91Ó°ÊÓ

There is a square loop which carries a steady current Iwith Rdistance from center to side.

There is a regular n-sided polygon carrying a steady current Iwith Rdistance from the center to any side.

The magnetic field at a distance R from a straight wire carrying current l is$\mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{l}}{\mathbf{4}\mathbf{Ï€¸é}}\left({\mathrm{²õ¾\pm ²Ôθ}}_2-{\mathrm{²õ¾\pm ²Ôθ}}_1\right)$

Here, ${\mathrm{μ}}_0$is the permeability of free space and ${\mathrm{θ}}_2$and ${\mathrm{θ}}_1$are the angles made by the ends of the wire with the point at which the field is calculated.

For a square, the angles made by the vertices with the center is $45°$.

Thus, from equation (1),

$$\begin{gathered} B=\frac{{\mathrm{μ}}_{0}l}{4\mathrm{π}R}\left(\sin 45°-\sin \left(-45°\right)\right) \\ =\frac{{\mathrm{μ}}_{0}l}{4\mathrm{π}R}\sqrt{2} \end{gathered}$$

Including four sides of the square, the net field is

$$\begin{gathered} \mathit{B}\mathbf{=}\mathbf{4}\mathbf{×}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{l}}{\mathbf{4}\mathbf{π}\mathbf{R}}\sqrt{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\sqrt{\mathbf{2}}{\mathbf{μ}}_{\mathbf{0}}\mathbf{l}}{\mathbf{4}\mathbf{π}\mathbf{R}} \end{gathered}$$

Thus, the field is $\frac{\sqrt{\mathbf{2}}{\mathbf{μ}}_{\mathbf{0}}\mathbf{l}}{\mathbf{4}\mathbf{Ï€¸é}}$.

For a regular n sided polygon, the angles made by the vertices with the center is $\frac{\mathrm{Ï€}}{n}$. Thus, from equation (1),

$$\begin{gathered} B=\frac{{\mathrm{μ}}_{0}l}{4\mathrm{π}R}\left(-\sin \left(\frac{\mathrm{π}}{n}\right)\right) \\ =\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}R}sin\frac{\mathrm{π}}{n} \end{gathered}$$

Including n sides of the polygon, the net field is

$$\begin{gathered} B=n×\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}R}\sin \frac{\mathrm{π}}{n} \\ =\frac{n{\mathrm{μ}}_{0}l}{2\mathrm{π}R}\sin \frac{\mathrm{π}}{n} \end{gathered}$$

Thus, the field is $\frac{n{\mathrm{μ}}_{0}l}{2\mathrm{π}R}\sin \frac{\mathrm{π}}{n}$.

As n becomes infinitely large, $\frac{\mathrm{Ï€}}{n}$becomes infinitely small and thus,

$\sin \frac{\mathrm{π}}{n}≈\frac{\mathrm{π}}{n}$

Equation (2) thus becomes,

$$\begin{gathered} B=\frac{n{\mathrm{μ}}_{0}l}{2\mathrm{π}R}\sin \frac{\mathrm{π}}{n} \\ ≈\frac{n{\mathrm{μ}}_{0}l}{2\mathrm{π}R}\sin \frac{\mathrm{π}}{n} \\ =\frac{{\mathrm{μ}}_{0}l}{2R} \end{gathered}$$

This is the magnetic field at the centre of a circle.