Question

A loop is formed by joining two semicircular wires of radii $R_1$ and $R_2$ carrying a current as shown in the figure. The current in the loop is $I$ and is placed in the uniform magnetic field $B$ which is parallel to the plane of loop. What will be the torque acting on the loop.

• A. $\frac{\pi IB(R_1^2+R_2^2)}{2}$
• B. $\pi IB(R_1^2+R_2^2)$
• C. $2\pi IB(R_1^2+R_2^2)$
• D. $\frac{3\pi IB(R_1^2+R_2^2)}{2}$

Answer 1

The correct option is A $\frac{\pi IB(R_1^2+R_2^2)}{2}$

The torque acting on a current carrying loop is given by, $$\tau =\overrightarrow{\mu }\times \overrightarrow{B}....\left(1\right)$$
$\Rightarrow \tau =\mu B\sin \theta =IAB\sin \theta [\because \mu =IA]$



From the figure,

$\theta ={90}^{\circ }$ and Area, $A=\frac{\pi R_1^2}{2}+\frac{\pi R_2^2}{2}$

$\therefore \tau =\frac{\pi IB(R_1^2+R_2^2)}{2}$

Hence, option (a) is the correct answer.