Name: Magnets and Magnetism
Uploaded: 2019-07-19
Duration: 3 min 46 s
Description: In this video, we will learn about the different types of magnets and the magnetic field created by different types of magnets. Learn more.

Question

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

Accepted Answer

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 $d B$ be the magnetic field due to a small length dl of the ring,

As per biot-savart rule,

$d B = \frac{μ_{0} I d l}{4 π r^{2}}$

We know that $r = \sqrt{x^{2} + R^{2}}$

This field have vertical component $d B_{y} = d B \sin ϕ$ and horizontal component $d B_{x} = d B \cos ϕ$

Vertical component will cancel out each other, only horizontal components are responsible.

So, $B = d B_{1} + d B_{2} + . . . . .$

$d B = \frac{μ_{0} I d l}{4 π r^{2}} \sin ϕ$

$d B = \frac{μ_{0} I d l}{4 π r^{2}} \times \frac{R}{r}$

$d B = \frac{μ_{0} I R}{4 π (x^{2} + R^{2})^{\frac{3}{2}}} d l$

Total field $B = \int d B = \int \frac{μ_{0} I R}{4 π (x^{2} + R^{2})^{\frac{3}{2}}} d l$

$B = \frac{μ_{0} I R}{4 π (x^{2} + R^{2})^{\frac{3}{2}}} \int d l$

But $\int d l =$ length of the loop $= 2 π R$

$B = \frac{μ_{0} I R}{4 π (x^{2} + R^{2})^{\frac{3}{2}}} \times 2 π R = \frac{μ_{0} I R^{2}}{2 (x^{2} + R^{2})^{\frac{3}{2}}}$

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

Important Questions on Magnetism and Matter

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