Question

The torque experienced by a given current carrying loop in a uniform magnetic field $\overrightarrow{B}=B_0(\hat{i}-\hat{j})$ would have magnitude

• A. $-2I{a}^2{B}_0\hat{k}$
• B. $2I{a}^2{B}_0\hat{k}$
• C. $0$
• D. $2I{a}^2{B}_0\hat{k}$

Answer 1

The correct option is C $0$

The torque acting on a current carrying loop is given by, $$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}....\left(1\right)$$

From the figure,

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

$\Rightarrow \mu =I{a}^2(\hat{j}-\hat{i})$

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

Using the equation $\left(1\right)$,

$\overrightarrow{\tau }=I{a}^2(\hat{j}-\hat{i})\times B_0(\hat{i}-\hat{j})$

$=I{a}^2{B}_0\left[\right(\hat{j}\times \hat{i})+(\hat{j}\times (-\hat{j}))-(\hat{i}\times \hat{i})+(\hat{i}\times \hat{j}\left)\right]$

$=I{a}^2{B}_0[-\hat{k}+0+0+\hat{k}]$

$=0$

Hence, option (c) is the correct answer.