91Ó°ÊÓ

The radius of the circuit is$r$.

The uniform line charge is$\mathrm{λ}$.

Electric field due to charge $\mathit{q}$at a distance $\mathit{r}$is proportional to the charge $\mathit{q}$and inversely proportional to the square of the distance $\mathit{r}$as,

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}$

The center line of the circular loop of radius $r$is drawn. At a point z on the center line the edges of the loop make the angle$\mathrm{θ}$, as shown below:

From the above, using Pythagoras theorem in left right triangle, we can write$R^2=r^2+;$

$\cos \mathrm{θ}=\frac{z}{\sqrt{r^2+z^2}}$

It can be considered that the circular loop is made up of symmetrical elements of small lengths $dx$, which are diagonally opposite then the differential field at the point P due to a pair of diagonally opposite elements can be written as

$dE=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{dq}{R^2}\cos \mathrm{θ}$.

Substitute $r^2+z^2$for$R^2$, $\frac{z}{\sqrt{r^2}+z^2}$for $\cos \mathrm{θ}and\mathrm{λ}dSfordq$into the equation.

$$\begin{gathered} dE=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{dq}{r^2+z^2}\left(\frac{z}{\sqrt{r^2+z^2}}\right) \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{z\mathrm{λ}dS}{{\left(r^2+z^2\right)}^{3/2}} \end{gathered}$$

Integrate above differential integral as,

$$\begin{gathered} E=∫dE \\ =∫\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{z\mathrm{λ}dS}{{\left(r^s+z^2\right)}^{3/2}} \\ =\frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{z}{{\left(r^s+z^2\right)}^{3/2}}\right)\left(2\mathrm{Ï€}r\right) \\ =\frac{\mathrm{λ}}{2\mathrm{Ï€}\mathrm{Î\mu }}\left(\frac{zr}{{\left(r^s+z^2\right)}^{3/2}≠}\right) \end{gathered}$$

Thus, the electric fieldat a distance zabove the center of a circular loop

is$E=\frac{\mathrm{λ}}{2{\mathrm{Î\mu }}_0}\left(\frac{zr}{{\left(r^s+z^2\right)}^{3/2}≠}\right)$.