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A short bar magnet of magnetic moment m is equal to 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?

5.4. 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?

Answers (1)
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Given,

Magnetic moment of magnet, m = 0.32 JT^{-1}

Magnetic field strength, B = 0.15 T

(a) Stable equilibrium: When the magnetic moment is along the magnetic field i.e. \theta = 0^{\circ}

(b) Unstable equilibrium: When the magnetic moment is at 180° with the magnetic field i.e. \theta = 180^{\circ}

(c) We know that,

        U = - m.B = -mBcos\theta

By putting the given values:

U = (-0.32)(0.15)(cos0^{\circ}) = -0.048 J

Therefore, Potential energy of the system in stable equilibrium is -0.048 J

Similarly,

U = (-0.32)(0.15)(cos180^{\circ}) = 0.048 J

Therefore, Potential energy of the system in unstable equilibrium is 0.048J.

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