Question

A long bar magnet of time period $T$ is cut into four equal parts by cutting it perpendicular to both length and breadth. The time period of each part is

• A. $\frac{T}{4}$
• B. $\frac{T}{16}$
• C. $\frac{T}{32}$
• D. $\frac{T}{2}$

Answer 1

The correct option is D $\frac{T}{2}$

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

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 =$\frac{\left(\frac{m}{4}\right)(\frac{l}{2}{)}^2}{12}=\frac{m{l}^2}{16\times 12}=\frac{I}{16}$

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

$\therefore T^{\prime }=2\pi \sqrt{\frac{\frac{I}{16}}{\frac{M}{4}B}}$

$=\frac{1}{2}\times 2\pi \sqrt{\frac{I}{MB}}$

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