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QPages 277-278: Q2: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q2 Solution: Two long bare magnets are placed with their axes coinciding in…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently forget to convert the distance from centimeters to meters before applying Coulomb's law for magnetism, leading to a $10^4$ error in the force calculation. Always ensure $r$ is in SI units (meters) when using $ rac{mu_0}{4pi} = 10^{-7} ext{ N/A}^2$.

Two long bare magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2.0 cm from the south pole of the second. If both the magnets have a pole strength of 10 Am, find the force exerted by one magnet of the other.

Step-by-Step Verified Solution

**Answer:** F=2.5×10−2 NF = 2.5 \times 10^{-2} \text{ N}. **Step-by-Step Explanation:** Given: m1=m2=10 Amm_1 = m_2 = 10 \text{ Am}, r=2 cm=0.02 mr = 2 \text{ cm} = 0.02 \text{ m}. F=μ04πm1m2r2=10−7×1004×10−4=2.5×10−2 NF = \frac{\mu_0}{4\pi} \frac{m_1 m_2}{r^2} = 10^{-7} \times \frac{100}{4 \times 10^{-4}} = 2.5 \times 10^{-2} \text{ N} [citation:17].

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