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A long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$ . The spiral lies in the $x y$ -plane and a steady current $I$ flows through the wire. The $z -$ component of the magnetic field at the centre of the spiral is
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$(a) \frac{μ_{0} N I}{2 (b - a)} \ln (\frac{b}{a})$

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

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

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

answer is A.

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Detailed Solution

If we take a small strip of $d r$ at a distance $r$ from centre, then number of turns this strip would be

$d n = (\frac{N}{b - a}) d r$

Magnetic field due to this element at the centre of the coil will be

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

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