Question

A thin magnetic needle oscillates in the horizontal plane with a time period of $2.0$ sec. If the needle is broken into $4$ equal parts perpendiculars to its length, then the time period (in second) of each part will be

Answer 1

Step 1: cutting of magnet

when a magnet is cut perpendicular to its length , the pole strength remains same

$m=$ pole strength of original magnet

$\mu =mL=$ magnetic moment of original magnet

for each one-fourth cut part :

Pole strength $=m$

Magnetic moment of each part $=m\frac{L}{4}$

$=\frac{\mu }{4}$

Step 2: moment of inertia

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

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

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

$I^{\prime }=\frac{\left(\frac{M}{4}\right)(L/4{)}^2}{12}$

$=\frac{1}{64}\cdot \frac{m{L}^2}{12}$

$=\frac{I}{64}$

Step 3: Time period

time period of oscillation of magnet is given by $T=2\pi \sqrt{\frac{\text{moment of inertia}}{\left(\text{magnetic moment ) (magnetic field}\right)}}$

for original magnet

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

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

$I\text{'}=\frac{I}{64}$

for each cut part , new time period $=2\pi \sqrt{\frac{\left(\frac{I}{64}\right)}{\left(\frac{\mu }{4}\right)B}}=\frac{T}{4}=\frac{2s}{4}$

$=0.5sec$⁡