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Step 1: Understand the Problem
We have a coil with N turns wound in the shape of a spiral, having inner radius $a$ and outer radius $b$. A current $I$ flows through this coil. We need to find the magnetic field $B$ at the center of this spiral.
Step 2: Express the Magnetic Field Due to a Single Circular Turn
For a single flat circular ring carrying current $I$ and having radius $r$, the magnetic field at the center is given by
$$ B_{\text{ring}} = \frac{\mu_0 \, I}{2 \, r} \,. $$
Step 3: Consider Differential Rings from Radius $a$ to $b$
In the spiral, each turn can be approximated as a circular ring with a slightly varying radius from $a$ to $b$. We assume the winding is uniform, so the turn density (turns per unit radial length) is
$$ \text{Turn density} = \frac{N}{b - a} \,. $$
Let us denote a small element of radius $r$, where $a \leq r \leq b$. Each small ring has a thickness $dr$ and carries a fraction of the total N turns proportional to $dr$.
Step 4: Write the Magnetic Field Due to a Differential Ring
For a small ring of radius $r$ with $dN$ turns, carrying the same current $I$, the magnetic field at the center is
$$ dB = \frac{\mu_0 \, I \, dN}{2 \, r} \,. $$
Because $dN = \left(\frac{N}{b - a}\right) \, dr$, we get
$$ dB = \frac{\mu_0 \, I \, \bigl(\frac{N}{b - a}\bigr) \, dr}{2 \, r} \,. $$
Step 5: Integrate from $r = a$ to $r = b$
The total magnetic field at the center is obtained by integrating $dB$ over $r$ from $a$ to $b$:
$$ B = \int_a^b \frac{\mu_0 \, I \, \bigl(\frac{N}{b - a}\bigr)}{2 \, r} \, dr
= \frac{\mu_0 \, I \, N}{2 \, (b - a)} \int_a^b \frac{dr}{r}
= \frac{\mu_0 \, I \, N}{2 \, (b - a)} \bigl[\ln r\bigr]_a^b \,. $$
Hence,
$$ B = \frac{\mu_0 \, I \, N}{2 \, (b - a)} \ln\left(\frac{b}{a}\right) \,.$$
This matches the given correct answer.
Step 6: Final Answer
$$ B = \frac{\mu_0 \, I \, N}{2 \, (b - a)} \ln\left(\frac{b}{a}\right). $$
