Potential energy of magnetic dipole:
Work done to rotate a magnetic dipole in a uniform magnetic field is stored as potential energy of the magnetic dipole.
If a magnetic dipole of magnetic dipole moment \( \vec{m} \) is placed at an angle \( \theta \) with respect to uniform magnetic field of strength \( \vec{B} \), then torque experienced by it is given by:
\[\tau = mB \sin \theta\]
If the dipole rotates through an angle \( d\theta \), then work done on it is:
\[dW = \tau d\theta = mB \sin \theta d\theta\]
Total work done to rotate the dipole from \( \theta_1 \) to \( \theta_2 \) position is:
$W = \int_{\theta_1}^{\theta_2} mB \sin \theta d\theta$
$W= mB \int_{\theta_1}^{\theta_2} \sin \theta d\theta$
$W= mB [-\cos \theta]_{\theta_1}^{\theta_2}$
$W= -mB [\cos \theta_2 - \cos \theta_1]$
This work done is stored as the potential energy \( U \) of the magnetic dipole:
$\therefore U = W = -mB (\cos \theta_2 - \cos \theta_1)$
$U= mB (\cos \theta_1 - \cos \theta_2)$
If \( \theta_1 = 90^\circ \) and \( \theta_2 = \theta \), then \( \cos \theta_1 = \cos 90^\circ = 0 \) and \( \cos \theta_2 = \cos \theta \).
Therefore:
$U = W = -mB \cos \theta = -\vec{m} \cdot \vec{B}$
Special Cases:
1. When \( \vec{m} \) and \( \vec{B} \) are anti-parallel (\( \theta = 180^\circ \)):
$U = mB$
The magnetic dipole has maximum potential energy and is in unstable equilibrium.
2. When \( \vec{m} \) and \( \vec{B} \) are parallel (\( \theta = 0^\circ \)):
$U = -mB$
The dipole has minimum potential energy and is in stable equilibrium.