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Question
The work done for rotating a magnet with magnetic dipole moment m, through 90° from its magnetic meridian is n times the work done to rotate it through 60°. Find the value of n.
Solution
Data: θ0 = 0°, θ1 =90°, θ2 = 60°, W1 = nW2
The work done by an external agent to rotate the magnet from θ0 to θ is
W = MB (cos θ0 - cos θ)
∴ W1 =MB (cos θ0 - cos θ1)
= MB (cos 0° - cos 90°)
= MB (1 - 0)
= MB
∴ W2 = MB (cos 0° - cos 60°)
= MB `(1 - 1/2)`
= 0.5 MB
∴ W1 = 2W2 = MB
Given W1 = nW2 . Therefore n = 2
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