# A short bar magne

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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