Question

(a) A long, straight solenoid has N turns, uniform cross-sectional area A, and length l. Show that the inductance of this solenoid is given by the equation $\mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{A}{\mathbf{N}}^{\mathbf{2}}}{\mathbf{l}}$. Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because B is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) A metallic laboratory spring is typically $\mathbf{5}\mathit{c}\mathit{m}\mathbf{}$long and $\mathbf{0}\mathbf{.}\mathbf{15}\mathit{c}\mathit{m}$in diameter and has $\mathbf{50}$coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

Short answer

a)$L=\frac{{\mathrm{\mu }}_0{\mathrm{AN}}^2}{\mathrm{l}}$

b)role="math" localid="1664194101205" $L=1.11\times {10}^{-7}H$

Step-by-step solution

Magnetic field in a solenoid

For a long solenoid of$\mathit{N}$turns, length$\mathit{i}$and current$\mathit{i}$, the magnetic field is given by$B={\mu }_{0}ni={\mu }_0\frac{N}{I}i$.

For the area of cross-section $\mathit{A}$, the magnetic flux is given by role="math" localid="1664194375207" $\varphi =BA=\frac{{\mathrm{\mu }}_0\mathrm{NAi}}{\mathrm{l}}$.

The self-inductance is defined as $L=\frac{N{\varphi }_B}{i}$.

Clearly, after substituting the flux equation, we will get$L=\frac{{\mu }_0{N}^{2}A}{l}$ .

Calculate the self-inductance

Given that $I=5cm$and $D=.015cm$$andN=50$.

role="math" localid="1664194642248" $$\begin{gathered} A=\frac{\pi D^2}{4} \\ A=\frac{\pi \times 0.{0015}^2}{4} \\ A=1.767\times {10}^{-6}{m}^2 \end{gathered}$$

The self-inductance is given by

$$\begin{gathered} L=\frac{4\pi \times {10}^{-7}\times {50}^2\times 1.767\times {10}^{-6}}{0.05} \\ L=1.11\times {10}^{-7}H \end{gathered}$$

So, the self-inductance is$1.11\times {10}^{-7}H$.