A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at `22^(@)` with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth's magnetic field at the place.

Explanation

Angle of dip, `delta = 22^@` and the horizontal component of the magnetic field `B_H = 0.35 G`
Let B be the resultant magnetic field at the place.
Since `B = B cos delta`
` therefore B = (B_H)/(cos delta) = (0.35)/(cos 22^@)`
` rArr B = (0.35)/(0.9272) = 0.38 G`