Question

A wire loop formed by joining two semicircular wires of radii $R_1$ and $R_2$ carries a current as shown in the adjoining diagram. The current in the loop is $I$ and it is placed in a uniform magnetic field $B$, parallel to its plane. What will be the torque acting on the loop?

• A. $\pi IB(R{{}_1}^2+R{{}_2}^2)$
• B.
  $\frac{\pi IB(R{{}_1}^2+R{{}_2}^2)}{2}$
• C. $\pi IB(R{{}_2}^2-R{{}_1}^2)$
• D. $\frac{\pi IB(R{{}_2}^2-R{{}_1}^2)}{2}$

Answer 1

The correct option is B

$\frac{\pi IB(R{{}_1}^2+R{{}_2}^2)}{2}$

From the diagram, it is clear that the direction of the current is clockwise. So, the direction of the magnetic moment of the entire loop is normal into the plane.

Torque,

$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$

$\Rightarrow \tau =\mu B\sin {90}^{\circ }=IAB\sin {90}^{\circ }$

$=I(\frac{\pi R{{}_2}^2}{2}+\frac{\pi R{{}_1}^2}{2})B$

$=\frac{\pi IB(R{{}_1}^2+R{{}_2}^2)}{2}$

Hence, option $\left(B\right)$ is the correct answer.

Why this Question ? To understand the concept of torque experienced by a current carrying loop in a uniform magnetic field.