1. Idea of the problem

A tightly wound spiral can be treated as a collection of many closely spaced circular turns whose radii vary continuously from $a$ to $b$.

The magnetic field at the centre due to one circular turn of radius $r$ carrying current $I$ is

$dB=\frac{\mu_0 I}{2r}$

If the spiral has total $N$ turns spread uniformly from radius $a$ to $b$, then the number of turns per unit radial length is

$n=\frac{N}{b-a}$

Hence, in a small radial thickness $dr$, the number of turns is

$dN = n\,dr = \frac{N}{b-a}dr$

2. Field due to elemental turns

The contribution to magnetic field at the centre from turns between $r$ and $r+dr$ is

$dB = \frac{\mu_0 I}{2r} \, dN$

Substitute $dN$:

$dB = \frac{\mu_0 I}{2r} \cdot \frac{N}{b-a}dr$

$dB = \frac{\mu_0 I N}{2(b-a)}\frac{dr}{r}$

3. Integrate from inner radius to outer radius

$B = \int_a^b dB = \frac{\mu_0 I N}{2(b-a)}\int_a^b \frac{dr}{r}$

Using

$\int_a^b \frac{dr}{r} = \ln\left(\frac{b}{a}\right)$

we get

$B = \frac{\mu_0 I N}{2(b-a)}\ln\left(\frac{b}{a}\right)$

4. Match with options

This matches:

$\boxed{\frac{\mu_0 I N}{2(b-a)}\ln\left(\frac{b}{a}\right)}$

So the correct option is A.

5. Check other options briefly

• B and D do not have the factor $N$, so they cannot represent the field of an $N$-turn spiral.
• C has incorrect dimensional form and also lacks $N$.

Therefore, only A is correct.