Magnetic field produced at P due to current in the element is given by Biot Savart'
dB = \(\frac{\mu_0}{4\pi}\frac{Idlsin\theta}{r^2}\) \(\theta = 90^\circ, sin\theta =1\)
dB = \(\frac{\mu_0}{4\pi}\frac{Idl}{r^2}\)
dB is resolved into two components as dB cos\(\theta\) along the axis and dBsin\(\theta \perp\) to the axis. Magnetic field produced at P due to current in the ful loop is given by
\(\Sigma dB = \Sigma dBcos\theta\)
B = \(\Sigma \frac{\mu_0}{4\pi}\frac{Idl}{r^2}cos\theta=\frac{\mu_0}{4\pi}\frac{I}{r^2}cos\theta\Sigma dl\) (\(\Sigma dl = 2\pi R\))
B = \(\frac{\mu_0}{4\pi}\frac{I}{r^2}cos\theta\times2\pi R\) In \(\bigtriangleup \) OAP, cos\(\theta\) = R/r
B= \(\frac{\mu_0}{4\pi}\frac{I}{r^2}\times\frac{R}{r}\times2\pi R\) \(r^2 = R^2+ x^2 \implies r = (R^2 + x^2)^{1/2}\)
B = \(\frac{\mu_0}{4\pi} \frac{I}{r^3}2\pi R^2\) \(\therefore r^3 = (R^2+x^2)^{3/2}\)
B = \(\frac{\mu_0}{4\pi}. \frac{2\pi R^2nI}{(R^2+x^2)^{3/2}}\)
At the centre of the coil , B = \(\frac{\mu_0}{2}\frac{nI}{R}\)
