First slide

Magnetic dipole in a uniform magnetic field

Question

A bar magnet of moment 2 A-m² is free to rotate about a Vertical axis passing through its centre. The magnet is released form rest from east west direction. The K.E of the magnet as it takes north-south direction is (B_H = 25 × 10^–6 T)

Moderate

A

25 × 10^–6 J
B

50 × 10^–6 J
C

75 × 10^–6 J
D

100 × 10^–6 J

Solution

$∆ KE = ∆ PE = MB (1 - cosθ) = 2 x 25 x 10^{- 6} (1 - 0) = 50 x 10^{- 6} J$

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