An insulated wire is wound so that it forms a flat coil with N =...

Let the number of turns per unit radial width be $n = \dfrac{N}{r_2 - r_1}$.

Consider an elemental ring of radius $r$ and width $dr$. The number of turns in this element is $dN = n \, dr$.

The magnetic moment of this elemental ring is $dM = I \, dN \, A = I \left( \dfrac{N}{r_2 - r_1} \, dr \right) (\pi r^2)$.

Integrating from $r_1$ to $r_2$ to find the total magnetic moment $M$:

$M = \int_{r_1}^{r_2} \dfrac{I \pi N}{r_2 - r_1} r^2 \, dr = \dfrac{I \pi N}{3(r_2 - r_1)} (r_2^3 - r_1^3)$

Substituting the given values: $I = 20 \times 10^{-3}$ A, $N = 200$, $r_1 = 0.03$ m, and $r_2 = 0.06$ m:

$M = \dfrac{20 \times 10^{-3} \times \pi \times 200}{3(0.06 - 0.03)} \left( (0.06)^3 - (0.03)^3 \right)$

$M = \dfrac{4\pi}{0.09} (216 \times 10^{-6} - 27 \times 10^{-6})$

$M = \dfrac{4\pi}{0.09} \times 189 \times 10^{-6} = 4\pi \times 2100 \times 10^{-6}$

$M = 84\pi \times 10^{-4} \approx 84 \times 3.14 \times 10^{-4} = 263.76 \times 10^{-4} \approx 2.64 \times 10^{-2}$ A.m$^2$

Comparing with $\alpha \times 10^{-2}$ A.m$^2$, we get $\alpha = 2.64$.

Answer: $2.64$