Question

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $T$. The magnet is cut along its length into six parts and these parts are then placed together as shown in figure. The time period of this combination will be

• A. Zero
• B. $T$
• C. $\frac{T}{\sqrt{3}}$
• D. $\frac{T}{2\sqrt{3}}$

Answer 1

The correct option is D $\frac{T}{2\sqrt{3}}$

When a magnet is cut lengthwise, the pole strength remains same

$m=$ pole strength of original magnet

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

for each cut part:

Pole strength $=m$

Magnetic moment of each part

$=m\frac{L}{6}=\frac{\mu }{6}$

let $M=$ mass of original magnet,

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

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

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

$I^{\prime }=\frac{\left(\frac{M}{6}\right){\left(\frac{L}{6}\right)}^2}{12}=\frac{1}{6^3}.\frac{M{L}^2}{12}=\frac{I}{6^3}$

moment of inertia of system

$=6\times \frac{I}{6^3}=\frac{I}{6^2}$

it can be observed that upper two magnets are placed in $S-N$ direction.
Net magnetic moment

$=\frac{4\mu }{6}-\frac{2\mu }{6}=\frac{\mu }{3}$

for original magnet

time period of oscillation of magnet is given by

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

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

Final time period

$T^{\prime }=2\pi \sqrt{\frac{\frac{I}{6^2}}{\frac{\mu }{3}\times B_H}}=\frac{T}{\sqrt{12}}=\frac{T}{2\sqrt{3}}$

Hence, the correct option id (c)