A bar magnet oscillates with time period 'T' in a uniform magnetic field B. Now the magnet is cut into two equal parts by cutting it perpendicular to its length. The time period of each part in the same magnetic field:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{T}{2} \)
A bar magnet oscillating in a uniform field is a rotational oscillator with time period \( T = 2\pi\sqrt{I/(MB)} \), where M is the magnetic moment and I the moment of inertia about the axis through its centre. The magnetic moment is M = (pole strength) × (length).
Cutting the magnet perpendicular to its length halves its length and mass, but the pole strength of each piece stays the same. For a bar of mass m and length l, \( I = ml^2/12 \).
\( M' = M/2 \)
\( \displaystyle I' = \frac{1}{12}\cdot\frac{m}{2}\cdot\left(\frac{l}{2}\right)^{2} = \frac{I}{8} \)
\( \displaystyle \frac{T'}{T} = \sqrt{\frac{I'}{I}\cdot\frac{M}{M'}} = \sqrt{\frac{1}{8}\times2} = \frac{1}{2} \)
\( \displaystyle T' = \frac{T}{2} \)