Q. The length of a magnet is large compared to it’s width and breadth. The time period of its oscillation in a vibration magnetometer is $T$. The magnet is cut along it’s length into six parts and these parts are then placed together as shown in the figure. The time period of this combination will be
A) $T$
B) $\frac{T}{\sqrt{3}}$
C) $\frac{T}{2\sqrt{3}}$
D) Zero
Explanation
(c) $T = 2\pi\sqrt{\frac{I}{MH}};MI$ of each part $= \frac{I}{6^{3}}$
and magnetic moment of each part $= \frac{M}{6}$
so net $MI$ of system $= \frac{I}{6^{3}} \times 6 = \frac{I}{6^{2}}$
and net magnetic moment $= \frac{4M}{6} - \frac{2M}{6} = \frac{M}{3}$
$\therefore$ Time period of the system
$T^{'} = 2\pi\sqrt{\frac{I/36}{(M/3)H}} = \frac{1}{2\sqrt{3}}2\pi\sqrt{\frac{I}{MH}} = \frac{T}{2\sqrt{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.