A bar magnet of magnetic moment
Question:

A bar magnet of magnetic moment 1.5 J T−1 lies aligned with the direction of a uniform magnetic field of 0.22 T.

(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?

Solution:

(a)Magnetic moment, M = 1.5 J T−1

Magnetic field strength, B = 0.22 T

(i)Initial angle between the axis and the magnetic field, θ1 = 0°

Final angle between the axis and the magnetic field, θ2 = 90°

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

$W=-M B\left(\cos \theta_{2}-\cos \theta_{1}\right)$

$=-1.5 \times 0.22\left(\cos 90^{\circ}-\cos 0^{\circ}\right)$

$=-0.33(0-1)$

$=0.33 \mathrm{~J}$

(ii) Initial angle between the axis and the magnetic field, θ1 = 0°

Final angle between the axis and the magnetic field, θ2 = 180°

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

$W=-M B\left(\cos \theta_{2}-\cos \theta_{1}\right)$

$=-1.5 \times 0.22\left(\cos 180-\cos 0^{\circ}\right)$

$=-0.33(-1-1)$

$=0.66 \mathrm{~J}$

(b) For case (i): $\theta=\theta_{2}=90^{\circ}$

$\therefore$ Torque, $\tau=M B \sin \theta$

$=1.5 \times 0.22 \sin 90^{\circ}$

$=0.33 \mathrm{~J}$

For case (ii): $\theta=\theta_{2}=180^{\circ}$

$\therefore$ Torque, $\tau=M B \sin \theta$

$=M B \sin 180^{\circ}=0 \mathrm{~J}$

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