Question

A square loop lies in $\text{x-y}$ plane as shown. It carries a steady current $I$ and is subjected to a magnetic field given $\overrightarrow{B}=(\frac{B_{0}z}{L}\hat{j}+\frac{B_{0}y}{L}\hat{k})$, where $B_0$ is a positive constant. The force acting on the side $OA$ is

• A. $\frac{1}{2}{B}_{0}IL\hat{j}$
  
• B. $\frac{1}{2}{B}_{0}IL\hat{i}$
  
• C. $B_{0}IL\hat{j}$
  
• D. $B_{o}IL\hat{i}$
  

Answer 1

The correct option is B $\frac{1}{2}{B}_{0}IL\hat{i}$



For side $OA,$ $\overrightarrow{dl}=dy\hat{j}$

Now, force on $dl$, $\overrightarrow{dF}=I(\overrightarrow{dl}\times \overrightarrow{B})$

Here,

$\overrightarrow{dl}\times \overrightarrow{B}=dy\hat{j}\times [\frac{B_{0}Z}{L}\hat{j}+\frac{B_{0}y}{L}\hat{k}]=\frac{B_{0}ydy}{L}\hat{i}$

$\therefore dF=|\overrightarrow{dF}|=\frac{B_{0}Iydy}{L}$

Now, total force on $OA$,

$F_{OA}=\int dF$

$\Rightarrow F_{OA}={\int }_o^L\frac{B_{0}Iydy}{L}=\frac{B_{o}I}{L}{\int }_o^{L}ydy$

$\Rightarrow F_{OA}=\frac{B_{o}I}{L}{\left[\frac{y^2}{2}\right]}_o^L=\frac{B_{o}I}{L}\frac{L^2}{2}$

$\therefore F=\frac{1}{2}{B}_{o}IL$ and $\overrightarrow{F}=\frac{1}{2}{B}_{o}IL\hat{i}$

Hence, option (b) is the correct answer.