Question

A magnetic needle free to rotate in a vertical plane parallel to themagnetic meridian has its north tip pointing down at 22° with thehorizontal. The horizontal component of the earth’s magnetic fieldat the place is known to be 0.35 G. Determine the magnitude of theearth’s magnetic field at the place.

Answer 1

Given: The horizontal component of earth’s magnetic field is B H =0.35 G, and the angle of the dip is 22°.

Let B be the magnitude of earth’s magnetic field.

The horizontal component of magnetic field is given as,

B H =Bcosδ B= B H cosδ

By substituting the given values, we get,

B= 0.35 cos22° = 0.35 0.9272 =0.38 G

Thus, the magnitude of earth’s magnetic field is 0.38 G.