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Q.Derive the expression for the magnetic field on the axis of a circular current loop, using Biot-Savart's law.

Karnataka PUCKarnataka II PUC Board 2018Subjective· 5mImportance★★★★★
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On the axis of a circular loop, B=μ0IR22(R2+x2)3/2B = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}.

Biot–Savart law: A current element I dl⃗I\,d\vec l produces at a point, at position r⃗\vec r from it, a field

dB=μ04πI dl sin⁡θr2.dB = \frac{\mu_0}{4\pi}\frac{I\,dl\,\sin\theta}{r^2}.

Setup: Consider a circular loop of radius RR carrying current II. Take a point P on its axis at distance xx from the centre O. Each element dl⃗d\vec l is perpendicular to r⃗\vec r (so sin⁡θ=1\sin\theta = 1), and the distance from every element to P is

r=R2+x2.r = \sqrt{R^2 + x^2}.

Thus

dB=μ04πI dl(R2+x2).dB = \frac{\mu_0}{4\pi}\frac{I\,dl}{(R^2 + x^2)}.

Resolving components: dB⃗d\vec B makes an angle with the axis. For diametrically opposite elements, the components perpendicular to the axis cancel, and only the axial components dBcos⁡αdB\cos\alpha add up, where

cos⁡α=RR2+x2.\cos\alpha = \frac{R}{\sqrt{R^2 + x^2}}.

So the net axial field is …

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