The length of a solenoid is ell,whose winding | vedclass

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(B) The inductance of a solenoid is given by $L = \mu_{0} N^{2} \frac{A_{s}}{\ell}$,where $A_{s}$ is the cross-sectional area of the solenoid and $N$

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$\frac{\mu_{0}}{4 \pi \ell} \left( \frac{R m}{\rho D} \right)$

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(B) The inductance of a solenoid is given by $L = \mu_{0} N^{2} \frac{A_{s}}{\ell}$,where $A_{s}$ is the cross-sectional area of the solenoid and $N$ is the number of turns.Let $x$ be the total length of the wire used for windings and $a$ be the cross-sectional area of the wire.The resistance $R$ is given by $R = \frac{ho x}{a}$.The mass $m$ of the wire is $m = a x D$.Multiplying these,we get $R m = \left( \frac{ho x}{a} ight) (a x D) = ho x^{2} D$.Thus,$x^{2} = \frac{R m}{ho D}$.The total length of the wire $x$ is also equal to $N 	imes (2 \pi r)$,where $r$ is the radius of the solenoid.Therefore,$N = \frac{x}{2 \pi r}$.Substituting $N$ into the inductance formula: $L = \mu_{0} \left( \frac{x}{2 \pi r} ight)^{2} \frac{\pi r^{2}}{\ell} = \mu_{0} \frac{x^{2}}{4 \pi^{2} r^{2}} \frac{\pi r^{2}}{\ell} = \frac{\mu_{0} x^{2}}{4 \pi \ell}$.Substituting $x^{2} = \frac{R m}{ho D}$,we get $L = \frac{\mu_{0}}{4 \pi \ell} \left( \frac{R m}{ho D} ight)$.

The length of a solenoid is $\ell$,whose windings are made of a material of density $D$ and resistivity $\rho$. The winding resistance is $R$. The inductance of the solenoid is (where $m$ = mass of winding wire,$\mu_{0}$ = permeability of free space).

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