A bar magnet of magnetic movement 1.5j/telsha lies in direction of one from magnetic field of .22 telsha what is amt of work
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(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 = - MB ( cos θ2 - cos θ1)
= -1.5 x 0.22 ( cos 90° - cos 0°)
= -0.33 (0 - 1)
= 0.33 J
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