Q. 44.3( 20 Votes )

A short bar magne

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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