A deflection magnetometer is adjusted and a magnet of magnetic moment M is placed on it in the usual manner and the observed deflection is θ . The period of oscillation of the needle before settling of the deflection is

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Correct answer: A. T^2=T_0^2 θ

In deflection magnetometer, field due to magnet

F

and horizontal component

B_H

of earth's field are perpendicular to each other.

\therefore

Net field is

\sqrt{F^2+B_H^2}

So the time period

T=2 \pi \sqrt{\frac{1}{M \sqrt{F^2+B_H}}}

When magnet is removed

T_0=2 \pi \sqrt{\frac{1}{M B_H}}

Also,

\quad \frac{F}{B_H}=\tan \theta

Dividing Eqs. (i) by (ii), we get

\begin{aligned} \frac{T}{T_0} & =\sqrt{\frac{B_H}{\sqrt{F^2+B_H^2}}} \\ & =\sqrt{\frac{B_H}{\sqrt{B_H^2+\tan ^2 \theta+B_H^2}}}=\sqrt{\frac{B_H}{B_H \sqrt{\sec ^2 \theta}}} \\ & =\sqrt{\cos \theta} \\ \Rightarrow \quad T^2 & =T_0^2 \cos \theta\end{aligned}

A deflection magnetometer is adjusted and a magnet of magnetic...

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