Question

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)?

Answer 1

t=−MB(cosθ2−cosθ1)
t=−MB(cos⁡θ2−cos⁡θ1)=−1.5×0.22(cos180∘−cosθ∘)
=−1.5×0.22(cos⁡180∘−cos⁡θ∘)But cos180∘=−1
cos⁡180∘=−1and cos0∘=1=−1.5×0.22×=−2

Answer : 0.66J

Answer 2

Answer:-

(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

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(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

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(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