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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Magnetism and Matter
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A bar magnet of magnetic moment $4 A-m^2$ is free to rotate about a vertical axis through its centre. The magnet is released from rest from east-west position. Kinetic energy of the magnet in the north- south position will be $(H = 25 \mu T)$

$\begin {array} {1 1} (a)\;10^{-2}J & \quad (b)\;10^{-4}J \\ (c)\;10^{-6}J & \quad (d)\;0 \: J \end {array}$


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Loss in $PE = \Delta (MB\: cos\theta )$
$ \Rightarrow 4 \times 25 \times 10^{-6} = 10^{-4}$
Ans : (b)
answered Feb 21, 2014 by thanvigandhi_1

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