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.

Answer

`0*8G`

Explanation

Here, `delta=60^@`, `H=0*4G`, `R=?`
`R=(H)/(cos delta)=(0*4)/(cos 60^@)=(0*4)/(1//2)=0*8G`