A bar magnet of magnetic moment M and moment of inertia I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscillation of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field `vecB`. What would be the similar period `T^'` for each piece?

Explanation

Moment of inertia of bar magnet (rectangular rod) about an axis passing through its centre and perpendicular to its length is:
` I = (ml^2)/(12)`
Where l is the length of the bar magnet and m is its mass. Time period of the bar magnet oscillating in an external magnetic field is given by
`T = 2pi sqrt((I)/(MB))`
Where M is the magnetic moment of bar magnet and B is the external magnetic field. Moment of inertia of bar magnet when it is cut into two equal pieces will become
`I = (m/2(l/2)^2)/(12) = (ml^2)/(12) xx 1/8 = I/8`

Magnetic dipole moment of each piece will also become half. New magnetic dipole moment will be
M. = M/2
Time period of oscillation will become
` T = 2pi sqrt( (I.)/(M.B)) = 2pi sqrt( (I/8)/(M/2 B) ) = T/2`