Question

Derive the expression for magnetic field at a point on the axis of a circular current carrying loop.

Solution

In the above diagram,Let O be the centre of the ring, R be the radius of the ring. The point P lie outside the ring on the axis, let the distance between the P and o be $$x$$, where we have to calculate the magnetic field.Let $$\vec{dB}$$ be the magnetic field due to a small length $$dl$$ of the ring,As per bio-severt rule,$$\vec{dB}=\dfrac{\mu_o Idl}{4\pi r^2}$$We know that $$r=\sqrt{(x^2+R^2)}$$So $$\vec{db}=\dfrac{\mu_o I dl}{4\pi \sqrt{x^2+R^2}}$$This field have vertical$$\vec{dB}_y=\vec{dB}sin\theta$$ and horizontal component$$\vec{dB}_x=\vec{dB} cos\theta$$,Vertical component will cancel out each other, only horizontal components are responsible, So, $$\vec{B}=\vec{dB}_1+\vec{dB}_2+.....$$$$\vec{dB}=\dfrac{\mu_o I dl sin\theta}{4\pi \sqrt{R^2+x^2}}$$$$\vec{dB}=\dfrac{\mu_o I x}{4\pi (R^2+x^2)^{\dfrac{3}{2}}}$$The magnetic field due to the circular current loop of radius a at a point which is a distance R away, and is on its axis,So $$\vec{B}=\dfrac{\mu_o I x^2}{2(R^2+x^2)^{\dfrac{3}{2}}}$$Physics

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