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QPages 277-278: Q14: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q14 Solution: A magnetic needle is free to rotate in a vertical plane which…

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⁡−123)\tan^{-1}2\sqrt{3}) with the horizontal, what would be the dip at that place?

Step-by-Step Verified Solution

**Answer:** δ=tan⁡−1(23)\delta = \tan^{-1}(2\sqrt{3}). **Step-by-Step Explanation:** Given: plane at 60� to magnetic meridian, apparent dip δ′=tan⁡−1(23)\delta' = \tan^{-1}(2\sqrt{3}). Using tan⁡δ′=tan⁡δcos⁡θ\tan \delta' = \frac{\tan \delta}{\cos \theta} with θ=60�\theta = 60�: tan⁡δ=tan⁡δ′cos⁡60�=23×0.5=3\tan \delta = \tan \delta' \cos 60� = 2\sqrt{3} \times 0.5 = \sqrt{3}. δ=60�\delta = 60� [citation:4].

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