A bar magnet of magnetic dipole moment `10 Am^(2)` is in stable equilibrium . If it is rotated through `30^(@)` , find its potential energy in new position [Given , B = 0.2 T]

A bar magnet of magnetic dipole moment 8 Am^(2) has poles separated by 0.4 m . Find the pole strength of bar magnet .

A bar magnet of magnetic dipole moment 8 Am^(2) has poles separated by 0.4 m . Find the pole strength of bar magnet .

A bar magnet of magnetic moment 2.0 A m^(2) is free to rotate about a vertical axis passing through its centre, the magnet is released from rest from east-west position. Then, the kinetic energy of the magnet as it takes north-south position is (horizontal component of earth field is 25 mu T)

A magnet of magnetic dipole moment 5.0 Am^(2) is lying in a uniform magnetic field of 7 xx 10^(-4) T such that its dipole moment vector makes an angle of 30° with the field. The work done in increasing this angle from 30° to 45° ls about ......... J.

The work done in rotating a bar magnet of magnetic moment M from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field B is

STATEMENT-1 : If potential energy of a dipole in stable equilibrium position is zero, its potential energy is unstable equilibrium position will be 2pE, where p and E represent the dipole moment and electric field respectively and STATEMENT-2 : The potential energy of a dipole is minimum in stable equilibruim position and maximum in unstable equilibrium position.