Magnetic Moment of Flat Spiral Coil

To find the magnetic moment of the flat spiral coil, we consider an elemental concentric circular strip of radius $r$ and infinitesimal width $dr$.

The total number of turns $N$ is uniformly distributed over the radial distance from $r_1$ to $r_2$. Therefore, the number of turns per unit radial length, $n$, is given by: $n = \frac{N}{r_2 - r_1}$

The number of turns in the elemental ring of thickness $dr$ is: $dN = n \, dr = \frac{N}{r_2 - r_1} dr$

The magnetic moment $dM$ contributed by these $dN$ turns, each carrying a current $I$ and enclosing an area $A = \pi r^2$, is: $dM = dN \cdot I \cdot (\pi r^2) = \frac{\pi N I}{r_2 - r_1} r^2 dr$

To find the total magnetic moment $M$ of the coil, we integrate $dM$ from $r = r_1$ to $r = r_2$: $M = \int_{r_1}^{r_2} \frac{\pi N I}{r_2 - r_1} r^2 dr$

$M = \frac{\pi N I}{r_2 - r_1} \left[ \frac{r^3}{3} \right]_{r_1}^{r_2} = \frac{\pi N I (r_2^3 - r_1^3)}{3(r_2 - r_1)}$

Using the algebraic identity $r_2^3 - r_1^3 = (r_2 - r_1)(r_2^2 + r_1 r_2 + r_1^2)$, the expression simplifies to: $M = \frac{1}{3} \pi N I (r_2^2 + r_1 r_2 + r_1^2)$

Given parameters:

• $N = 200$
• $I = 20 \text{ mA} = 20 \times 10^{-3} \text{ A}$
• $r_1 = 3 \text{ cm} = 3 \times 10^{-2} \text{ m}$
• $r_2 = 6 \text{ cm} = 6 \times 10^{-2} \text{ m}$

Calculating the term $(r_2^2 + r_1 r_2 + r_1^2)$: $r_1^2 = (3 \times 10^{-2})^2 = 9 \times 10^{-4} \text{ m}^2$ $r_2^2 = (6 \times 10^{-2})^2 = 36 \times 10^{-4} \text{ m}^2$ $r_1 r_2 = (3 \times 10^{-2})(6 \times 10^{-2}) = 18 \times 10^{-4} \text{ m}^2$ $r_2^2 + r_1 r_2 + r_1^2 = (36 + 18 + 9) \times 10^{-4} = 63 \times 10^{-4} \text{ m}^2$

Now, substituting these values into the formula for $M$: $M = \frac{1}{3} \pi (200) (20 \times 10^{-3}) (63 \times 10^{-4})$ $M = \frac{1}{3} \pi \cdot (4) \cdot (63 \times 10^{-4})$ $M = 84 \pi \times 10^{-4} \text{ A}\cdot\text{m}^2$

Taking $\pi \approx 3.1416$: $M = 84 \times 3.1416 \times 10^{-4} \approx 263.89 \times 10^{-4} \text{ A}\cdot\text{m}^2 = 2.64 \times 10^{-2} \text{ A}\cdot\text{m}^2$

Comparing this with $M = \alpha \times 10^{-2} \text{ A}\cdot\text{m}^2$, we get: $\alpha = 2.64$

Thus, the correct option is B.