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 90^o-cos 0^o)

= -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 180^o-cos 0^o)

= -0.33 Jx(-1-1)

= 0.66 J

(b) For case i) θ = θ₂ = 90^o

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 = 180^o

τ = MBsinθ = 1.5 J T⁻¹ x.22 Tx0 = 0 J