Question

The wire loop $PQRS$ formed by joining two semi circular wires of radii $R_1$ and $R_2$ carries a current $I$ as shown the figure. The magnitude of magnetic induction at the centre $O$ is

• A. $\frac{{\mu }_{0}I}{4}\left(\frac{1}{R_2}-\frac{1}{R_1}\right)$
• B. $\frac{{\mu }_{0}I}{4}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
• C. ${\mu }_{0}I\left(\frac{1}{R_2}-\frac{1}{R_1}\right)$
• D. ${\mu }_{0}I\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Answer 1

Answer: (B)

The magnetic induction at $O$ due to current through $QP$ and $SR$ will be zero.
Thus, the magnetic induction at $O$ will be only due to current through semicircles of radii $R_1$ and $R_2$ respectively.
Magnetic field at $O$ due to semicircular arc $PS$ of radius $R_1$ will be
$B_1=\frac{{\mu }_0}{4\pi }\frac{\pi I}{R_1}$.
Its direction is perpendicular to coil directed outwards. Magnetic field at $O$ due to semicircular arc $RQ$ of radius $R_2$ will be,
$B_2=\frac{{\mu }_0}{4\pi }\frac{\pi I}{R_2}$.
Its direction is perpendicular to coil directed inwards.
The resultant magnetic field at $O$, $\vec{B}=\left(B_1-B_2\right)$ outwards
$\therefore B=\frac{{\mu }_0}{4\pi }I\pi \left[\frac{1}{R_1}-\frac{1}{R_2}\right]=\frac{{\mu }_0}{4}I\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$