Flux Through a Circle - 3 - Corrected Version

Consider the line:

$\frac{x-5}{2}=\frac{y-3}{4}=\frac{z}{9} = p$

When $p=0$ and $p=1$ , two points passing through the line are obtained which are $(5,3,0)$ and $(7,7,9)$ . From here, a unit vector in the direction of the line can be obtained. That unit vector is:

$\hat{L} = \frac{2\hat{i} + 4\hat{j} +9\hat{k}}{\sqrt{101}}$

An elementary length of the wire is:

$d\vec{L} = (dp) \hat{L}$

The position vector of any point on the line can be parameterized as:

$\vec{r}_l = \left(\frac{2p}{\sqrt{101}}+5\right)\hat{i}+\left(\frac{4p}{\sqrt{101}}+3\right)\hat{j}+\left(\frac{9p}{\sqrt{101}}\right)\hat{k}$

An arbitrary point on the unit circle is defined as:

$\vec{r}_c = x\hat{i}+y\hat{j}$

$\vec{r} = \vec{r}_c - \vec{r}_l$

Using Biot Savart Law:

$d\vec{B} = \frac{\mu_oI}{4\pi} \left(\frac{d\vec{L} \times \hat{r}}{\mid \vec{r} \mid^2}\right)$

The elementary area vector of the unit circle is:

$d\vec{A} = (dx) (dy)\hat{k}$

The magnetic flux is:

$d\phi = d\vec{B}.d\vec{A}$

And after simplifying, we get:

$\phi = \int_{-\infty}^{\infty} \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{\sqrt{101}\,\left(\frac{y}{101}-\frac{2\,x}{101}+\frac{7}{101}\right)}{{\left({\left(2\,p-x+5\right)}^2+{\left(4\,p-y+3\right)}^2+81\,p^2\right)}^{3/2}}(dy) (dx) (dp)$

Numerically integrating the above monster gives the result of $\boxed{0.14941518577}Tm^2$