Question

Time period for a magnet is T. If it is divided in four equal parts along its axis and perpendicular to its axis as shown then time period for each part will be

• A. 4T
• B. T/4
• C. T/2
• D. T

Answer 1

Answer: (C)

T/2

$T=2\pi \sqrt{\frac{I}{MB}}$

Where I is moment of inertia = $\frac{m{l}^2}{12}$

$m$, is the mass and l is the length of the rod

$M$ is the magnetic dipole moment = $ql$

Where $q$ is pole strength

$\therefore$ New inertia = $\frac{\frac{m}{4}(\frac{l}{2}{)}^2}{12}=\frac{m{l}^2}{16\times 12}\Rightarrow \frac{I}{16}$

New magnetic dipole moment = $\frac{q}{2}\times \frac{l}{2}=\frac{ql}{4}=\frac{M}{4}$

$\therefore$ $T^{\prime }=2\pi \sqrt{\frac{\frac{I}{16}}{\frac{MB}{4}}}=\frac{1}{2}\times 2\pi \sqrt{\frac{I}{MB}}$

$\therefore$ $T^{\prime }=\frac{T}{2}$