Question

The time period of a freely suspended magnet is $2sec$. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period (in seconds) will be

Answer 1

when a magnet is cut along perpendicular bisector, the magnetic length of each part is half of original length but the pole strength remains same

$m=$ pole strength of original magnet
$\mu =mL=$ magnetic moment of original magnet
for each half cut part :
Pole strength $=m$
Magnetic moment of each part $=mL$
$=\frac{m\left(2L\right)}{2}=\frac{\mu }{2}$

let $M$ = mass of original magnet,

$I=$ moment of inertia of original magnet about perpendicular bisector axis

$I=\frac{\left(M\right){\left(2L\right)}^2}{12}=\frac{\left(M\right){\left(L\right)}^2}{3}$

Let $I^{\prime }$ be the new moment of inertia of each cut part

$I^{\prime }=\frac{\left(\frac{M}{2}\right){\left(L\right)}^2}{12}=\frac{1}{8}.\frac{M{L}^2}{3}=\frac{I}{8}$

for original magnet

time period of oscillation of magnet is given by

$T=2\pi \sqrt{\frac{I}{\mu B_H}}=2sec$

(I=moment of inertia of oscillating magnet, $\mu =$ magnetic moment, $B_H=$magnetic field)

New time period for each half part,

$T^{\prime }=2\pi \sqrt{\frac{\frac{I}{8}}{\left(\frac{\mu }{2}\right)\left(B_H\right)}}=\frac{1}{2}.2\pi \sqrt{\frac{I}{\mu B_H}}$

$=\frac{1}{2}\times 2sec$

$=1sec$

Hence the new time period is 1 sec