A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at `60^@` with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth's magnetic field at the place.

Explanation

As per question, a magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at `60^@` with the horizontal, hence the angle of dip at given place is `60^@` (i.e. , `delta = 60^@`) .
Moreover horizontal component of the earth.s magnetic field at the place
`B_H = 0.4 G = 0.4 xx 10^(-4) T ` and `B_H = B_E cos delta`
`therefore ` Earth magnetic field at the place `B_E =(B_H)/(cos delta) = (0.4xx10^(-4)T)/(cos 60^@) = 8.0 xx 10^(-5) T `