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 :

Engineering

Physics

Magnetic Field

Question

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

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

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

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

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(b – a) $\to$ N

$dr \to (\frac{N}{b - a}) dr$

$B = \frac{μ_{0} IN}{2 (b - a)} ℓn (b / a)$

$dB = (\frac{μ_{0} I}{2 r}) (\frac{N}{b - a}) dr$

$B = \frac{μ_{0} IN}{2 (b - a)} \int_{a}^{b} \frac{dr}{r}$

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