Q. A bar magnet of magnetic moment $220 \,Am ^{2}$ is suspended in a magnetic field of intensity $0.25\, N / Am$. The couple required to deflect it through $30^{\circ}$ is:
Jharkhand CECEJharkhand CECE 2003
Solution:
A bar magnet suspended in a uniform magnetic field sets itself with its axis parallel to the field. A magnet placed in the magnetic field experiences a torque which rotates the magnet to a position in which the axis of the magnet is parallel to the field.
The magnitude of torque acting on a current loop placed in a magnetic field
$\vec{B}$ with its axis at an angle $\theta$ with the direction of $\vec{B}$ is given by
$\tau=i A B \sin \theta$
Here, magnitude of dipole moment, $M=i A$
$\therefore \tau=M B \sin \theta$
Putting the numerical values, we have
$M=220\, Am ^{2}, B=0.25 \,N / A m, \theta=30^{\circ}$
$\therefore \tau=220 \times 0.25 \times \cos 30^{\circ}$
$=220 \times 0.25 \times \frac{\sqrt{3}}{2}$
$=47.63 \,Nm$
