A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $I$ passes through the coil, the magnetic field at the centre is

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$\frac{2 μ_{0} N I}{a}$

$\frac{μ_{0} N I}{b}$

$\frac{μ_{0} N I}{2 (b - a)} \ln \frac{b}{a}$

$\frac{μ_{0} I}{2 (b - a)} \ln \frac{b}{a}$

answer is C.

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Detailed Solution

$d N = (\frac{N}{b - a}) d r N o w, d B = \frac{μ_{0} (d N) I}{2 r} a n d B = \int_{a}^{b} \frac{μ_{0} (\frac{N}{b - a}) d r I}{2 r} = \frac{μ_{0} N I}{2 (b - a)} \ln (\frac{b}{a})$

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