Question

The time period of a freely suspended magnet is $4$ seconds. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be

• A. $4$ sec
• B. $0.5$ sec
• C. $2$ sec
• D. $0.25$ sec

Answer 1

The correct option is C $2$ sec

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


$m$ = pole strength of original magnet
$\mu =m\left(2L\right)=$ magnetic moment of original magnet

For each half cut part:
Pole strength $=m$
Magnetic moment of each part
${\mu }^{\prime }=mL=\frac{m\left(2L\right)}{2}=\frac{\mu }{2}$

Let,

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

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

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

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

$\Rightarrow I^{\prime }=\frac{1}{8}\frac{M{L}^2}{3}$

$\Rightarrow I^{\prime }=\frac{I}{8}$

Time period of oscillation of magnet is given by

$T=2\pi \sqrt{\frac{I}{M{B}_H}}$

For original magnet,

$T=2\pi \sqrt{\frac{I}{\mu B_H}}=4sec$

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

New time period for each half part,

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

$\Rightarrow T^{\prime }=\frac{1}{2}.2\pi \sqrt{\frac{I}{\mu B_H}}sec$

$\Rightarrow T^{\prime }=\frac{1}{2}\times 4sec$

$\Rightarrow T^{\prime }=2sec$

Hence, the correct option is (B)