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Question

A bar magnet of magnetic moment 6 J/T is aligned at 600 with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and
(b) the torque on the magnet in the final orientation
in case (ii).

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Solution

Given, magnetic momentum, m = 6 JT1

External magnetic field, B = 0.44 T

θ1=600cos θ1=cos 600=12.

(a) Work done in turning the magnet normal to the field,

W=mB(cos θ2cos θ1)

(i) Here, θ2=900

W=+mB cosθ1

= 6×0.44×12=1.32 J

(ii) Here θ2=1800

W=mB(cos θ2cos θ1)

W=6×0.44(112)

= 3.96 J

(b) Torque on magnet when moment is aligned opposite to the field,

τ=mB sin θ

putting the values,

= 6×0.44×sin 180

= 0(sin 180=0)

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