1.

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 ?

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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