Question

A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one-piece is made to oscillate freely in the same field. If its period of oscillation is T’, the ratio $\dfrac{T’}{T}$ is
A. $\dfrac{1}{{2\sqrt 2 }}$
B. $\dfrac{1}{2}$
C. $2$
D. $\dfrac{1}{4}$

Answer 1

To find the ratio of time period after and before breaking the magnet, we can use the formula for the time period of the oscillating magnet. The magnetic field intensity is kept constant, so we can obtain a relationship between time period, magnetic moment, and moment of inertia. We need to find the moment of inertia and the magnetic moment after the magnet is broke into two halves.
Formula used:
\eqalign{ & I = \dfrac{1}{{12}}m{L^2} \cr & M = m \times 2l \cr & T = 2\pi \sqrt {\dfrac{I}{{MB}}} \cr}

Complete step-by-step solution:
Initially, the magnet has a length, say, $l$. When it is broken into half the length will be $\dfrac{l}{2}$. This also changes the moment of inertia of the magnet. If we take the initial moment of inertia to be $I$, it is mathematically given by
$I = \dfrac{1}{{12}}m{L^2}$
Where m is the mass of the magnet and L is the geometric length of the magnet.
When the magnet is broken into two pieces, the mass is also halved. The new moment of inertia, $I'$ of the magnet, will be given by
\eqalign{ & I' = \dfrac{1}{{12}} \times \dfrac{m}{2} \times {\left( {\dfrac{L}{2}} \right)^2} \cr & \Rightarrow I' = \dfrac{1}{8} \times \dfrac{1}{{12}}m{L^2} \cr & \Rightarrow I' = \dfrac{I}{8} \cr}
Similarly, the magnetic dipole moment is initially given by
$M = m \times 2l$
Where,
M is the magnetic dipole moment
m is the magnetic pole strength
l is the magnetic length of the magnet
When the magnet is broken into two halves the magnetic pole strength remains the same but the magnetic length will change. The new magnetic dipole strength, say, M’ will be
\eqalign{ & M' = m \times \left( {\dfrac{{2l}}{2}} \right) \cr & \Rightarrow M' = \dfrac{1}{2} \times \left( {m \times 2l} \right) \cr & \Rightarrow M' = \dfrac{M}{2} \cr}
The time period for an oscillating magnet is given by
$T = 2\pi \sqrt {\dfrac{I}{{MB}}}$
Where,
T is the time period of the oscillation
I is the moment of inertia of magnet
M is the magnetic dipole moment
B is the magnetic field to which the magnet is subjected
Since the magnetic field B is constant, we have
$T \propto \sqrt {\dfrac{I}{M}}$
From, this relationship we can write
\eqalign{ & \dfrac{{T'}}{T} = \sqrt {\dfrac{{I'}}{{M'}} \times \dfrac{M}{I}} \cr & \Rightarrow \dfrac{{T'}}{T} = \sqrt {\dfrac{{\dfrac{I}{8}}}{{\dfrac{M}{2}}} \times \dfrac{M}{I}} \cr & \Rightarrow \dfrac{{T'}}{T} = \sqrt {\dfrac{2}{8}} = \dfrac{1}{2} \cr & \therefore \dfrac{{T'}}{T} = \dfrac{1}{2} \cr}
Therefore, the correct option is B.

Note: The magnetic pole strength of a magnet is the force exerted by one of the faces on the face of another similar magnet. Though the magnet is broken into half, the magnetic pole strength will remain constant. In a bar magnet, the magnetic poles are situated a smaller distance than the actual length of the magnet. Therefore, it is different from the geometric length and called magnetic length.