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A magnetic dipole of moment $2.5 \mathrm{Am}^2$ is free to rotate about a vertical axis passing through its centre. It is released from East-West direction. Its kinetic energy at the moment, it takes North-South position is $\left(B_H=3 \times 10^{-5} \mathrm{~T}\right)$
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The correct answer is:
$75 \mu \mathrm{J}$
When magnetic dipole is released from $\mathrm{E}-\mathrm{W}$ a torque acts on it. So, in the displacement from E-W to N-S work is done by the torque.
$\begin{aligned}
& \mathrm{KE}=\text { work done } \& \mathrm{~W}=\int_{\theta_1}^{\theta_2} \tau . d \theta \\
& =\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\
& \therefore \mathrm{KE}=\mathrm{MB} \cos 0^{\circ} \\
& =2.5 \times 3 \times 10^{-5} \\
& =7.5 \times 10^{-5} \mathrm{~J}=75 \times 10^{-6} \mu \mathrm{J}
\end{aligned}$
$\begin{aligned}
& \mathrm{KE}=\text { work done } \& \mathrm{~W}=\int_{\theta_1}^{\theta_2} \tau . d \theta \\
& =\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\
& \therefore \mathrm{KE}=\mathrm{MB} \cos 0^{\circ} \\
& =2.5 \times 3 \times 10^{-5} \\
& =7.5 \times 10^{-5} \mathrm{~J}=75 \times 10^{-6} \mu \mathrm{J}
\end{aligned}$
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