Q. A small bar magnet $A$ oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60{^\circ}$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be
A) $\frac{T}{\sqrt{2}}$
B) $T$
C) $\sqrt{2}T$
D) $2T$
Explanation
(a) $T = 2\pi\sqrt{\frac{1}{MB}} \Rightarrow \frac{T}{T^{'}} = \sqrt{\frac{B^{'}}{B}} = \sqrt{\frac{B}{B_{H}}}$
$\frac{T}{T^{'}} = \sqrt{\frac{1}{\cos\phi}} = \sqrt{\frac{1}{\cos{60{^\circ}}}} = \sqrt{2} \Rightarrow T^{'} = \frac{T}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.