure. If the current flowing in the loop is $i$, then the magnetic induction at the point $O$

• A. $\frac{\mu_{0} i}{4}\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right]$
• B. $\frac{\mu_{0} i}{4}\left[\frac{1}{r_{1}}+\frac{1}{r_{2}}\right]$
• C. $\frac{\mu_{0} i}{2}\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right]$
• D. $\frac{\mu_{0} i}{2}\left[\frac{1}{r_{1}}+\frac{1}{r_{2}}\right]$

Answer 1

Answer: (A)

$B_{P Q}=B_{R A}=0$
$B_{Q R}=\frac{\mu_{0}}{4} \cdot \frac{i}{r_{1}}$ (vertically upward)
$B_{S P}=\frac{\mu_{0}}{4} \cdot \frac{i}{r_{2}}$ (vertically downward)
So, $B_{ net }=B_{Q R}-B_{S P}$
$=\frac{\mu_{0}}{4}\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right) i$