The horizontal component of the earth's magnetic field at a place is `1/(sqrt(3))` times its vertical component there. Find the value of the angle of dip at that place. What is the ratio of the horizontal component to the total magnetic field of the earth at that place?

Explanation

As per question `B_H = 1/sqrt3 B_V ` or `(B_V)/(V_H) = sqrt3`
But `(B_V)/(B_H)= tan delta` , where `delta` is the angle of dip at given place. Hence,
`tan delta = sqrt3 ` or ` delta = tan^(-1) sqrt3 = pi/3 ` or `60^@`
Again `B_H = D_E cos delta = B_E cos 60^@ = (B_E)/(2) implies (B_H)/(B_E) =1/2`.