Question

A circular current carrying loop of radius $R$ is bent about its diameter by ${90}^{\circ }$ and placed in a magnetic field $\overrightarrow{B}=B_0(\hat{i}+\hat{j})$ as shown in figure.

• A. The torque acting on the loop is zero.
• B. The magnetic moment of the loop is $\frac{I\pi R^2}{2}(-\hat{i}-\hat{j})$.
• C. The angular acceleration of the loop is non-zero.
• D. The magnetic moment of the loop is $\frac{I\pi R^2}{2}(-\hat{i}+\hat{j})$.

Answer 1

Answer: (A), (B)

The magnetic moment of the loop is $\frac{I\pi R^2}{2}(-\hat{i}-\hat{j})$.

The problem given can be made into two parts, as shown in the diagrams.

$1.$ For loop lying in $y-z$ plane:

$\overrightarrow{\mu }=I{\overrightarrow{A}}_1=\frac{I\pi R^2}{2}(-\hat{i})$


$2.$ For loop lying in $x-z$ plane:

$\overrightarrow{\mu }=I{\overrightarrow{A}}_2=\frac{I\pi R^2}{2}(-\hat{j})$


Net magnetic moment:

$\overrightarrow{\mu }={\overrightarrow{\mu }}_1+{\overrightarrow{\mu }}_2=\frac{I\pi R^2}{2}(-\hat{i}-\hat{j})$

Given, $\overrightarrow{B}=B_0(\hat{i}+\hat{j})$

Now, Torque,

$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}=\frac{I{B}_0\pi R^2}{2}(-\hat{k}+\hat{k})$

$\therefore \overrightarrow{\tau }=0$

Further, angular acceleration,

$I\alpha =\tau$

As, $\tau =0$

$\therefore \alpha =0$

Hence, options $\left(A\right)$ and $\left(B\right)$ are correct.