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A magnet suspended at30^@ with magnetic meridian makes an angle of45^@ with the horizontal . What shall be the actual value of the angle of dip ? |
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Answer» Solution :Here , `theta = 30^@` Apparent value of DIP ,` delta_1 = 45^@` Actual value of dip ` delta=?` If H is horizontal component of earth.s magnetic fieldin the MAGNETICMERIDIAN , then `tan delta=(V)/(H)` LET `H_1` be component of H at ` 30^@` to magnetic meridian , then ` tan delta_1 = (V)/(H_1) = (V)/(H cos theta ) = ( tan delta)/( cos theta )` ` or tan delta = tan delta_1xx cos theta =tan 45^@ xx cos 30^@` ` = 1 xx (sqrt(3))/(2) = (1.732)/(2) = 0.866` |
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