Question

4-2 Find the inductance of a closely wound solenoid of radius \( R \) and length \( L \) having \( N \) turns when \( R \ll L \). 4-2 The inductance of the solenoid may be found by equating the energy of the enclosed field to \( \frac{1}{2} L I^{2} \) or by differentiating the flux at any turn with respect to \( I \) and summing over the turns. Since the magnetic induction field is nearly constant through the volume of the solenoid and nearly zero outside the solenoid, either method ought to work. For either method we need the field of a solenoid of length \( \ell \) and \( N \) turns: \( B=\mu_{0} N I \ell \). Using energy: \[ \begin{aligned} W & =\frac{1}{2} L I^{2}=\int \frac{B^{2}}{2 \mu_{0}} d^{3} r \\ & =\left(\frac{\mu_{0} N I}{\ell}\right)^{2} \frac{\pi R^{2} \ell}{2 \mu_{0}}=\frac{\pi \mu_{0} I^{2} N^{2} R^{2}}{2 \ell} \end{aligned} \] from which we deduce \[ L=\frac{\pi \mu_{0} N^{2} R^{2}}{\ell} \] From the flux: \[ L=N \frac{\partial \Phi}{\partial I}=N \frac{\partial}{\partial I} \int \frac{\mu_{0} N I}{\ell} d S=\frac{\mu_{0} \pi R^{2} N^{2}}{\ell} \]

          4-2 Find the inductance of a closely wound solenoid of radius \( R \) and length
\( L \) having \( N \) turns when \( R \ll L \).
4-2 The inductance of the solenoid may be found by equating the energy of the enclosed field to \( \frac{1}{2} L I^{2} \) or by differentiating the flux at any turn with respect to \( I \) and summing over the turns. Since the magnetic induction field is nearly constant through the volume of the solenoid and nearly zero outside the solenoid, either method ought to work. For either method we need the field of a solenoid of length \( \ell \) and \( N \) turns: \( B=\mu_{0} N I \ell \).
Using energy:
\[
\begin{aligned}
W & =\frac{1}{2} L I^{2}=\int \frac{B^{2}}{2 \mu_{0}} d^{3} r \\
& =\left(\frac{\mu_{0} N I}{\ell}\right)^{2} \frac{\pi R^{2} \ell}{2 \mu_{0}}=\frac{\pi \mu_{0} I^{2} N^{2} R^{2}}{2 \ell}
\end{aligned}
\]
from which we deduce
\[
L=\frac{\pi \mu_{0} N^{2} R^{2}}{\ell}
\]

From the flux:
\[
L=N \frac{\partial \Phi}{\partial I}=N \frac{\partial}{\partial I} \int \frac{\mu_{0} N I}{\ell} d S=\frac{\mu_{0} \pi R^{2} N^{2}}{\ell}
\]

4-2 Find the inductance of a closely wound solenoid of radius R and length
L having N turns when R ≪ L.
4-2 The inductance of the solenoid may be found by equating the energy of the enclosed field to (1)/(2) L I^2 or by differentiating the flux at any turn with respect to I and summing over the turns. Since the magnetic induction field is nearly constant through the volume of the solenoid and nearly zero outside the solenoid, either method ought to work. For either method we need the field of a solenoid of length ℓ and N turns: B=μ0 N I ℓ.
Using energy:

W =(1)/(2) L I^2=∫(B^2)/(2 μ0) d^3 r
=((μ0 N I)/(ℓ))^2(π R^2ℓ)/(2 μ0)=(πμ0 I^2 N^2 R^2)/(2 ℓ)

from which we deduce

L=(πμ0 N^2 R^2)/(ℓ)

From the flux:

L=N (∂Φ)/(∂ I)=N (∂)/(∂ I)∫(μ0 N I)/(ℓ) d S=(μ0π R^2 N^2)/(ℓ)

Added by Aaron M.

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