Question

A bar magnet of magnetic moment $M$ and moment of inertia $I$ (about centre and perpendicular to length) is cut into two equal pieces, perpendicular to its length. Let $T$ be the period of oscillations of the original magnet about an axis through the mid-point, perpendicular to length, in a magnetic field $\vec{B}$. If the similar period $T^{\prime }$ for each piece is $\frac{T}{n}$, then calculate $n$.

Answer 1

Answer: 2

The period of oscillations of the original magnet is given by
$T=2\pi \sqrt{\frac{I}{MB}}$
Let $M^{\prime }$ be the magnetic moment and $I$ be the moment of inertia of a piece of the magnet. Then, its period of oscillations is given by
$T^{\prime }=2\pi \sqrt{\frac{I^{\prime }}{M^{\prime }B}}$
As the bar magnet is cut into two equal pieces, a piece has the same pole strength but length equal to one half of that original magnet. Therefore, magnetic moment of the piece is
$\therefore M^{\prime }=\frac{M}{2}$
Further, the mass of one piece of the magnet is equal to one half of that original magnet. Further, as the moment of inertia depends on the square of the length of an object in the form of a bar and its mass,
$\therefore I^{\prime }=\frac{1}{2}\times \frac{1}{4}I=\frac{I}{8}$
It may be pointed out that the factor $\frac{1}{2}$. appears due to its half mass and the factor $\frac{1}{4}$ appears due to its half length in reference to the original magnet.
Hence, $T^{\prime }=2\pi \sqrt{\frac{I/8}{(M/2)B}}=\frac{T}{2}$