To manufacture a solenoid of length 1 ,m and | vedclass

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(B) The inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$, where $l$ is the length of the solenoid and $A = \pi r^2$ is the cross-secti

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$0.10 \,km$

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(B) The inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$, where $l$ is the length of the solenoid and $A = \pi r^2$ is the cross-sectional area.Let $x$ be the total length of the wire required. The wire is wound into $N$ turns, so $x = (2 \pi r) N$, which implies $N = \frac{x}{2 \pi r}$.Substituting $N$ into the inductance formula:$L = \frac{\mu_0 (x / 2 \pi r)^2 	imes \pi r^2}{l} = \frac{\mu_0 x^2}{4 \pi l}$.Rearranging for $x$:$x^2 = \frac{4 \pi L l}{\mu_0} \implies x = \sqrt{\frac{4 \pi L l}{\mu_0}}$.Given $L = 1 \,mH = 10^{-3} \,H$, $l = 1 \,m$, and $\mu_0 = 4 \pi 	imes 10^{-7} \,T \cdot m/A$:$x = \sqrt{\frac{4 \pi 	imes 10^{-3} 	imes 1}{4 \pi 	imes 10^{-7}}} = \sqrt{10^4} = 100 \,m$.Converting to kilometers: $100 \,m = 0.10 \,km$.

To manufacture a solenoid of length $1 \,m$ and inductance $1 \,mH$, the length of thin wire required is (cross-sectional diameter of a solenoid is considerably less than the length).

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