Question

A square loop of wire carrying current $I$ is lying in the plane of paper as shown in figure. The magnetic field is present in the region as shown. The loop will tend to rotate: (Take length of the side of the square as $l$ and assume $PQ$ is along $+x$ axis and $RS$ is along $+y$ axis)

• A. about $PQ$ with $KL$ coming out of the page , with the net torque $IB{l}^2\hat{i}$.
• B. about $PQ$ with $KL$ coming out of the page , with the net torque $IB{l}^2\hat{j}$.
• C. about $RS$ with $MK$ coming out of the page , with the net torque $IB{l}^2\hat{j}$.
• D. about $RS$ with $MK$ going into the page , with the net torque $IB{l}^2\hat{i}$.

Answer 1

The correct option is A about $PQ$ with $KL$ coming out of the page , with the net torque $IB{l}^2\hat{i}$.


Force on a current carrying wire of length $l$ is, $$\overrightarrow{F}=I(\overrightarrow{l}\times \overrightarrow{B})$$$\Rightarrow \overrightarrow{F_{MK}}=0$, since current in $MK$ is along $\overrightarrow{B}.$

$\Rightarrow \overrightarrow{F_{LN}}=0$, since current in $LN$ is opposite to $\overrightarrow{B}.$

$\Rightarrow \overrightarrow{F_{KL}}=I(l\hat{i}\times B\hat{j})=IBl\hat{k}$

$\Rightarrow \overrightarrow{F_{MN}}=I\left[l\right(-\hat{i})\times B\hat{j}]=IBl(-\hat{k})$

The line of action is not the same for the two forces.

Hence, the loop will rotate about $PQ$ with $KL$ coming out of the page.

So, the net torque about $PQ$,

${\overrightarrow{\tau }}_{net}=\frac{l\hat{j}}{2}\times \overrightarrow{F_{KL}}+\frac{l\widehat{(}-j)}{2}\times \overrightarrow{F_{MN}}$

$\Rightarrow {\tau }_{net}=IB{l}^2\hat{i}$

Hence, option (a) is the correct answer.

Why this question? To teach the concept that, for a closed rectangular current carrying loop, the net force is zero, but the net torque is not zero.