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Q.

A bar magnet is at right angles to a uniform magnetic field. The couple acting on the magnet is to be one fourth by rotating it from the position. The angle of rotation is

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a

sin^–1(0.25)

b

90°–sin^–1(0.25)

c

cos^–1(0.25)

d

90°–cos^–1(0.25)

answer is B.

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Detailed Solution

$τ = MBsin [\frac{π}{2} - θ] w h e r e θ i s t h e a n g l e o f r o t a t i o n .$

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