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Step-by-Step Solution
Step 1: Write the general expression for the time period of a bar magnet
When a bar magnet oscillates in a uniform magnetic field, its time period $T$ is given by
$T = 2\pi \sqrt{\frac{I}{M B}},$
where:
$I$ is the moment of inertia of the magnet about its center.
$M$ is the magnetic moment of the magnet.
$B$ is the magnetic field strength.
For a thin, long magnet of mass $m$ and length $\ell$, the moment of inertia (about the center perpendicular to its length) is
$I = \frac{1}{12} m \ell^2.$
Step 2: Understand the modification – cutting the magnet into three equal parts
The original magnet is cut along its length into three equal pieces and these pieces are placed one over the other with like poles together.
Step 3: Analyze the new magnetic moment ($M'$)
The magnetic moment $M$ of a bar magnet can be written as:
$M = \text{pole strength (}m\text{)} \times \ell.$
After cutting into three pieces each of length $\ell/3$, if these pieces are placed so that their like poles coincide, the total effective length for the vector part remains the same overall (since we stack them aligned in the same orientation). Therefore, the net magnetic moment remains:
$M' = m \times \left(\frac{\ell}{3}\right) \times 3 = m \ell = M.$
Step 4: Calculate the new moment of inertia ($I'$)
Each piece has mass $m/3$ and length $\ell/3$. The moment of inertia of one piece about its midpoint is
$I_{\text{piece}} = \frac{1}{12}\left(\frac{m}{3}\right)\left(\frac{\ell}{3}\right)^2.$
Since there are three such pieces stacked together, and they share the same pivot axis, the total new moment of inertia is
$I' = 3 \times I_{\text{piece}}
= 3 \times \frac{1}{12}\left(\frac{m}{3}\right)\left(\frac{\ell}{3}\right)^2 = \frac{I}{9},$
where $I = \frac{1}{12} m \ell^2$ is the original moment of inertia.
Step 5: Determine the new time period ($T'$)
Since the magnetic moment remains $M$, but the moment of inertia becomes $I/9$, the new time period is:
$T' = 2\pi \sqrt{\frac{I'}{M B}}
= 2\pi \sqrt{\frac{I/9}{M B}}
= \frac{1}{3} \left(2\pi \sqrt{\frac{I}{M B}}\right)
= \frac{T}{3}.$
Given the original time period $T = 2\ \text{s}$, we have
$T' = \frac{2\ \text{s}}{3} = \frac{2}{3}\ \text{s}.$
Final Answer
$\displaystyle \frac{2}{3}\,\text{s}$
