Question

A circular coil with area $A$ and $N$ turns is free to rotate about a diameter that coincides with the $x-$ axis . Current $I$ is circulating in the coil. There is a uniform magnetic field $\overrightarrow{B}$ on the positive $y-$ direction. Calculate the magnitude and direction of the torque $\overrightarrow{\tau }$.

• A. $NIAB;\hat{j}$
• B. $NIAB;-\hat{j}$
• C. Zero
• D. $NIAB;\hat{i}$

Answer 1

The correct option is C Zero

Torque acting on the circular coil $$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$$But for $N$turns, $\mu =NIA$

Thus, $\tau =NIAB\sin \theta$

Area vector of the circular coil is antiparallel to direction of magnetic field. So $\theta ={180}^{\circ }$.

$\therefore \tau =NIAB\sin {180}^{\circ }=0$

Direction of magnetic moment is along the same direction of area vector (i.e along $-\hat{j}$)

From the data given in the question, direction of magnetic field is $\hat{j}$

Thus, direction of torque vector is $\hat{\tau }=-\hat{j}\times \hat{j}=\overrightarrow{0}$ (No direction)

Hence, option (c) is the correct answer.