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(B) The magnetic moment $M$ of a coil is given by $M = N I A$. For a flat spiral coil where the turns are uniformly distributed between radii $r_1$ an

Vedclass · Accepted Answer

$2.64$

Vedclass · Answer

(B) The magnetic moment $M$ of a coil is given by $M = N I A$. For a flat spiral coil where the turns are uniformly distributed between radii $r_1$ and $r_2$,the effective area $A$ is calculated by integrating the area of each turn: $A = \int_{r_1}^{r_2} \pi r^2 \frac{N}{r_2 - r_1} dr = \frac{N \pi}{r_2 - r_1} [\frac{r^3}{3}]_{r_1}^{r_2} = N \pi \frac{r_2^3 - r_1^3}{3(r_2 - r_1)} = N \pi \frac{r_1^2 + r_1 r_2 + r_2^2}{3}$.Given $N = 200$,$r_1 = 0.03	ext{ m}$,$r_2 = 0.06	ext{ m}$,and $I = 20	ext{ mA} = 0.02	ext{ A}$.Substituting these values: $M = 200 	imes 0.02 	imes \pi 	imes \frac{(0.03)^2 + (0.03)(0.06) + (0.06)^2}{3}$.$M = 4 	imes \pi 	imes \frac{0.0009 + 0.0018 + 0.0036}{3} = 4 	imes \pi 	imes \frac{0.0063}{3} = 4 	imes \pi 	imes 0.0021 = 0.0084 	imes 3.14159 \approx 0.02639	ext{ A.m}^2$.This can be written as $2.639 	imes 10^{-2}	ext{ A.m}^2$. Rounding to two decimal places,$\alpha = 2.64$.

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