Verified Solution • 2025–26 Session moderate

QPages 230-234: Q20: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q20 Solution: A straight horizontal wire of mass 10 mg and length 1.0 m…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often write $F = i l B$ and ignore the $sin\theta$ term, forgetting that the magnetic force is $F = i l B\sin\theta$ and that the minimum field is achieved only when $B$ is perpendicular to the wire; always include the sine factor and set it to 1 only after confirming the geometry.

A straight horizontal wire of mass 10 mg and length 1.0 m carries a current of 2.0 A. What minimum magnetic field BB should be applied in the region, so that the magnetic force on the wire may balance its weight?

Step-by-Step Verified Solution

**Answer:** B=4.9×10−5 TB = 4.9 \times 10^{-5} \text{ T}. **Step-by-Step Explanation:** For the magnetic force to balance weight: ilB=mgi l B = m g. B=mgil=10×10−6×9.82.0×1.0=4.9×10−5 TB = \frac{m g}{i l} = \frac{10 \times 10^{-6} \times 9.8}{2.0 \times 1.0} = 4.9 \times 10^{-5} \text{ T}.

All Questions in this Chapter

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