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(cosθ2 - cosθ1)
= -1.5 J T−1 x0.22 T(cos 90o-cos 0o)
= -0.33 Jx(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, θ 2 = 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 J T−1 x.22 T(cos 180o-cos 0o)
= -0.33 Jx(-1-1)
= 0.66 J
(b) For case i) θ = θ2 = 90o
τ = MBsinθ = 1.5 J T−1 x 0.22 T x 1 = .33 J
(Remember the torque and the work applied by the torque are not the same thing)
For case ii) θ = θ2 = 180o
τ = MBsinθ = 1.5 J T−1 x.22 Tx0 = 0 J