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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 $$B =  \dfrac{\mu_o i R}{4\pi  (R^2 + x^2)^{3/2}}(2\pi R)$$
$$\implies$$  $$B =  \dfrac{\mu_o i R^2}{2  (R^2 + x^2)^{3/2}}$$

658736_623501_ans_fea84153ab1848bc815e7158d77bc5eb.png

Physics

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