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At a place on earth, horizontal componen...

At a place on earth, horizontal component of earth's magnetic field is `B_(1)` and vertical component of earth's magnetic field si `B_(2)`. If a magnetic needle is kept vertica, 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

A

`(1)/(sqrt(B_(1) cos alpha)`

B

`(1)/(sqrtB_(2))`

C

`(1)/(sqrt((B_(1)cosalpha)^(2))+B_(2)^(2))`

D

infinite

Text Solution

Verified by Experts

The correct Answer is:
C
`B=sqrt((B_(1) cos alpha)^(2)+B_(2)^(2))`
Time period, `T=2pisqrt((1)/(MB))`
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