5.10. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22 \degree 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.

Answers (1)

Given,

The horizontal component of earth’s magnetic field, B_{H} = 0.35 G

Angle made by the needle with the horizontal plane at the place = Angle of dip = \delta = 22 \degree

We know, B_{H} = B cos\delta, where B is earth's magnetic field

B = B_{H}/cos\delta = 0.35/(cos22 \degree) = 0.377 G

The earth’s magnetic field strength at the place is 0.377 G.

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