Question

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. $T^2=T_0\cos \theta$
• C. $T=\frac{T_0}{\cos \theta }$
• D. $T^2=\frac{T_0^2}{\cos \theta }$

Answer 1

The correct option is A $T^2=T_0^2\cos \theta$

In the usual setting of deflection magnetometer, field due to magnet $\left(F\right)$ and horizontal component $\left(H\right)$ of earth's field are perpendicular to each other. Therefore, the net field on the magnetic needle is $\sqrt{F^2+H^2}$
$\therefore T=2\pi \sqrt{\frac{I}{M\sqrt{F^2+H^2}}}....\left(i\right)$
When the magnet is removed,
$T_0=2\pi \sqrt{\frac{I}{MH}}....\left(ii\right)$
Also, $\frac{F}{H}=\tan \theta$
Dividing (i) by (ii), we get
$\frac{T}{T_0}=\sqrt{\frac{H}{\sqrt{F^2+H^2}}}$
$=\sqrt{\frac{H}{\sqrt{H^2{\tan }^2\theta +H^2}}}=\sqrt{\frac{H}{H\sqrt{{\sec }^2\theta }}}=\sqrt{\cos \theta }$
$\Rightarrow \frac{T^2}{T_0^2}=\cos \theta \therefore T^2=T_0^2\cos \theta$.