A magnetic needle free to rotate in a vertical plane parallel to magnetic meridian has its north tip down at `60^@` with the horizontal. The horizontal component of earth's magnetic field at that place is `0*4G`. Determine the magnitude of earth's magnetic field at the place.

Explanation

`delta=60^@,H=0.4G,R=?`
As, `H=Rcosdelta`
`R=H/(cosdelta)=0.4/(cos60^@)=0.4/(1//2)=0.4xx2R'=0.8G`.