Question: Q. 6. A short bar magnet of magnetic moment $m=0.32 \mathrm{JT}^{-1}$ is placed in a uniform magnetic field of $0.15 \mathrm{~T}$. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and unstable equilibrium? (b) What is the potential energy of the magnet in each case?
U] [Foreign 2013]
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Solution:
Ans. External magnetic field,
$$ B=0.15 \mathrm{~T} $$
Magnetic moment of the bar magnet,
$$ m=0.32 \mathrm{~J} \mathrm{~T}^{-1} $$
(a) In stable equilibrium, the orientation would be such that the direction of magnetic moment and magnetic field is parallel to each other (angle between them $=0^{\circ}$ )
In unstable equilibrium, the orientation would be such that the direction of magnetic moment and magnetic field is parallel to each other (angle between them $=180^{\circ}$ )
(b) Potential energy stored in a magnet,
$$ \mathrm{U}=-m B \cos \theta $$
In case of stable equilibrium
$$ \begin{aligned} & \mathrm{U}=-0.32 \times 0.15 \cos 0^{\circ} \mathrm{J} \ & \mathrm{U}=-0.048 \mathrm{~J} \end{aligned} $$
(-ve sign shows Work (energy) is required to re-align the magnetic dipole in an external $B$ field) 1 In case of unstable equilibrium
$$ \begin{aligned} & \mathrm{U}=-0.32 \times 0.15 \cos 180^{\circ} \mathrm{J} \ & \mathrm{U}=-0.32 \times 0.15 \times-1 \ & \mathrm{U}=0.048 \mathrm{~J} \end{aligned} $$
(It has maximum energy in this orientation). 1
