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Q.Derive an expression for the intensity of magnetic field at any point on the axis of a circular current loop.

Karnataka PUCKarnataka II PUC Board 2019Subjective· 5mImportance★★★★★
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The axial field of a circular current loop is B=μ0IR22(R2+x2)3/2B = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}, reducing to μ0I2R\dfrac{\mu_0 I}{2R} at the centre.

Set-up. Consider a circular loop of radius RR carrying current II, and a point PP on its axis at distance xx from the centre OO. A current element I dl⃗I\,d\vec{l} is at distance r=R2+x2r = \sqrt{R^2 + x^2} from PP, and dl⃗⊥r⃗d\vec{l} \perp \vec{r}.

Biot–Savart law. The field due to the element has magnitude

dB=μ04πI dl sin⁡90∘r2=μ04πI dlR2+x2.dB = \frac{\mu_0}{4\pi}\frac{I\,dl\,\sin 90^\circ}{r^2} = \frac{\mu_0}{4\pi}\frac{I\,dl}{R^2 + x^2}.

Resolving components. Each dBdB can be resolved into an axial component dBcos⁡ϕdB\cos\phi (along the axis) and a perpendicular component dBsin⁡ϕdB\sin\phi. By symmetry the perpendicular components from diametrically opposite elements cancel, leaving only the axial components, where cos⁡ϕ=RR2+x2\cos\phi = \dfrac{R}{\sqrt{R^2 + x^2}}.

Integrate over the loop. …

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