Q. 44.5( 17 Votes )

# 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 :

Given:

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

Strength of external magnetic field, B = 0.15T

(a) When the magnetic moment is aligned (0°) with the magnetic field then we consider this as stable equilibrium, if the magnet is rotated, then it will have the tendency to come back in this position.

θ = 0°

We know that,

U = - **M**.**B**

U = -M × B × Cos θ …(1)

Where, U = potential energy

M = magnetic moment

B = magnetic field

θ = Angle between filed and magnetic moment

By putting the given values in the equation (1), we have,

U = -0.32 × 0.15 × Cos 0°

U = -0.048 J

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

(b) Whereas, in case of unstable equilibrium the moment is at 180° with the magnetic field. If the magnet is rotated then it will never come back in the initial position.

θ = 180°

Now, by putting values in equation (1), we get,

U = -0.32 × 0.15 × Cos 180°

U = 0.048 J

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

__Note: The direction of the magnetic moment is from south pole to north pole inside the magnet.__

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