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 pass through the coil, the magnetic field at the centre is

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$\frac{μ_{0} N I}{b}$

$\frac{2 μ_{0} N I}{a}$

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

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

answer is C.

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Detailed Solution

Number of turns per unit width $h = \frac{N}{b - a}$

Consider an elemental ring of radius x and with thickness dx Number of turns in the ring $= d N = \frac{N d x}{b - a}$

Magnetic field at the centre due to the ring element

$d B = \frac{μ_{0} (d N) i}{2 x} = \frac{μ_{0} i}{2} \cdot \frac{N d x}{(b - a)} \cdot \frac{1}{x}$

$∴$ Field at the centre

$= \int d B = \frac{μ_{0} N i}{2 (b - a)} \int_{a}^{b} \frac{d x}{x}$

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

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