Verified Solution • 2025–26 Session moderate

QPages 249-253: Q49: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q49 Solution: A long, cylindrical wire of radius carries a current…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the total current $i$ into Ampère's law for a point inside the wire, yielding $B=\mu_0 i/(2\pi a)$. Remember that only the current within the Amperian loop contributes: $i_{enc}=i\,\frac{a^2}{b^2}$ for uniform distribution, so $B=\mu_0 i_{enc}/(2\pi a)=\mu_0 i a/(2\pi b^2)$.

A long, cylindrical wire of radius

b

carries a current

i

distributed uniformly over its cross section. Find the magnitude of the magnetic field at a point inside the wire at a distance

a

from the axis.

Step-by-Step Verified Solution

**Answer:**

B = \frac{\mu_0 i a}{2\pi b^2}

. **Step-by-Step Explanation:** For a point inside the wire at distance

a

:

B \times 2\pi a = \mu_0 i \frac{a^2}{b^2} \Rightarrow B = \frac{\mu_0 i a}{2\pi b^2}

.

All Questions in this Chapter

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