To manufacture a solenoid of length ell and i | vedclass

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(A) The inductance $L$ of a long solenoid is given by $L = \frac{\mu_0 N^2 A}{\ell}$,where $N$ is the number of turns,$A$ is the cross-sectional area,

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$\left[\frac{4 \pi \ell L}{\mu_0}\right]^{\frac{1}{2}}$

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(A) The inductance $L$ of a long solenoid is given by $L = \frac{\mu_0 N^2 A}{\ell}$,where $N$ is the number of turns,$A$ is the cross-sectional area,and $\ell$ is the length.Let $r$ be the radius of the solenoid,so $A = \pi r^2$. Thus,$L = \frac{\mu_0 N^2 \pi r^2}{\ell}$.The total length of the wire $W$ is the circumference of one turn multiplied by the number of turns: $W = N(2 \pi r)$.From this,$N = \frac{W}{2 \pi r}$.Substituting $N$ into the inductance formula: $L = \frac{\mu_0 (W / 2 \pi r)^2 \pi r^2}{\ell} = \frac{\mu_0 W^2 \pi r^2}{\ell (4 \pi^2 r^2)} = \frac{\mu_0 W^2}{4 \pi \ell}$.Solving for $W$: $W^2 = \frac{4 \pi \ell L}{\mu_0}$,which gives $W = \left[\frac{4 \pi \ell L}{\mu_0}\right]^{\frac{1}{2}}$.

To manufacture a solenoid of length $\ell$ and inductance $L$,the length of the thin wire required is (Diameter of the solenoid is very small compared to its length,$\mu_0$ is the permeability of free space).

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