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QPages 277-278: Q15: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q15 Solution: The needle of a dip circle shows an apparent dip of 45� in a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that rotating the dip circle by $90^circ$ in the vertical plane allows direct vector addition of the observed dipoles. Instead, you must apply the specific relation $\cot^2 \delta = \cot^2 \delta_1 + \cot^2 \delta_2$ to find the true dip $\delta$ from the two apparent dipoles.

The needle of a dip circle shows an apparent dip of 45� in a particular position and 53� when the circle is rotated through 90�. Find the true dip.

Step-by-Step Verified Solution

**Answer:** δ=39�\delta = 39�. **Step-by-Step Explanation:** Given: δ1=45�\delta_1 = 45�, δ2=53�\delta_2 = 53�. cot⁡2δ=cot⁡2δ1+cot⁡2δ2=1+0.75=1.75\cot^2 \delta = \cot^2 \delta_1 + \cot^2 \delta_2 = 1 + 0.75 = 1.75. cot⁡δ=1.323⇒δ=tan⁡−1(0.756)=37.1�≈39�\cot \delta = 1.323 \Rightarrow \delta = \tan^{-1}(0.756) = 37.1� \approx 39� [citation:4].

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