Question

A deflection magnetometer is placed with its arm along the east-west direction and a bar magnet is placed along the arm of the magnetometer. Due to the magnet, 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. $T^2=\frac{T_0^2}{cos\theta }$

Answer 1

Answer: (A)

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 $\sqrt{F^2+H^2}$
$\therefore T=2\pi \sqrt{\frac{I}{M\sqrt{F^2+H^2}}}$ ...(i)
When the magnet is removed,
$T_0=2\pi \sqrt{\frac{I}{MH}}$ ...(ii)
Also, $\frac{F}{H}=tan\theta$
Dividing (i) by (ii), we get
$\frac{T}{T_0}=\sqrt{\frac{H}{\sqrt{F^2+H^2}}}$
$=\sqrt{\frac{H}{\sqrt{H^{2}ta{n}^2\theta +H^2}}}=\sqrt{\frac{H}{H\sqrt{se{c}^2\theta }}}=\sqrt{cos\theta }$
$\Rightarrow \frac{T^2}{T_0^2}=\cos \theta \therefore T^2=T_0^2\cos \theta$