Q. Two identical bar magnets are placed one above the other such that they are mutually perpendicular and bisect each other. The time period of this combination in a horizontal magnetic field is $T$. The time period of each magnet in the same field is
A) $\sqrt{2}\ T$
B) $2^{\frac{1}{4}}\ T$
C) $2^{\frac{1}{4}}\ T$
D) $2^{\frac{1}{2}}\ T$
Explanation
(c) Time period of combination
$T = 2\pi\sqrt{\frac{2I}{\sqrt{2}M.H}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
and time period of each magnet
$T^{'} = 2\pi\sqrt{\frac{I}{MH}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From (i) and (ii), we get
$T^{'} = \frac{T}{2^{1/4}} = 2^{- 1/4}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.