91Ó°ÊÓ

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

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),

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}}·\frac{r}{a^2}$

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

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

Magnetic flux,

${\mathrm{Ï•}}_B=Li$

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

${\mathrm{Ï•}}_B=âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{A}$

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

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

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,

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

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,

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

Magnetic flux due tofield B is,

${\mathrm{Ï•}}_B={\mathrm{Ï•}}_{B1}+{\mathrm{Ï•}}_{B2}=âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{A}={∫}_0^{d/2}B·dA+{∫}_{d/2}^{d}B·dA={∫}_0^{d/2}B·dA$

$$\begin{gathered} {\mathrm{ϕ}}_B=2{∫}_0^a{B}_{in}·dA+2{∫}_0^{d/2}{B}_{out}·dA \\ {\mathrm{ϕ}}_B==2{∫}_0^a\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}}·\frac{r}{a^2}+\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}\left(d-r\right)}\right)·dA+2{∫}_a^{d/2}\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}+\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}\left(d-r\right)}\right)·dA \end{gathered}$$

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

$$\begin{gathered} \frac{{\mathrm{ϕ}}_B}{l}=2{∫}_0^a\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}}·\frac{r}{a^2}+\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}\left(d-r\right)}\right)·dr+2{∫}_a^{d/2}\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}+\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}\left(d-r\right)}\right)·dr \\ \frac{{\mathrm{ϕ}}_B}{l}=\frac{{\mathrm{μ}}_{0}i}{\mathrm{π}}\left[{∫}_0^a\left(\frac{r}{a^2}+\frac{1}{\left(d-r\right)}\right)·dr+{∫}_0^{d/2}\left(\frac{1}{r}+\frac{1}{\left(d-r\right)}\right)·dr\right] \\ \frac{{\mathrm{ϕ}}_B}{l}=\frac{{\mathrm{μ}}_{0}i}{\mathrm{π}}\left[\frac{1}{2}-\ln \left(\frac{d-a}{d}\right)+\ln \left(\frac{d-a}{a}\right)\right] \end{gathered}$$

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

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

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