Q. A deflection magnetometer is adjusted in the usual way. When a magnet is introduced, the deflection observed is$\ \theta$, and the period of oscillation of the needle in the magnetometer is T. When the magnet is removed, the period of oscillation is $T_{0}.$ The relation between T and $T_{0}$ is
A) $T^{2} = T_{0}^{2}\cos\theta$
B) $T = T_{0}\cos\theta$
C) $T = \frac{T_{0}}{\cos\theta}$
D) $T^{2} = \frac{T_{0}^{2}}{\cos\theta}$
Explanation
(a) For first case
$T = 2\pi\sqrt{\frac{I}{M\sqrt{F^{2} + H^{2}}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
When magnet is removed
$T_{0} = 2\pi\sqrt{\frac{T}{MH}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)\ $
Also, $\frac{F}{H} = \tan\theta$
From Eqs. (i) and (ii) we have
$\frac{T}{T_{0}} = \sqrt{\frac{H}{\sqrt{F^{2} + H^{2}}}}$
$\frac{T^{2}}{\ T_{0}^{2}} = \cos{\theta\ T^{2} = T_{0}^{2}\cos\theta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.