Question

the period of oscillation of a bar magnet in a vibration magnetometer is measured to be T. the bar magnet is now carefully cut into two equal pieces without affecting strength what will be the period of oscillation of one of the pieces

Answer 1

time period of bar magnet is given by,
$T=2\pi\sqrt{\frac{I}{MB}}$

where T is time period of oscillation of bar magnet, I is moment of inertia of bar magnet, M is magnetic moment and B is magnetic field intensity of bar magnet.

Let I is initial moment of inertia ,
M is magnetic moment
B is magnetic field intensity .

now bar magnet is cut into two equal piece without affecting strength
so, magnetic moment of each piece , M' = M/2
moment of inertia of each piece , I'= I/8 [ as we know, moment of inertia is directly proportional to square of length of rod and mass ]

now, $T'=2\pi\sqrt{\frac{I'}{M'B}}$

$T'=2\pi\sqrt{\frac{I/8}{M/2B}}$

$T'=2\pi\sqrt{\frac{I}{4MB}}$

hence, T' = T/2

hence, time period of each piece of bar magnet is half of time period of bar magnet.