Question

A square loop of wire carrying current $I$ is lying in the plane of paper as shown in fig. The magnetic field is present in the region as shown. The loop will tend to rotate :

• A. about $PQ$ with $KL$ coming out of the page
• B. about $PQ$ with $KL$ going into the page
• C. about $RS$ with $MK$ coming out of the page
• D. about $RS$ with $MK$ going into the page

Answer 1

The correct option is A about $PQ$ with $KL$ coming out of the page

Let the $X$ and $Y$ axes are along $OQ$ and $OR$ respectively.
${\overrightarrow{F}}_{MK}=0$ since current in MK is along $\overrightarrow{B}$.
${\overrightarrow{F}}_{LN}=0$ since current in LN is opposite $\overrightarrow{B}$.
${\overrightarrow{F}}_{KL}=i(KL\hat{i}\times B\hat{j})=iB\hat{k}$
${\overrightarrow{F}}_{MN}=i(MN\hat{i}\times B\hat{j})=ilB(-\hat{k})$
The line of action is not same for the two forces ${\overrightarrow{F}}_{KL}$ and ${\overrightarrow{F}}_{MN}$
Perpendicular distance between ${\overrightarrow{F}}_{KL}$ and ${\overrightarrow{F}}_{MN}$ is $l$.
$\overrightarrow{\tau }=ilB\times l=i{l}^{2}B$
Hence, the loop will rotate about $PQ$ with $KL$ coming out of the page