Question

The time period of a thin magnet in earth's magnetic field is $T$. If the magnet is cut into two equal parts perpendicular to its length, the time period of each part in the same field will be

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

Answer 1

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


Magnetic moment of original magnet,
$\mu =ml$
$(m$= pole strength)

Magnetic moment of each cut parts

${\mu }^{\prime }=\frac{ml}{2}=\frac{\mu }{2}$

(When magnet is cut perpendicular to its length, pole strength remains unchanged)

If $M$ = mass of original magnet,

Mass of each cut part,

$M^{\prime }=\frac{M}{2}$

Moment of inertia of original magnet,

$I=\frac{M{l}^2}{12}$

(about perpendicular bisector axis)

Moment of inertia of cut part magnet,

$I^{\prime }=\frac{\left(M/2\right)(l/2{)}^2}{12}=\frac{I}{8}$

Time period of a magnet is given by,

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

Where, $I$ = moment of inertia of oscillating magnet, $\mu$= magnetic moment, $B_H$ = horizontal component of earth's field

If initial period is $T$, then time period of cut part,

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

$\therefore$ Time period of each part is half of initial time period

Hence, option (A) is correct