In a certain place ,the horizontal component of magnetic field is `(1)/(sqrt(3))` times to vertical component . The angle of dip at this place is

Options

A. a)zero
B. b)`pi//3`
C. c)`pi//2`
D. d)`pi//6`

Answer

B

Explanation

Given that
`B_(H)=(1)/(sqrt(3))B_(V)`
Also `tantheta=(B_(V))/(B_(H))=sqrt(3)=tan60^(@)`
`thereforetheta60^(@)=(pi)/(3)`