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

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

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

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

answer is A.

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Detailed Solution

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

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

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