Verified Solution • 2025–26 Session moderate

QPages 249-253: Q21: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q21 Solution: A wire of length l is bent in the form of an equilateral…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often take the wire length $l$ as the side of the polygon or use the altitude instead of the perpendicular distance from the centre to a side. Remember that for a regular $n$‑gon the side length is $l/n$, and the distance $d$ from the centre (centroid) to each side is $d=\frac{\text{side}}{2\tan(\pi/n)}$ (e.g., $d=\frac{\sqrt3}{6}\,\text{side}$ for a triangle). Using the correct $d$ ensures the $\sin\theta$ terms are evaluated with the half‑angle ($\theta=30^circ$ for a triangle, $45^circ$ for a square) and the field contributions add properly.

A wire of length l is bent in the form of an equilateral triangle and carries an electric current i. 1. Find the magnetic field B at the centre. 2. If the wire is bent in the form of a square, what would be the value of B at the centre?

Step-by-Step Verified Solution

**Answer:** (a)

B = \frac{3\mu_0 i}{2\pi l}

. (b)

B = \frac{2\sqrt{2}\mu_0 i}{\pi l}

. **Step-by-Step Explanation:** (a) Equilateral triangle of side

l/3

. Distance from centre to each side =

\frac{l}{6\sqrt{3}}

.

B = 3 \times \frac{\mu_0 i}{4\pi d}(\sin 30� + \sin 30�) = \frac{3\mu_0 i}{2\pi l}

. (b) Square of side

l/4

. Distance from centre to each side =

l/8

.

B = 4 \times \frac{\mu_0 i}{4\pi (l/8)}(\sin 45� + \sin 45�) = \frac{2\sqrt{2}\mu_0 i}{\pi l}

.

All Questions in this Chapter

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