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QPages 277-278: Q19: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q19 Solution: The magnetometer of the previous problem is used with the same…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the Tan-B position formula with the Tan-A position, incorrectly using $d^2$ in the denominator instead of the correct $(d^2 + l^2)^{3/2}$ (which simplifies to $d^3$ for a short magnet). Always verify the geometric orientation: Tan-B requires the cube of the distance, not the square.

The magnetometer of the previous problem is used with the same magnet in Tan-B position. Where should the magnet be placed to produce a 37� deflection of the needle?

Step-by-Step Verified Solution

**Answer:** d=7.9 cmd = 7.9 \text{ cm}. **Step-by-Step Explanation:** Using M/BH=3.75×103M/B_H = 3.75 \times 10^3 and Tan-B position formula with θ=37�\theta = 37�: tan⁡37�=μ04πM(d2+l2)3/2BH\tan 37� = \frac{\mu_0}{4\pi} \frac{M}{(d^2 + l^2)^{3/2} B_H}. For short magnet: 0.75=10−7×3.75×103d30.75 = 10^{-7} \times \frac{3.75 \times 10^3}{d^3}. d3=3.75×10−40.75=5×10−4⇒d=0.079 m=7.9 cmd^3 = \frac{3.75 \times 10^{-4}}{0.75} = 5 \times 10^{-4} \Rightarrow d = 0.079 \text{ m} = 7.9 \text{ cm} [citation:4].

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