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A bar magnet of magnetic moment is free to rotate about a vertical axis passing through its centre. The magnet is released from rest from east-west position. Then the KE of the magnet as it takes N-S position is
a. b. c. d.

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Sure, here are the step-by-step solutions: Given: To solve for , we need to isolate it on one side of the equation. We can start by dividing both sides of the equation by : Next, we can divide both sides of the equation by \left({{\cos{\theta}}_{{{1}}}-}{\cos{\theta}}_{{{2}}}}\right): \dfrac{W}{M\left({{\cos{\theta}}_{{{1}}}-}{\cos{\theta}}_{{{2}}}}\right)} = B Therefore, the solution is: B = \dfrac{W}{M\left({{\cos{\theta}}_{{{1}}}-}{\cos{\theta}}_{{{2}}}}\right)}
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Question Text
A bar magnet of magnetic moment is free to rotate about a vertical axis passing through its centre. The magnet is released from rest from east-west position. Then the KE of the magnet as it takes N-S position is
a. b. c. d.
TopicMagnetism, EMI and AC
SubjectPhysics
ClassClass 11
Answer TypeText solution:1