Potential Energy of Magnetic Dipole in Magnetic Field || Magnetism Class 12 Physics Notes
JEE Mains & Advanced
Potential Energy of magnetic dipole in a magnetic field is defined as the amount of work done in rotating the dipole from zero potential energy position to any desired position.
A current loop does not experience a net force in a magnetic field. It, however, experiences a torque. This is very similar to the behavior of an electric dipole in an electric field. A current loop, therefore, behaves like a magnetic dipole.
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Potential Energy of a Bar Magnet in Uniform Magnetic Field
When a bar magnet of dipole moment M is kept in a uniform magnetic field B it experiences a torque $\tau=M B \sin \theta$ which tries to align it parallel to the direction of the field.
If the magnet is to be rotated against this torque work has to be done.
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The work done in rotating dipole by small-angle d$\theta $ is $d W =\tau d \theta$
Total work done in rotating it from angle $\theta_{1}$ to $\theta_{2}$ is
$\mathrm{W}=\int \mathrm{dW}=\int_{\theta_{1}}^{\theta_{2}} \tau \mathrm{d} \theta=\mathrm{MB} \int_{\theta_{1}}^{\theta_{2}} \sin \theta \mathrm{d} \theta$
$=\operatorname{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$
This work done in rotating the magnet is stored inside the magnet as its potential energy.
So U = MB $\left(\cos \theta_{1}-\cos \theta_{2}\right)$
The potential energy of a bar magnet in a magnetic field is defined as work done in rotating it from a direction perpendicular to the field to any given direction.
$U = W _{ \theta }- W _{\frac{\pi}{2}}=- MB \cos \theta=-\overrightarrow{ M } \cdot \overrightarrow{ B }$ 
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