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)?
(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
(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 = - MB (cos θ2 - cos θ1)
= -1.5 x 0.22 (cos 180° - cos 0°)
= -0.33 (-1 - 1)
= 0.66 J
(b)For case (i): θ = θ2 = 90°
∴Torque, T = MB sinθ
= 1.5 x 0.22 sin 90°
=0.33 J
Forcase (ii): θ = θ2 = 180°
∴Torque, T = MB sin θ
=MB sin 180° = 0
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