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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

In a vertical plane at `30^@` from the magnetic meridian, the horizontal component is
`H'=Hcos30^@`

While verticale component is still `V`. Therefore, apparent dip will be given by
`tantheta'=V/H'=V/(Hcos30^@)`
`But, V/H=tantheta(where theta=true angle of dip)`
`: tantheta'=(tantheta)/(cos30^@)`
`:. theta=tan^-1[tantheta'cos30^@]`
`=tan^-1[(tan45^@)(cos30^@)]`
`=41^@`
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