Explanation

(b) As shown figure take an element $dl\ at\ C\ of\ wire,\ where\ OC = l.\ \ Let\ PC = r\ \ and\ \ \angle OPC = \ \phi$.
According to Biot Savarts law; magnitude of magnetic field induction at $P$ due to current element at $C$ is
[image:4_image116.jpeg]
$dB = \frac{\mu_{0}}{4\pi}\ \frac{idl\sin\theta}{r^{2}}$ (i)
Here, $\theta\ \ \ = 90{^\circ} + \phi;r = a\sec\phi$
And $l\ \ \ \ = a\tan{\phi;dl = a\sec^{2}{\phi\ d\ \phi}}$
$\therefore\ \ dB = \frac{\mu_{0}}{4\pi}\ \frac{i(\sec^{2}{\phi\ d\ \phi)\sin{(90{^\circ} + \phi)}}}{a^{2}\sec^{2}\phi}$
$\ \ = \frac{\mu_{0}}{4\pi}\frac{1}{a}\cos{\phi\ d\ \phi}$
Total magnetic field induction at $P$ is
$B\ \ \ = \ \int_{\phi_{1}}^{}{90{^\circ}\ \frac{\mu_{0}}{4\pi}\frac{1}{a}\cos{\phi\ d\ \phi = \ \ \frac{\mu_{0}}{4\pi}\frac{1}{a}\ (\sin{\phi){_{\phi}}_{1}^{90}{^\circ}}}}\ $
$= \ $ $\frac{\mu_{0}}{4\pi}\frac{1}{a}(1 - \sin{\phi_{1}) =}\frac{\mu_{0}}{4\pi}\frac{1}{a}\ \left( 1 - \frac{b}{\sqrt{a^{2} + b^{2}}} \right)$