Question

A loop carrying current $I$ lies in the $xy$-plane as shown in figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic field $B=B_0\hat{i}$ exists where $B_0$ is a constant. The torque acting on the current loop is:

• A. $B_{0}I{a}^2(1+\frac{\pi }{2})\hat{j}$
• B. $B_{0}I{a}^2\pi \hat{j}$
• C. $-B_{0}I{a}^2\pi \hat{k}$
• D. $B_{0}I{a}^2(1+\frac{3\pi }{2})\hat{i}$

Answer 1

The correct option is A $B_{0}I{a}^2(1+\frac{\pi }{2})\hat{j}$

Given, $\overrightarrow{B}=B_0\hat{i}$

The given semicircular parts will form effectively two circular shapes each with radius, $r=\frac{a}{2}$

Total area of the given loop is,

$A=A_{\text{circular}}+A_{\text{square}}$

$A=[\pi {\left(\frac{a}{2}\right)}^2+\pi {\left(\frac{a}{2}\right)}^2]+a^2$

$A=\frac{\pi a^2}{2}+a^2=a^2(1+\frac{\pi }{2})$

Now the current is in anticlockwise direction, hence magnetic moment of loop will be,

$\overrightarrow{\mu }=I\overrightarrow{A}=I(1+\frac{\pi }{2})a^2\hat{k}$

Torque experienced by coil is; $\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$

$\Rightarrow \overrightarrow{\tau }=I(1+\frac{\pi }{2})a^2\hat{k}\times B_0\hat{i}$

$\Rightarrow \overrightarrow{\tau }=B_{0}I(1+\frac{\pi }{2})a^2\hat{j}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.