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^2 = (T_0^2)/(cos theta)`
C. `T = T_0 cos theta `
D. `T = (T_0)/(cos theta )`

Answer

A

Explanation

Let the magnetic field of magnet be B and `B_H` be the Earth.s horizontal magnetic field. Net magnetic field, `B. =sqrt( B_H^2+B)`
Time period of needle
`T = 2pi sqrt((I)/(MB))`
`T = 2pi sqrt( (I)/(MB.))` ....(i)
When the small magnet is removed, B = 0,
`T_0 = 2pi sqrt( (I)/(MB_H) rArr T_0^2 = (4pi^2 I)/(MB_H)` ...(ii)
From magnetometer
`B = B_H tan theta `
`B. = sqrt(B_H^2 + B_H^2 tan^2 theta) =sqrt(B_H^2 sec^2 theta)`
`B. = B_H sec theta `
From eqn (i)
`T = 2pi sqrt((I)/(MB_H sec theta)`
`T^2 = (4pi^2 I)/(MB_H) xx (1)/(sec theta)`
from eqn (ii)
`T^2 = T_0^2 cos theta`