Question

A square coil of $N$ turns with length of each side equal to $L$, carring a current $i$ is placed in a uniform magnetic fileld $\overrightarrow{B}=B_0\hat{j}$ as shown in figure. What is the torque acting on coil?

• A. $\left(B_{0}Ni{L}^2\right)\hat{k}$
• B. $(-B_{0}Ni{L}^2)\hat{k}$
• C. $\left(B_{0}Ni{L}^2\right)\hat{j}$
• D. $(-B_{0}Ni{L}^2)\hat{j}$

Answer 1

The correct option is B $(-B_{0}Ni{L}^2)\hat{k}$

The area vector for coil is,

$\overrightarrow{A}=L^2(-\hat{i})$

Now, $\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$

$\overrightarrow{\tau }=Ni\overrightarrow{A}\times \overrightarrow{B}$

$\overrightarrow{\tau }=Ni\left[\right(L^2(-i)\times B_0\left(j\right)]$

$\overrightarrow{\tau }=Ni{L}^2{B}_0[-\hat{i}\times \hat{j}]$

$\overrightarrow{\tau }=-Ni{L}^2{B}_0\hat{k}$

$\therefore \overrightarrow{\tau }=-Ni{L}^2{B}_0\hat{k}$

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