Verified Solution • 2025–26 Session moderate

QPages 277-278: Q17: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q17 Solution: A moving-coil galvanometer has a 50-turn coil of size 2 cm _ 2…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently insert the coil area as $2\times2=4$ (cm$^2$) directly into the torque formula, forgetting to convert to square metres ($4\times10^{-4}\,$m$^2$); this inflates the torque by $10^4$. Always convert all dimensions to SI units before using $\tau=n i A B\sin\theta$ (with $\sin\theta=1$ for a perpendicular orientation).

A moving-coil galvanometer has a 50-turn coil of size 2 cm _ 2 cm. It is suspended between the magnetic poles producing a magnetic field of 0.5 T. Find the torque on the coil due to the magnetic field when a current of 20 mA passes through it.

Step-by-Step Verified Solution

**Answer:** τ=2×10−4 N m\tau = 2 \times 10^{-4} \text{ N m}. **Step-by-Step Explanation:** Given: n=50n = 50, A=4×10−4 m2A = 4 \times 10^{-4} \text{ m}^2, i=0.02 Ai = 0.02 \text{ A}, B=0.5 TB = 0.5 \text{ T}. τ=niAB=50×0.02×4×10−4×0.5=2×10−4 N m\tau = n i A B = 50 \times 0.02 \times 4 \times 10^{-4} \times 0.5 = 2 \times 10^{-4} \text{ N m} [citation:12].

All Questions in this Chapter

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