Question

# Derive an expression for magnetic field strength at any point on the axis of a circular current carrying loop using Biot-Savart's law.

Solution

## Biot-Savart's law : Magnetic field at an axial point P due to a current element of the ring  $$dB = \dfrac{\mu_o}{4\pi} \dfrac{i(\vec{dl}\times \hat{r})}{r^2}$$where $$r = \sqrt{R^2+ x^2}$$ $$\therefore$$ We get $$B = \dfrac{\mu_o}{4\pi} \dfrac{idl}{(R^2 + x^2)}$$We resolve $$dB$$ into vertical and horizontal components. Now all the vertical components cancel out each other and so only the horizontal components survive which results in the net magnetic field at P in the horizontal direction.Net magnetic field $$B = \int dB \sin\theta$$$$B = \int \dfrac{\mu_o i dl}{4\pi (R^2 + x^2)} \dfrac{R}{(R^2 + x^2)^{1/2}}$$Or $$B = \dfrac{\mu_o i R}{4\pi (R^2 + x^2)^{3/2}} \int dl$$Or $$B = \dfrac{\mu_o i R}{4\pi (R^2 + x^2)^{3/2}}(2\pi R)$$$$\implies$$  $$B = \dfrac{\mu_o i R^2}{2 (R^2 + x^2)^{3/2}}$$Physics

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