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QPages 277-278: Q8: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q8 Solution: A magnetic dipole of magnetic moment is placed horizontally…

A magnetic dipole of magnetic moment 1.44 A m21.44 \text{ A m}^2 is placed horizontally with the north pole pointing towards north. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18 _T.

Step-by-Step Verified Solution

**Answer:** d=20 cmd = 20 \text{ cm}. **Step-by-Step Explanation:** Given: M=1.44 A m2M = 1.44 \text{ A m}^2, BH=18μT=18×10−6 TB_H = 18 \mu\text{T} = 18 \times 10^{-6} \text{ T}. Neutral point in broadside-on: B=μ04πMd3=BHB = \frac{\mu_0}{4\pi} \frac{M}{d^3} = B_H. 18×10−6=10−7×1.44d3⇒d3=8×10−3⇒d=0.2 m=20 cm18 \times 10^{-6} = 10^{-7} \times \frac{1.44}{d^3} \Rightarrow d^3 = 8 \times 10^{-3} \Rightarrow d = 0.2 \text{ m} = 20 \text{ cm} [citation:17].

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