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

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

`tan 45^(@) = (tan delta)/(cos 30^(@))`
`1 = (2 tan delta)/(sqrt3)`
`delta = tan^(-1) ((sqrt3)/(2))`
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