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QPages 277-278: Q21: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q21 Solution: A bar magnet takes second to complete one oscillation in an…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the Earth's horizontal magnetic field from microtesla ($\mu\text{T}$) to tesla (T) before substitution, leading to a magnetic moment error by a factor of $10^{12}$. Always ensure $B_H$ is expressed in SI base units (Tesla) when using the period formula $T = 2\pi \sqrt{I/MB_H}$.

A bar magnet takes π/10\pi/10 second to complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is 1.2×10−4 kg m21.2 \times 10^{-4} \text{ kg m}^2 and the earth's horizontal magnetic field is 30 _T. Find the magnetic moment of the magnet.

Step-by-Step Verified Solution

**Answer:** M=1600 A m2M = 1600 \text{ A m}^2. **Step-by-Step Explanation:** Given: T=π/10 sT = \pi/10 \text{ s}, I=1.2×10−4 kg m2I = 1.2 \times 10^{-4} \text{ kg m}^2, BH=30μTB_H = 30 \mu\text{T}. T=2πIMBH⇒π10=2π1.2×10−4M×30×10−6T = 2\pi \sqrt{\frac{I}{MB_H}} \Rightarrow \frac{\pi}{10} = 2\pi \sqrt{\frac{1.2 \times 10^{-4}}{M \times 30 \times 10^{-6}}}. Squaring and solving gives M=1600 A m2M = 1600 \text{ A m}^2 [citation:12].

All Questions in this Chapter

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