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What is the angle of dip at a place where the horizontal component of the earth's magnetic field is 1/sqrt3 times the vertical component ? |
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Answer» Solution :If `theta` is the angle of dip, `TANTHETA = B_V/B_H = (B_Vsqrt3)/B_V = SQRT3 ` `THEREFORE ` angle of dip `theta = tan^-1 sqrt3 = 60^@` |
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