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A short bar magnet of magnetic moment m = 0.32 JT–1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
A short bar magnet of magnetic moment m = 0.32 JT–1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
CBSE | Class 12 | Magnetism and Matter | NCERT | Physics
A short bar magnet of magnetic moment m = 0.32 JT–1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Answer –

Moment of the bar magnet is given by M = 0.32 JT–1 

External magnetic field, is given by B = 0.15 T

  • The magnetic field is aligned with the bar magnet. This system is said to be in a state of stable equilibrium. As a result, the angle Θ between the bar magnet and the magnetic field is 0.

Potential energy of the system is given by –

\[-MB\cos \theta =-0.32\times 0.15\times \cos {{0}^{\bullet }}\]

\[=-4.8\times {{10}^{-2}}J\]

  • In this case, the bar magnet is oriented at 180° to the magnetic field. Hence, it is in unstable equilibrium. Here , θ=180°

Potential energy of the system is given by –            

\[-MB\cos \theta =-0.32\times 0.15\times \cos {{180}^{\bullet }}\]

\[=4.8\times {{10}^{-2}}J\]

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