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

Given I=moment of inertia of the bar magnet
m=mass of bar magnet
l= magnet of manget about an any passin g through its centre and perpendicular to its length.
M=magnetic moment of the magnet
B=uniform magnetic field in which magnet is oscillating. we get time period of
oscillation is, `T=2pi sqrt((T)/(MB))`
Here, `I=(ml^(2))/(12)`
When magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is
`I^(@)=(m)/(2)((l//2)^(2))/(12)=(ml^(2))/(12)xx(1)/(B)=(l)/(B)`
Magnetic dipole moment `m =M//2`
Its time period of oscillation is
`T'=2pi sqrt( (I)/(M//B))=2pisqrt((I//B)/((M//2)B))=(2pi)/(2)sqrt((T)/(MB))`