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) cos alpha_(1)))`
B. `(1)/(sqrtB_(2))`
C. `(1)/(sqrt(B_(1) cos alpha)^(2)+B_(2)^(2))`
D. infinite

Answer

c

Explanation

(c) Resultant magnetic field in the place,
`B=sqrt((B_(1) cos alpha))^(2)+B_(2)^(2)`
Time period, `T=2pisqrt((I)/(MB))`