# 5.7 a) 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.  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

Given.

Magnetic moment, M= 1.5 $\dpi{80} J T^{-1}$

Magnetic field strength, B= 0.22 T

Now,

The initial angle between the axis and the magnetic field, $\theta_1$= 0°

Final angle, $\theta_2$= 90°

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

$\dpi{100} W = - MB ( cos \theta_2 - cos \theta_1)$

$\dpi{100} \implies W = - (1.5)(0.22) ( cos 90^{\circ} - cos 0^{\circ}) = -0.33(0-1)$
= 0.33 J

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