A bar magnet of magnetic moment `4Am^(2)` is free to rotate about a vertical axis passing through it centre.The magnet is released from rest from east-west position.Then K.E of the magnet as it takes north-south position is `B_(H)=20 mu T`

A bar magnet of magnetic moment 2Am^(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 KE of the magnet as it takes N-S position is (B_(H)=25muT)

49.A bar magnet of magnetic moment 2Am^(2) is free to rotate about a vertical axis passing through its centre.The magnet is released from rest east-west position.Then the K.E of the magnet as it takes north-south position is (2n+40)mu J then "n" is ( B_(H)=25 mu T )

A bar magnet of moment 40 A-m^(2) is free to rotate about a vetical axis passing through its centre. The magnet is released from rest from east west direction. The kinetic energy of the magnet as it takes north-south direction is (B_(H)=30muT)

A bar magnet of magnetic moment 2.0 A m^2 is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east-west position. Find the kinetic energy of the magnet as it takes the north-south position. The horizontal component of the earth's magnetic field is B = 25 muT .

A bar magnet of magnetic moment 2.0 A-m^2 is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east west position. Find the kinetic energy of the magnet as it takes the north south position. The horizontal component of the earth's magnetic field as B=25muT . Earth's magnetic field is from south to north.

A bar of magnetic moment 2.5 Am^(2) is free to rotate about a vertical axis through its centre . The magnet is released from the rest from the east - west direction . Find the kinetic energy of the magnet as it aligns itself in the north - south direction . The horizontal component of earth's magnetic field is 0.3 G

A bar magnet of magnetic moment 20 Am^(2) is divided into two equal parts by cuting it along its axis. The magnetic moment of each part will be

A short bar magnet of magnetic moment 3.2 Am^(2) is placed on a horizontal table with its axis along the magnetic E-W direction.The resultant horizontal magnetic induction on its equator at a distance of 20 cm from its centre is (B_(H)=4 xx10^(-5) T)

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]