91影视

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

$B=\frac{{渭}_{0}i}{2蟺}路\frac{r}{a^2}$

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

$B=\frac{{渭}_{0}i}{2蟺r}$

Magnetic flux,

${蠒}_B=Li$

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

${蠒}_B=鈭�\stackrel{鈫�}{B}路d\stackrel{鈫�}{A}$

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

$B_{in}=\frac{{渭}_{0}i}{2蟺}路\frac{r}{a^2}+\frac{{渭}_{0}i}{2蟺\left(d-r\right)}$

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,

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

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,

$B_{in}+B_{out}=\left(\frac{{渭}_{0}i}{2蟺}路\frac{r}{a^2}+\frac{{渭}_{0}i}{2蟺\left(d-r\right)}+\frac{{渭}_{0}i}{2蟺r}+\frac{{渭}_{0}i}{2蟺\left(d-r\right)}\right)$

Magnetic flux due tofield B is,

${蠒}_B={蠒}_{B1}+{蠒}_{B2}=鈭�\stackrel{鈫�}{B}路d\stackrel{鈫�}{A}={鈭�}_0^{d/2}B路dA+{鈭�}_{d/2}^{d}B路dA={鈭�}_0^{d/2}B路dA$

$$\begin{gathered} {蠒}_B=2{鈭�}_0^a{B}_{in}路dA+2{鈭�}_0^{d/2}{B}_{out}路dA \\ {蠒}_B==2{鈭�}_0^a\left(\frac{{渭}_{0}i}{2蟺}路\frac{r}{a^2}+\frac{{渭}_{0}i}{2蟺\left(d-r\right)}\right)路dA+2{鈭�}_a^{d/2}\left(\frac{{渭}_{0}i}{2蟺r}+\frac{{渭}_{0}i}{2蟺\left(d-r\right)}\right)路dA \end{gathered}$$

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

$$\begin{gathered} \frac{{蠒}_B}{l}=2{鈭�}_0^a\left(\frac{{渭}_{0}i}{2蟺}路\frac{r}{a^2}+\frac{{渭}_{0}i}{2蟺\left(d-r\right)}\right)路dr+2{鈭�}_a^{d/2}\left(\frac{{渭}_{0}i}{2蟺r}+\frac{{渭}_{0}i}{2蟺\left(d-r\right)}\right)路dr \\ \frac{{蠒}_B}{l}=\frac{{渭}_{0}i}{蟺}\left[{鈭�}_0^a\left(\frac{r}{a^2}+\frac{1}{\left(d-r\right)}\right)路dr+{鈭�}_0^{d/2}\left(\frac{1}{r}+\frac{1}{\left(d-r\right)}\right)路dr\right] \\ \frac{{蠒}_B}{l}=\frac{{渭}_{0}i}{蟺}\left[\frac{1}{2}-\ln \left(\frac{d-a}{d}\right)+\ln \left(\frac{d-a}{a}\right)\right] \end{gathered}$$

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

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

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