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 θ and the period of oscillation

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Quantrex Academy · Accepted Answer

Correct answer: A. T^ 2 = T _ 0 ^ 2 cos θ

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}}

∴ T = 2\pi\sqrt{\dfrac{I}{M\sqrt{F^{2} + H^{2}}}}

...(i)

When the magnet is removed,

{T}_{0} = 2\pi\sqrt{\dfrac{I}{MH}}

...(ii)

Also,

\dfrac{F}{H} = tan \theta

Dividing (i) by (ii), we get

\dfrac{T}{T0} = \sqrt{\dfrac{H}{F^{2} + H^{2}}}

= \sqrt{\dfrac{H}{H^{2}\tan^{2} \theta + H^{2}}} = \sqrt{\dfrac{H}{H\sqrt{\sec^{2} \theta}}} = \sqrt{cos \theta}

\begin{cases}⇒ T^{2}{T}_{0}^{2} = cos \theta & ∴ T^{2} = {T}_{0}^{2}cos \theta\end{cases}

A deflection magnetometer is placed with its arm along the...

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