Question
An insulated wire is wound so that it forms a flat coil with $N = 200$ turns. The radius of the innermost turn is $r_{1} = 3\text{ }cm$, and of the outermost turn $r_{2} = 6\text{ }cm$. If 20 mA current flow in it then the magnetic moment will be $\alpha \times 10^{- 2}\text{ }A.m^{2}$. The value of $\alpha$ is $\_\_\_\_$。
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Solution
Sol. [Key Concept: $dM = I\ dN\ \pi\ r^{2}$ with $\frac{N}{r_{2} - r_{1}}$ turns per unit radius; integrate $r$ from $r_{1}$ to $r_{2}$.]
$${dM = (dI)\pi r^{2} = (IdN)\pi r^{2} }{M = \int_{i}^{f}\mspace{2mu} I\left( \frac{N}{r_{ex} - r_{i}} \right)dr\pi r^{2} }{M = \frac{I\pi N}{r_{ex} - r_{i}}\left\lbrack \frac{r^{3}}{3} \right\rbrack_{ri}^{r_{ex}} }{M = \frac{I\pi N}{3\left( r_{ex} - r_{in} \right)}\left( r_{ex}\ ^{3} - r_{i}\ ^{3} \right) }{M = \frac{20 \times 10^{- 3} \times 3.14 \times 200\left\lbrack 6^{3} - 3^{3} \right\rbrack 10^{- 6}}{3\lbrack 3\rbrack \times 10^{- 2}} }{M = \frac{2373.84 \times 10^{- 6}}{9 \times 10^{- 2}} }{M = 263.76 \times 10^{- 4} }{M = 2.64 \times 10^{- 2}{Am}^{2} }$$
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