Name: Fundamental Particles
Uploaded: 2019-07-19
Duration: 4 min 20 s
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Question

Using Biot and Savart's law, find the expression for magnetic field at an axial point of a circular current loop in vector form. Draw required diagram.

Accepted Answer

Consider a circular coil of radius $R$ with a current $I$ in Y-Z plane with the centre $O$ . Consider a point $P$ on the axis of the coil at a distance $x$ from the centre of the coil.

Consider two small elements of the coil each of length $d l$ , at $C$ and $D$ which are situated at diametrically opposite edges. The distance between $C$ and $P$ is $r$ and the angle between $d \vec{l}$ and $\vec{r}$ is $90 °$ .

From Biot-savart's law, the magnetic field at point $P$ due to the current element $I d \vec{l}$ is given by

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

$| d \vec{B} | = \frac{μ_{0}}{4 π} \frac{I d l s i n 90 °}{r^{2}}$

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

The direction of $d \vec{B}$ is in a plane perpendicular to $d \vec{l}$ and $\vec{r}$ , and is directed as given by right-hand screw rule, i.e., acting along $P E$ .

Similarly, the magnitude of magnetic field at $P$ due to the current element of length $d l$ at $D$ is given by

$| d \vec{B'} | = \frac{μ_{0}}{4 π} \frac{I d l}{r^{2}}$

The direction of $d \vec{B'}$ is in a plane perpendicular to $d \vec{l}$ and $\vec{r}$ , and is directed as given by right-hand screw rule, i.e., acting along $P F$ .

Therefore, $| d \vec{B} | = | d \vec{B'} | = \frac{μ_{0}}{4 π} \frac{I d l}{r^{2}}$

On resolving $| d \vec{B} |$ and $| d \vec{B'} |$ into two rectangular components, we have

$| d \vec{B} | c o s ϕ$ along $P Y$ and $| d \vec{B} | s i n ϕ$ along $P X$

$| d \vec{B'} | c o s ϕ$ along $P Y'$ and $| d \vec{B'} | s i n ϕ$ along $P X$

Since the cosine components are equal and opposite, they cancel each other. The sine components being equal and in the same direction are added up. The same is true for all the diametrically opposite pairs of current elements in the coil.

Thus the total magnetic field at $P$ due to the current through the whole circular coil is given by

$|\vec{B}| = \int | d \vec{B} | s i n ϕ$

$|\vec{B}| = \int \frac{μ_{0}}{4 π} \frac{I d l}{r^{2}} s i n ϕ$

From the diagram $\sin ϕ = \frac{R}{r}$

$|\vec{B}| = \frac{μ_{0}}{4 π} \frac{I R}{r^{3}} \int d l$

$\int d l =2 π R$

$|\vec{B}| = \frac{μ_{0}}{4 π} \frac{I R}{r^{3}} \times 2 π R$

$|\vec{B}| = \frac{μ_{0}}{4 π} \frac{2 π I R^{2}}{r^{3}}$

From Pythagorean theorem $r^{2} = R^{2} + x^{2}$

$r = {(R^{2} + x^{2})}^{\frac{1}{2}}$

$r^{3} = {(R^{2} + x^{2})}^{\frac{3}{2}}$

$|\vec{B}| = \frac{μ_{0}}{4 π} \frac{2 π I R^{2}}{{(R^{2} + x^{2})}^{\frac{3}{2}}}$ along $P X$

If the coil has $N$ turns

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

In vector form

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

Using Biot and Savart's law, find the expression for magnetic field at an axial point of a circular current loop in vector form. Draw required diagram.

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