Question

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

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

Answer 1

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

The magnetic field is$\overrightarrow{B}=B_0\hat{j}$ and the magnetic moment
$\overrightarrow{m}=i\overrightarrow{A}=-i\left(N{L}^2\hat{i}\right)$
The torque is given by $\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}$
$=-iN{L}^2\hat{i}\times B_0\hat{j}=-iN{B}_0{L}^2\hat{i}\times \hat{j}$
$=-iN{B}_0{L}^2\hat{k}$