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 $T_{0}$.Find 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) In the usual setting of deflecting magnetometer, field due to magnet $(F)$ and horizontal component $(H)$ of earth’s field are perpendicular to each other. Therefore, the net field on the magnetic needle is $\sqrt{F^{2} + H^{2}}$
$\therefore T = 2\pi\sqrt{\frac{I}{M\sqrt{F^{2} + H^{2}}}}$ …(i)
When magnet is removed,
$T_{0} = 2\pi\sqrt{\frac{I}{MH}}$ …(ii)
Also, $\frac{F}{H} = tan\theta$
Dividing Eq. (i) by Eq.(ii), we get
$\frac{T}{T_{0}} = \sqrt{\frac{H}{\sqrt{F^{2} + H^{2}}}}$
$= \sqrt{\frac{H}{\sqrt{H^{2}\tan^{2}{\theta + H^{2}}}}} = \sqrt{\frac{H}{H\sqrt{\sec^{2}\theta}}}$
$= \sqrt{cos\theta}\ or\ \frac{T_{2}^{2}}{T_{0}^{2}} = cos\theta$
$T^{2} = T_{0}^{2}cos\theta.$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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