Verified Solution • 2025–26 Session moderate

QPages 249-253: Q50: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q50 Solution: A solid wire of radius 10 cm carries a current of 5.0 A…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often apply the external formula $B=\mu_0 I/(2\pi r)$ for points inside the solid conductor, forgetting that only the fraction $I_{enc}=I\,(r^2/R^2)$ contributes; the correct interior field is $B=\mu_0 I r/(2\pi R^2)$ and grows linearly with $r$. Always convert cm to meters and use the appropriate Ampère‑law expression for $r<R$ versus $r\ge R$.

A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross section. Find the magnetic field BB at a point at a distance a) 2 cm b) 10 cm and c) 20 cm away from the axis. Sketch a graph BB versus xx for 0<x<20 cm0 < x < 20 \text{ cm}.

Step-by-Step Verified Solution

**Answer:** (a) B=2×10−6B = 2 \times 10^{-6} T (b) B=10−5B = 10^{-5} T (c) B=5×10−6B = 5 \times 10^{-6} T. **Step-by-Step Explanation:** i=5i = 5 A, b=0.1b = 0.1 m. (a) a=0.02a = 0.02 m: B=μ0ia2πb2=2×10−6B = \frac{\mu_0 i a}{2\pi b^2} = 2 \times 10^{-6} T. (b) a=0.1a = 0.1 m: B=μ0i2πb=10−5B = \frac{\mu_0 i}{2\pi b} = 10^{-5} T. (c) a=0.2a = 0.2 m: B=μ0i2πa=5×10−6B = \frac{\mu_0 i}{2\pi a} = 5 \times 10^{-6} T.

All Questions in this Chapter

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