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Question
A magnetic needle is free to rotate in a vertical plane which makes an angle of 60° with the magnetic meridian. If the needle stays in a direction making an angle of `tan^-1(2sqrt(3))` with the horizontal, what would be the dip at that place?
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Solution
Given :
Angle made by the magnetic meridian with the plane of rotation of the needle , `θ = 60^circ`
Angle made by the needle with the horizontal, `δ_1 = tan^-1(2/sqrt(3))`
If δ is the angle of dip , then
`tan δ_1 = tanδ/cosθ`
⇒ `tan δ = tan δ_1 cos θ`
⇒ `tan δ = tan (tan^-1 2/sqrt3) cos60^circ`
⇒ `tan δ = 2/sqrt(3) xx 1/2 = 1/sqrt(3)`
⇒ `δ = 30^circ`
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