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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 magnitude of the earth's magnetic field at the place. |
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Answer» Solution :Here angle of dip `delta=22^@ ` and horizontal COMPONENT of earth.s MAGNETIC FIELD `B_H = 0.35 G =0.35 XX 10^(-4) T` `B_H = B_E cos delta` `implies` Earth magnetic field `B_E = (B_H)/(cos delta) = (0.35xx10^(-4))/(cos 22^@) = (0.35xx10^(-4))/(0.9272) = 3.8 xx 10^(-5) T ` or0.38 G= `0.35 xx 10^(-4) T` `B_H = B_E = cos delta` `implies` Earth.s magneticfield `B_E = (B_H)/(cos delta) =(0.35xx10^(-4))/(0.9272)=3.8xx10^(-5) T ` or 0.38 G. |
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