The magnetic field at P for current I in dl element,
\[dB=\frac{\mu_{0}I}{4\pi }\frac{dlSin{90}^{0}}{r^{2}}=\frac{\mu_{0}I}{4\pi }\frac{dl}{r^{2}}\]
Sum of Y-component of dB for the whole loop will be zero. Now X-component of dB is,
\[dB_{x}=\frac{\mu_{0}I}{4\pi }\frac{dl}{r^{2}}Sin\theta =\frac{\mu_{0}I}{4\pi }\frac{dl}{r^{2}}\frac{R}{r}=\frac{\mu_{0}I}{4\pi }\frac{Rdl}{r^{3}}\]
From the figure, $r=(R^{2}+x^{2})^{\frac{1}{2}}$
Therefore,
\[dB_{x}=\frac{\mu_{0}I}{4\pi}\frac{Rdl}{({R}^{2}+{x}^{2})^{\frac{3}{2}}}\]
Total magnetic field at P for the current in the entire loop,
\[B=\oint dB_{x}=\oint \frac{\mu_{0}I}{4\pi}\frac{Rdl}{({R}^{2}+{x}^{2})^{\frac{3}{2}}}=\frac{\mu_{0}I}{4\pi}\frac{R(2\pi R)}{({R}^{2}+{x}^{2})^{\frac{3}{2}}}=\frac{\mu_{0}I}{2}\frac{R^{2}}{({R}^{2}+{x}^{2})^{\frac{3}{2}}}\]
Magnetic field at the center O (i.e. at x=0) for the current in the entire loop,
\[B=\frac{\mu _{0}I}{2R}\]
