Physics - Magnetism Question with Solution | TestHub

In the usual setting of deflection magnetometer, field due to magnet ($\mathrm{F}$) and horizontal component ($\mathrm{H}$) of earth's field are perpendicular to each other. Therefore, the net field on the magnetic needle is$\sqrt{{\mathrm{F}}^2+{\mathrm{H}}^2}$

$\therefore \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{M}\sqrt{{\mathrm{F}}^2+{\mathrm{H}}^2}}}$...(i)

When the magnet is removed,

${\mathrm{T}}_0=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{MH}}}$...(ii)

Also,$\frac{\mathrm{F}}{\mathrm{H}}=\tan \mathrm{\theta }$

Dividing (i) by (ii), we get

$\frac{\mathrm{T}}{{\mathrm{T}}_0}=\sqrt{\frac{\mathrm{H}}{\sqrt{{\mathrm{F}}^2+{\mathrm{H}}^2}}}$

$=\sqrt{\frac{\mathrm{H}}{\sqrt{{\mathrm{H}}^2{\tan }^2\mathrm{\theta }+{\mathrm{H}}^2}}}=\sqrt{\frac{\mathrm{H}}{\mathrm{H}\sqrt{{\sec }^2\mathrm{\theta }}}}=\sqrt{\cos \mathrm{\theta }}$

$\begin{array}{cc}\Rightarrow \frac{{\mathrm{T}}^2}{{\mathrm{T}}_0^2}=\cos \mathrm{\theta }& \therefore {\mathrm{T}}^2={\mathrm{T}}_0^2\cos \mathrm{\theta }\end{array}$