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
$\frac{T}{T_{0}}=\sqrt{\frac{H}{\sqrt{F^{2} + H^{2}}}}$
$=\sqrt{\frac{H}{\sqrt{H^{2} tan^{2} \theta + H^{2}}}}=\sqrt{\frac{H}{H \sqrt{sec^{2} ⁡ \theta }}}=\sqrt{cos ⁡ \theta }$
$\Rightarrow \frac{ T ^{2}}{ T _{0}^{2}}=\cos \theta \quad \therefore T ^{2}= T _{0}^{2} \cos \theta$