Magnetic Field at the Center of Spiral Wire

To find the Z-component of the magnetic field at the center of the spiral wire, we can divide the spiral into elemental concentric circular loops.

1. Turn Density:
The total number of turns $N$ is uniformly distributed over the radial width from inner radius $a$ to outer radius $b$. Thus, the number of turns per unit radial distance is:
$n = \frac{N}{b - a}$

2. Elemental Circular Loop:
Consider a thin circular ring element of radius $r$ and thickness $dr$, where $a \le r \le b$.

The number of turns $dN$ in this elemental width $dr$ is given by:
$dN = n \, dr = \frac{N}{b - a} \, dr$

3. Magnetic Field Due to the Element:
The magnetic field $dB$ at the center produced by a circular loop of radius $r$ carrying current $I$ with $dN$ turns is directed along the Z-axis (normal to the X-Y plane) and is given by Biot-Savart law as:
$dB = \frac{\mu_0 (dN) I}{2r}$

Substituting the expression for $dN$:
$dB = \frac{\mu_0 I}{2r} \left( \frac{N}{b - a} \, dr \right) = \frac{\mu_0 N I}{2(b - a)} \frac{dr}{r}$

4. Total Magnetic Field:
To find the total magnetic field $B$ at the center, we integrate $dB$ from $r = a$ to $r = b$:
$B = \int_a^b dB = \frac{\mu_0 N I}{2(b - a)} \int_a^b \frac{dr}{r}$

Evaluating the integral:
$\int_a^b \frac{dr}{r} = \left[ \ln r \right]_a^b = \ln b - \ln a = \ln\left(\frac{b}{a}\right)$

Substituting this back into the expression for $B$:
$B = \frac{\mu_0 N I}{2(b - a)} \ln\left(\frac{b}{a}\right)$

This corresponds to Option (A).