An insulated wire is wound so that it forms a flat coil with \(N = 200\) turns. The radius of the innermost turn is \(r_1 = 3\) cm, and of the outermost turn \(r_2 = 6\) cm. If \(20\) mA current flows in it then the magnetic moment will be \(\alpha \times 10^{-2}\) A.m\(^2\). The value of \(\alpha\) is _____.

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: \(2.64\)