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A magnetic needle is free to rotate in a vertical plane which makes an angle of 60 with the magnetic angle of tan^-1(2/√3) with the horizontal, what would be the dip at that place?

A magnetic needle is free to rotate in a vertical plane which makes an angle of 60 with the magnetic angle of tan^-1(2/√3) with the horizontal, what would be the dip at that place?

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago
Sol. If δ base 1 and δ base 2 be the apparent dips shown by the dip circle in the 2⊥r positions, the true dip δ is given by Cot^2 δ = Cot^2 δ base 1 + Cot^2 δ base 2 ⇒ Cot^2 δ = Cot^2 45° + Cot^2 53° ⇒ Cot^2 δ = 1.56 ⇒ δ = 38.6 ≈ 39°

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