At a place on earth, horizontal component of earth's magnetic field is `B_(1)` and vertical component of earth's magnetic field is `B_(2)`. If a magnetic needle is kept vertical, in a plane making angle `alpha` with the horizontal component of magnetic field, then square of time period of oscillation of needle when slightly distributed is proportional to

Options

A. `1/(sqrt(B_(1)cosalpha))`
B. `1/(sqrt(B_(2)))`
C. `1/(sqrt((B_(1)cosalpha)^(2))+B_(2)^(2))`
D. infinite

Answer

C

Explanation

Resultant magnetic field in the plane, `B_(1)`
`B=sqrt((B_(1)cosalpha)^(2)+B_(2)^(2))`
Time period, `T=2pisqrt(1/(MB))`