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?
Simran Bhatia , 10 Years ago
Grade 11
1 Answers
Aditi Chauhan
Last Activity: 10 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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