Question

A bar magnet of magnetic moment $M$ is placed in the magnetic field $B$. The torque acting on the magnet is:

Answer 1

Step 1: Given Data

Magnetic Moment of the bar magnet $=M$

External magnetic field $=B$

Let the angle made between the bar magnet and the direction of the external magnetic field be $\theta$.

Let $F$ be the magnitude of the coupling force.

Let $d$ is the perpendicular distance between the two couple force.

Let $i$ be the current.

Let $L$ be the length.

Step 2: Formula Used

We know that force can be given as,

$F=iLB$

We know that dipole moment is given as,

$M=iLd$

Step 3: Calculate the Torque

We know that torque is given as,

$\tau =Fd\sin \theta$

$\Rightarrow \tau =iLBd\sin \theta$

$\Rightarrow \tau =MB\sin \theta$

$\Rightarrow \overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$

Hence, the torque acting on the magnet is $\overrightarrow{M}\times \overrightarrow{B}$.