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A deflection magnetometer is adjusted in the usual way. When a magnet is introduced, the deflection observed is $θ,$ 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

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$T^{2} = T_{0}^{2} \cos θ$

$T = T_{0} \cos θ$

$T = \frac{T_{0}}{\cos θ}$

$T^{2} = \frac{T_{0}^{2}}{\cos θ}$

answer is A.

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Detailed Solution

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 $\sqrt{F^{2} + H^{2}}$

$\frac{T}{T_{0}} = \sqrt{\frac{H}{\sqrt{H^{2} \tan^{2} θ + H^{2}}}} = \sqrt{\frac{H}{H \sqrt{\sec^{2} θ}}} = \sqrt{\cos θ}$

$\Rightarrow \frac{T^{2}}{T_{0}^{2}} = \cos θ ∴ T^{2} = T_{0}^{2} \cos θ$

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