A bar magnet of magnetic moment `5Am^(2)` is suspended in a uniform magnetic field of strength "2T" making an angle `60^(0)` with the direction of the field.The potential energy of the magnet in that position is (A) 5J (B) -5J (C) 10J (D) -10J

A magnet of magnetic moment 20 A - m^(2) is freely suspended is a uniform magnetic field of strength 6 T . Work done to rotate it by 60^(@) will be .

The potential enargy of bar magnet of magnetic moment 8 Am ^(2) placed in a uniform magnetic field of 2 T at an angle of 120^(@) is equal to

A bar magnet of magnetic moment M is divided into 'n' equal parts cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength 2T and held making an angle 60^(@) with the direction of the field. When the magnet is released the K.E of the magnet in the equilibrium position is

A bar magnet of length 0.2 m and pole strength 5 Am is kept in a uniform magnetic induction field of strength 15 wb//m^(-2) making an angle of 30^(@) with the field. Find the couple acting on it.

A bar magnet of magnetic moment of 5*0Am^2 has poles 20cm apart. Calculate the pole strength.

A bar magnet of magnetic moment 0.4 Am^(2) is placed in a uniform magnetic field of induction 5 xx10^(-2) tesla. What is the torque acting on the magnet if the angle between the magnetic induction and the magnetic moment is 30^(@) .