Explanation

(d) Consider an element of thickness $dr$ at a distance $r$ from the centre of spiral coil.
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Number of turns in coil $= n$
Number of turns per unit length
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{n}{b - a}$
Number of turns in element $dr = dn$
Number of turns per unit length in element $dr$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{ndr}{b - a}$
$ie,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dn = \frac{ndr}{b - a}$
Magnetic field at its centre due to element $dr$ is
$\ \ dB = \frac{\mu_{0}Idn}{2r} = \frac{\mu_{0}I}{2}\frac{n}{(b - a)}\frac{dr}{r}$
$\therefore B = \int_{a}^{b}\frac{\mu_{0}Indr}{2(b - a)r} = \frac{\mu_{0}In}{2(b - a)}\ \int_{a}^{b}\frac{dr}{r}$
$\ \ \ \ \ \ \ = \frac{\mu_{0}In}{2(b - a)}\log_{e}\left( \frac{b}{a} \right)$