The horizontal component of the earth's magnetic field at a certain place is `2.6 xx 10^(-5) T` and the angle of dip is `60^(@)`. What is the magnetic field of the earth at this location ?

Explanation

Given `H_(E) = 2.6 xx 10^(-5)T`,
D (or) `delta = 60^(@)`
`B_(E) = (H_(E))/(cos D) = (2.6 xx 10^(-5))/(cos 60^(@)) = (2.6 xx 10^(-5))/((1//2)) = 5.2 xx 10^(-5)T`
`:. B_(E) = 5.2 xx 10^(-5) T`