A deflection magnetometer is adjusted and a magnet of magnetic moment M is placed on it in the usual manner and the observed deflection is theta. The period of oscillation of the needle before setting of the deflection is T. When the magnet is removed, the period of oscillation of the needle is ` T_(0)` before setting to `0^(@).` If the eart's induced magnetic field is `B_(H)`, 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)^(2))/("cos"theta`

Answer

A

Explanation

In deflection magnetometer, field due to magnet F and horizontal component B_(H) are perpendicular to each other so, the net field is given by,
`B_(H)=sqrt(f^(2)+B_(H)^(2))`
Time period of magnet
`2pisqrt((I)/(MB_(H)))`
`impliesTprop(1)/sqrt(B_(H))`
`implies(T)/(T_(0))=sqrt((B_(H))/(sqrt(F^(2)+B^(2)H`
`F=B_(H)"tan"theta`
`(T)/(T_(0)=sqrt((B_(H))/(B_(H)^(2)("tan"^(2)thetaH))`
`=sqrt((B_(H))/(B_(H)sqrtsec^(2)theta)`
`=sqrt"cos"theta`
therefore `T^(2)=T_(0)^(2)"cos"theta`