The magnetic moment of magnet is 1 joule per tesla. How much work is done in turning it through `90^(@)` from the magnetic meridian at a place where `B = 1.6 xx 10^(-5)T` ?

Magnetic moment of bar magnet is M. The work done to turn the magnet by 90^(@) of magnet in direction of magnetic field B will be

The work done in rotating a magnet of pole strength 1 A-m and length 1 cm through an angle of 60^(@) from the magnetic meridian is (H=30A//m)

A magnet of dipole moment 2Am^(2) is deflected through 30^(@) from magnetic meridian. The required deflecting torque is (B_(H)=0.4xx10^(-4)T)

The magnetic moment of ion is close to 36 xx 10^(-24) "joule"//"Tesla" . The number of unpaired electrons of the ions are

The work done in turning a magnet of magnetic moment M by an angle of 90^(@) from the magnetic meridian is n times the corresponding work done to turn through an angle of 60^(@) . What is the value of n ?

A magnet of moment M is lying in a magnetic field of induction B. W1 is the work done in turning it from 0^(@) to 60^(@) and W_(2) is the work done in turning it from 30^(@) to 90^(@) . Then

A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through 180^(@) to deflect the magnet by 30^(@) from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through 270^(@) to deflect the magnet 30^(@) from magnetic meridian. The ratio of magnetic moment of magnets is