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. $T^{ 2}=T^{ 2}_{0} cos\theta$
• B. $T =T_{0} cos\theta$
• C. $T =\frac{T_{0}}{cos\theta }$
• D. $T^{ 2} =\frac{T^{ 2}_{0}}{cos\theta }$

Answer 1

Answer: (A)

$T^{ 2}=T^{ 2}_{0} cos\theta$

Answer 2

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 $\sqrt{F^{2}+H^{2}}$ $\therefore\quad T=2\pi\sqrt{\frac{1}{M\sqrt{F^{2}+H^{2}}}}\quad\quad\quad\quad\quad\dots\left(i\right)$ When the magnet is removed, $T_{0}=2\pi\sqrt{\frac{1}{MH}}\quad \quad \quad \quad \quad\quad\quad\quad\quad\quad \dots \left(ii\right)$ Also, $\, \frac{F}{H} = tan\theta$ Dividing (i) by (ii), we get $\frac{T}{T_{0}}=\sqrt{\frac{H}{\sqrt{F^{2}+H^{2}}}}$ $=\sqrt{\frac{H}{\sqrt{H^{2}\,tan^{2}\,\theta+H^{2}}}}=\sqrt{\frac{H}{H\sqrt{sec^{2}\,\theta}}}=\sqrt{cos\theta}$ $\Rightarrow\quad \frac{T^{2}}{T^{2}_{0}}=cos\,\theta\quad\quad\quad\therefore\quad T^{2}=T^{2}_{0}\,cos\theta$