Question

Two identical long wires of radius $\mathit{a}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{}\mathbf{\text{mm}}$are parallel and carry identical currents in opposite directions. Their center-to-center separation is $\mathit{d}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{cm}$.Neglect the flux within the wires but consider the flux in the region between the wires. What is the inductance per unit length of the wires?

Short answer

Inductance per unit length of the wire,$\frac{L}{l}=1.81\times {10}^{-6}\text{H/m}$.

Step-by-step solution

Given

1. The radius of each long wire, $a=1.53\times {10}^{-3}\text{m}=0.153\times {10}^{-2}\text{m}$.
2. Center to center separation between two wires, $d=14.2\mathrm{cm}=14.2\times {10}^{-2}\mathrm{m}$
3. Unit length of wire, $I=1\mathrm{m}$.

Understanding the concept

The crucial step in this problem is to find the magnetic flux due to current-carrying wires. Using the flux, we can find the inductance of the given system; by dividing inductance by unit length, we will get the required answer.

Formula:

The magnitude of the magnetic field at a radial distance from the center of the long wire (inside),

$B=\frac{{\mu }_{0}i}{2\pi }·\frac{r}{a^2}$

The magnitude of the magnetic field at a radial distance r from the center of the long wire r (inside)

$B=\frac{{\mu }_{0}i}{2\pi r}$

Magnetic flux,

${\varphi }_B=Li$

Magnetic flux due to the field across the open surface attached to the closed loop is

${\varphi }_B=\oint \overrightarrow{B}·d\overrightarrow{A}$

Calculate inductance per unit length of the wire.

Let’s denote two wires as$w_1$and $w_2$.

Now we will find B field due to wire$w_1$and $w_2$at point p inside $w_1$ at distance r from the center of $w_1$which is calculated by using amperes law

$B_{in}=\frac{{\mu }_{0}i}{2\pi }·\frac{r}{a^2}+\frac{{\mu }_{0}i}{2\pi \left(d-r\right)}$

In the above expression, first term is due to $w_1$ and the second term is due to $w_2$.

Similarly, we will find B field due to wire $w_1$and $w_2$at point p outside, at distance r from the center of $w_1$which is calculated by using amperes law,

$B_{out}=\frac{{\mu }_{0}i}{2\pi r}+\frac{{\mu }_{0}i}{2\pi \left(d-r\right)}$

In the above expression first term is due to $w_1$and second term is due to $w_2$.

Total field in the region between $w_1$ and $w_2$ is then,

$B_{in}+B_{out}=\left(\frac{{\mu }_{0}i}{2\pi }·\frac{r}{a^2}+\frac{{\mu }_{0}i}{2\pi \left(d-r\right)}+\frac{{\mu }_{0}i}{2\pi r}+\frac{{\mu }_{0}i}{2\pi \left(d-r\right)}\right)$

Magnetic flux due tofield B is,

${\varphi }_B={\varphi }_{B1}+{\varphi }_{B2}=\oint \overrightarrow{B}·d\overrightarrow{A}={\int }_0^{d/2}B·dA+{\int }_{d/2}^{d}B·dA={\int }_0^{d/2}B·dA$

$$\begin{gathered} {\varphi }_B=2{\int }_0^a{B}_{in}·dA+2{\int }_0^{d/2}{B}_{out}·dA \\ {\varphi }_B==2{\int }_0^a\left(\frac{{\mu }_{0}i}{2\pi }·\frac{r}{a^2}+\frac{{\mu }_{0}i}{2\pi \left(d-r\right)}\right)·dA+2{\int }_a^{d/2}\left(\frac{{\mu }_{0}i}{2\pi r}+\frac{{\mu }_{0}i}{2\pi \left(d-r\right)}\right)·dA \end{gathered}$$

Here area element is $dA=ldr$ using this in the above equation we get,

$$\begin{gathered} \frac{{\varphi }_B}{l}=2{\int }_0^a\left(\frac{{\mu }_{0}i}{2\pi }·\frac{r}{a^2}+\frac{{\mu }_{0}i}{2\pi \left(d-r\right)}\right)·dr+2{\int }_a^{d/2}\left(\frac{{\mu }_{0}i}{2\pi r}+\frac{{\mu }_{0}i}{2\pi \left(d-r\right)}\right)·dr \\ \frac{{\varphi }_B}{l}=\frac{{\mu }_{0}i}{\pi }\left[{\int }_0^a\left(\frac{r}{a^2}+\frac{1}{\left(d-r\right)}\right)·dr+{\int }_0^{d/2}\left(\frac{1}{r}+\frac{1}{\left(d-r\right)}\right)·dr\right] \\ \frac{{\varphi }_B}{l}=\frac{{\mu }_{0}i}{\pi }\left[\frac{1}{2}-\ln \left(\frac{d-a}{d}\right)+\ln \left(\frac{d-a}{a}\right)\right] \end{gathered}$$

The first two terms represent flux inside the wire which we will neglect

$\begin{array}{rcl}\frac{{\varphi }_B}{l}& =& \frac{{\mu }_{0}i}{\pi }\left[\ln \left(\frac{d-a}{a}\right)\right]\\ \frac{L}{l}& =& \frac{{\varphi }_B}{il}=\frac{{\mu }_0}{l\pi }\left[\ln \left(\frac{d-a}{a}\right)\right]\\ \frac{L}{l}& =& \frac{{\varphi }_B}{il}=\frac{4{\mu }_0}{4\pi l}\left[\ln \left(\frac{d-a}{a}\right)\right]=\frac{4\times {10}^{-7}}{1.0}\ln \left(\frac{14.2-0.153}{0.153}\right)=4\times {10}^{-7}\times \ln \left(91.81\right)\\ & & \end{array}$

$\frac{L}{l}=1.81\times {10}^{-6}\text{H/m}$