Question

A wire loop formed by joining two semicircular wires of radii $R_1$ and $R_2$ carries a current $I$ as shown in the following diagram. The magnetic induction at the centre $O$ is :

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

Answer 1

The correct option is D $\frac{{\mu }_{0}I\pi }{4\pi }(\frac{1}{R_1}+\frac{1}{R_2})$

Magnetic field at $O$ due to segments $FA$ and $CD$ is equal to zero.
Magnetic field at $O$ due to semi-circle $ABC:\frac{1}{2}\frac{{\mu }_{0}I}{2R_2}(-\hat{k})$
Magnetic field at $O$ due to semi-circle $DEF:\frac{1}{2}\frac{{\mu }_{0}I}{2R_1}(-\hat{k})$
${\overrightarrow{B}}_{net}=\frac{1}{2}\frac{{\mu }_{0}I}{2R_2}(-\hat{k})+\frac{1}{2}\frac{{\mu }_{0}I}{2R_1}(-\hat{k})=\frac{{\mu }_{0}I}{4}(\frac{1}{R_1}+\frac{1}{R_2})(-\hat{k})=\frac{{\mu }_{0}I\pi }{4\pi }(\frac{1}{R_1}+\frac{1}{R_2})(-\hat{k})$where $\hat{k}$ is out of the plane of the loop and perpendicular to the loop