Question 13

A bar magnet of magnetic moment 6 J/T is aligned at 60° 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).

Answer

(a) (i)

M = 6 J/T

Q_{1} = 60º Q_{2} = 90º

B = 0.44T

W = MB(cosQ_{1} - cosQ_{2})

= 6 x 0.44(cos60º - cos90º)

= 6 x 0.44 x ½

= 1.32 J

(a) (ii)

Q_{1} = 60º Q_{2} = 180º (opposite to magnetic field)

W = MB(cosQ_{1} - cosQ_{2})

= 6 x 0.44(cos60º - cos90º)

= 6 x 0.44 x (½ - (-1))

= 6 x 0.44 x 3/2

= 3.96 J

(b) = M X B

= MB sin Ø

= 6 x 0.44 x sin 180º

= 6 x 0.44 x 0

= 0

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