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QPages 249-253: Q24: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q24 Solution: A regular polygon of sides is formed by bending a wire of total …

A regular polygon of nn sides is formed by bending a wire of total length 2πr2\pi r which carries a current ii. a) Find the magnetic field BB at the centre of the polygon. b) By letting n→∞n \to \infty, deduce the expression for the magnetic field at the centre of a circular current.

Step-by-Step Verified Solution

**Answer:** (a) B=μ0ni2πrtan⁡(π/n)B = \frac{\mu_0 n i}{2\pi r}\tan(\pi/n). (b) As n→∞n \to \infty, B→μ0i2rB \to \frac{\mu_0 i}{2r}. **Step-by-Step Explanation:** (a) Regular polygon of nn sides with total length 2πr2\pi r. Each side length = 2πr/n2\pi r/n. Field at centre due to one side: B1=μ0i4πd(sin⁡θ+sin⁡θ)B_1 = \frac{\mu_0 i}{4\pi d}(\sin\theta + \sin\theta) where dd is apothem. Total field: B=nB1=μ0ni2πrtan⁡(π/n)B = nB_1 = \frac{\mu_0 n i}{2\pi r}\tan(\pi/n). (b) For large nn, tan⁡(π/n)≈π/n\tan(\pi/n) \approx \pi/n, so B→μ0i2rB \to \frac{\mu_0 i}{2r} (circular loop).

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