Physics - Magnetism Question with Solution | TestHub

In the usual setting of deflection 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 $\sqrt{F^2+H^2}$
When the magnet is removed,
Also, $\frac{F}{H}=\tan \theta$ Dividing (i) by (ii), we get $\begin{array}{rl}& \frac{T}{T_0}=\sqrt{\frac{H}{\sqrt{F^2+H^2}}}\\ =& \sqrt{\frac{H}{\sqrt{H^2{\tan }^2\theta +H^2}}}=\sqrt{\frac{H}{H\sqrt{{\sec }^2\theta }}}=\sqrt{\cos \theta }\\ \Rightarrow & \frac{T^2}{T_0^2}=\cos \theta \therefore T^2=T_0^2\cos \theta \end{array}$