At a certain place, the horizontal component of earth's magnetic field is `sqrt(3)` times the vertical component. The angle of dip at that place is

Options

A. `(pi)/(6)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. 0

Answer

A

Explanation

Given `B_(H)= sqrt(3)B_(V) " "...(i)`
We know, `B_(H)=B cos theta`
`:. (B_(V))/(B_(H))= tan theta`
`rArr B_(H)=(B_(V))/(tan theta) " "....(ii)`
Comparing eq. (i) and (ii)
`(B_(V))/(tan theta)= sqrt(3) B_(V)`
`:. tan theta=(1)/(sqrt(3))`
`theta=tan^(-1)((1)/(sqrt(3)))`
`=30^(@)=(pi)/(6)` radians