The magnetic field due to the earth has a horizontal component of `26 muT` at a place where the dip is `60^@`. Find the vertical component and the magnitude of the field.

Explanation

Here `s(dip) = 60^(@)`
`Since B_(H) = B cos 60^(@)`
`rArr 26 xx 10^(-6) = B xx (1)/(2)`
`rArr B = 52 xx 10^(-6) = 52 muT`
`B_(1) = B sin delta = 52 xx 10^(-6) xx (sqrt(3)/(2)`
`= 64.98 muT = 45 mu T`