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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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Text solutionVerified
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. |
Topic | Magnetism, EMI and AC |
Subject | Physics |
Class | Class 11 |
Answer Type | Text solution:1 |