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# 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 mag

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.

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Given,

The horizontal component of earth’s magnetic field, $\dpi{100} B_{H}$ = 0.35 G

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

We know, $\dpi{100} B_{H}$ = B cos$\delta$, where B is earth's magnetic field

B = $\dpi{100} B_{H}$/cos$\delta$ = 0.35/(cos$22 \degree$) = 0.377 G

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

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