Question

A bar magnet of magnetic moment $M$ and moment of inertia $I$ (about centre perpendicular to length) is cut into two equal pieces perpendicular to length. Let . $T$ be the period of oscillations of the original magnet about an axis through the mid point perpendicular to length in magnetic field $B$. The similar period $T'$ for each piece would be

• A. $ \frac{T}{2}$
• B. $ \frac{3T}{4}$
• C. $ \frac{5T}{2}$
• D. $T$

Answer 1

Answer: (A)

The moment of inertia of the bar magnet of mass $m$, length $l$ about an axis passing through its centre and perpendicular to its length.
$I = \frac{ml^{2}}{12}$
and magnetic dipole moment is $M$.
When the magnet is cut into two equal pieces perpendicular to length then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre
$I' = \frac{m}{2} \frac{\left(l/ 2\right)^{2}}{12} = \frac{ml^{2}}{12} \times \frac{1}{8} = \frac{I}{8}$
and magnetic dipole moment, $M' = \frac{M}{2}$
Now, time period of oscillation, $T = 2\pi \sqrt{\frac{I}{MB}}$
and time period of oscillation on each piece of magnet
$T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{\frac{I}{8}}{\frac{M}{2}B}} \Rightarrow T' = \frac{2\pi}{2} \sqrt{\frac{I}{MB}} = \frac{T}{2}$