In a deflection magnetometer which 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

Options

A. `T^(2)=T_(0)^(2)/(cos theta)`
B. `T=T_(0)/(cos theta)`
C. `T=T_(0)cos theta`
D. `T^(2)=T_(0)^(2)cos theta`

Answer

D

Explanation

When deflection magnetometer is rest in usual way the field due to magnet (F) and horizontal component (H) of earth's field are perpendicular to each other. In this setting
`T=2pisqrt(I/(Msqrt(F^(2)+H^(2)))) ...(i)`
After removing the magnet
`T_(0)=2pisqrt(I/(MH)) ...(ii)`
We know, `F/H=tantheta`
Dividing (i) by (ii), we get,
`T/T_(0)=sqrt(H/(sqrtF^(2)+H^(2)))`
`=sqrt(H/sqrt(H^(2) tan^(2)theta+H^(2)))=sqrt(H/(Hsqrt(sec^(2)theta)))=sqrt(costheta)`
`implies T^(2)/T_(0)^(2)=cos theta :. T^(2)=T_(0)^(2) cos theta`