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In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is 0.26 G and the dip angle is 60^@ . What is the magnetic field of the earth at this location ? |
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Answer» SOLUTION :It is GIVEN that `H_E=0.26G` `COS60^@=H_E/B_E` `B_E=H_E/(cos60^@)=(0.26)/((1/2))=0.52G` |
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