Question: (a) Find the magnetic field at the center of a square loop, which carries a steady current I.Let Rbe the distance from center to side (Fig. 5.22).

(b) Find the field at the center of a regular n-sided polygon, carrying a steady current

I.Again, let Rbe the distance from the center to any side.

(c) Check that your formula reduces to the field at the center of a circular loop, in

the limit $\mathit{n}\mathbf{\to }\mathbf{\infty }$.

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$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{4}\mathbf{\pi R}}\left({\mathrm{sin\theta }}_2-{\mathrm{sin\theta }}_1\right)$

Here, ${\mu }_0$is the permeability of free space and ${\theta }_2$and ${\theta }_1$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),

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

Including four sides of the square, the net field is

$$\begin{gathered} \mathit{B}\mathbf{=}\mathbf{4}\mathbf{\times }\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{4}\mathbf{\pi }\mathbf{R}}\sqrt{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\sqrt{\mathbf{2}}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{4}\mathbf{\pi }\mathbf{R}} \end{gathered}$$

Thus, the field is $\frac{\sqrt{\mathbf{2}}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{4}\mathbf{\pi R}}$.

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

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

Including n sides of the polygon, the net field is

$$\begin{gathered} B=n\times \frac{{\mu }_{0}l}{2\pi R}\sin \frac{\pi }{n} \\ =\frac{n{\mu }_{0}l}{2\pi R}\sin \frac{\pi }{n} \end{gathered}$$

Thus, the field is $\frac{n{\mu }_{0}l}{2\pi R}\sin \frac{\pi }{n}$.

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

$\sin \frac{\pi }{n}\approx \frac{\pi }{n}$

Equation (2) thus becomes,

$$\begin{gathered} B=\frac{n{\mu }_{0}l}{2\pi R}\sin \frac{\pi }{n} \\ \approx \frac{n{\mu }_{0}l}{2\pi R}\sin \frac{\pi }{n} \\ =\frac{{\mu }_{0}l}{2R} \end{gathered}$$

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