The ratio of magnetic field at the centre of a current carrying coil of radius r to the magnetic field at distance r from the centre of coil on its…

The formula to calculate the magnetic field at any axial point situated at a distance $x$x from the centre of a circular current carrying conductor of radius $r$r can be written as

$$B=\frac{{\mu }_{0}I{r}^{2}}{2{({r}^{2}+{x}^{2})}^{3/2}}...(1)$$B=2(r2+x2)3/2μ0​Ir2​...(1)

Substitute $0$0 for $x$x in equation (1) to calculate the magnetic field $({B}_{c})$(Bc​) at the centre of the conductor.

$$\begin{matrix}{B}_{c} & = & \frac{{\mu }_{0}I{r}^{2}}{2{({r}^{2}+{0}^{2})}^{3/2}} \\ & = & \frac{{\mu }_{0}I{r}^{2}}{2{r}^{3}} \\ & = & \frac{{\mu }_{0}I}{2r}...(2)\end{matrix}$$Bc​​===​2(r2+02)3/2μ0​Ir2​2r3μ0​Ir2​2rμ0​I​...(2)​

Substitute $r$r for $x$x in equation (1) to calculate the magnetic field $({B}_{r})$(Br​) at the given distance from the centre of the conductor.

$$\begin{matrix}{B}_{r} & = & \frac{{\mu }_{0}I{r}^{2}}{2{({r}^{2}+{r}^{2})}^{3/2}} \\ & = & \frac{{\mu }_{0}I{r}^{2}}{4\sqrt{2}{r}^{3}} \\ & = & \frac{{\mu }_{0}I}{4\sqrt{2}r}...(3)\end{matrix}$$Br​​===​2(r2+r2)3/2μ0​Ir2​42​r3μ0​Ir2​42​rμ0​I​...(3)​

Divide equation (2) by equation (3) to calculate the required ratio of the magnetic fields.

$$\begin{matrix}\frac{{B}_{c}}{{B}_{r}} & = & \frac{\frac{{\mu }_{0}I}{2r}}{\frac{{\mu }_{0}I}{4\sqrt{2}r}} \\ & = & 2\sqrt{2} \\ & = & \sqrt{8}\end{matrix}$$Br​Bc​​​===​42​rμ0​I​2rμ0​I​​22​8​​