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A magnet suspended at 30^@ with magneti...

A magnet suspended at ` 30^@` with magnetic meridian makes an angle of ` 45^@` with the horizontal . What shall be the actual value of the angle of dip ?

Text Solution

Verified by Experts

Here , `theta = 30^@`
Apparent value of dip ,` delta_1 = 45^@`
Actual value of dip ` delta=?`
If H is horizontal component of earth.s magnetic field in the magnetic meridian , 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_1 xx cos theta = tan 45^@ xx cos 30^@`
` = 1 xx (sqrt(3))/(2) = (1.732)/(2) = 0.866`
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