$\mathrm{dF}=\frac{\mu_{0} \mathrm{i}_{1} \times \mathrm{i}_{1} \times \mathrm{i}}{2 \pi \mathrm{a}}(\mathrm{a}=2 \mathrm{R})$

$\because i_{1}=\frac{i}{2 \pi R} d x$

$\mathrm{dF}=\frac{\mu_{0}}{2 \pi \mathrm{a}}\left(\frac{\mathrm{i}}{2 \pi \mathrm{R}} \mathrm{dx}\right)^{2} \mathrm{I}=\frac{\mu_{0}}{2 \pi \times 2 \mathrm{R}} \cdot \frac{\mathrm{i}^{2}}{4 \pi^{2} \mathrm{R}^{2}} \cdot \mathrm{dx} . \mathrm{dx} . \mathrm{I}$

Pressure on the element $=\frac{\mu_{0}}{2 \pi \cdot 2 R} \cdot \frac{\mathrm{i}^{2} \mathrm{d} \mathrm{x}}{4 \pi^{2} \mathrm{R}^{2}} \cdot \frac{\mathrm{d} \mathrm{x} . \mathrm{l}}{\mathrm{d} \mathrm{x} . \mathrm{l}}$

Since area of element $= \mathrm{dxl}$

$\mathrm{dp}=\frac{\mu_{0} \mathrm{i}^{2}}{2 \pi \cdot 2 \mathrm{R} \cdot 4 \pi^{2} \cdot \mathrm{R}^{2}} \times \mathrm{d} \mathrm{x}$

Effective pressure $=\frac{\mu_{0} \mathrm{i}^{2}}{2 \pi \cdot 2 \mathrm{R} \cdot 4 \pi^{2} \cdot \mathrm{R}^{2}} \times 2 \pi \mathrm{r}$

Since $\int \mathrm{d} \mathrm{x}=2 \pi \mathrm{R}$

$P=\frac{\mu_{0} i^{2}}{8 \pi^{2} R^{2}}$