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CONTEXT

I am attempting to derive the magnetic field caused by a strait, infinitely long wire. My approach is to use the Biot–Savart law. The magnitude of the resultant magnetic field is well known and equal to $$ \left\| \vec{B}\left({\vec{r}}\right) \right\| = \frac{\mu_o I}{2\pi \sqrt{(r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2}} $$

MY RESULTS

From [1], the Biot–Savart law, $$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int_C \frac{I \, d\vec{l} \times\vec{\hat r'}}{|\vec{r'}|^2}.$$

Fix $\vec{r}$ in space as the point to measure the magnetic field. Let the position of the contour of the strait wire be $r_{l,x}\hat{e}_x + r_{l,y}\hat{e}_y + r_{l,z} \hat{e}_z$. So, $$\vec{r}{\bf{'}} = (\vec{r} - \vec{r}_l) = (r_{x} - r_{l,x}) \hat{e}_x + (r_{y} - r_{l,y} )\hat{e}_y + (r_{z} -r_{l,z} )\hat{e}_z.$$ As a consequence, $$\left\| \vec{r}{\bf{'}} \right\| = \sqrt{ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + (r_{z} -r_{l,z} )^2 }.$$ Without loss of generality, allow that a constant current moves along a contour whose differential length is $d\vec{l} = dr_{l,z}\vec{e}_z$. Plugging in to the Biot-Savart equation, I find \begin{equation} \vec{B}\left({\vec{r}}\right) = I\frac{\mu_o}{4\pi}\int_{-\infty}^{\infty} \frac{ dr_{l,z}\vec{e}_z \times \left[ (r_{x} - r_{l,x}) \hat{e}_x + (r_{y} - r_{l,y} )\hat{e}_y + (r_{z} -r_{l,z} )\hat{e}_z\right]}{ \sqrt{ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + (r_{z} -r_{l,z} )^2 }} \end{equation} From the cross product \begin{equation} \vec{B}\left({\vec{r}}\right) = I\frac{\mu_o}{4\pi}\int_{-\infty}^{\infty} \frac{ dr_{l,z} \left[ (r_{x} - r_{l,x}) \hat{e}_y - (r_{y} - r_{l,y} )\hat{e}_x \right]}{ \sqrt{ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + (r_{z} -r_{l,z} )^2 }} \end{equation} Lets expand. \begin{align} \vec{B}\left({\vec{r}}\right) &= I\frac{\mu_o}{4\pi}\int_{-\infty}^{\infty} \frac{ dr_{l,z} \left[ (r_{x} - r_{l,x}) \hat{e}_y - (r_{y} - r_{l,y} )\hat{e}_x \right]}{ \sqrt{\left[ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + r_{z}^2\right] -2r_{z}r_{l,z} + r_{l,z}^2 }} \end{align} By $X$, I denote the polynomial defined as $$X = a+ b r_{l,z} + cr_{l,z}^2;$$ by $q$, I denote the determinant defined as \begin{align*} q &= 4ac-b^2 \\ &= 4\left[ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + r_{z}^2\right]-(-2r_z)^2 \\ &= 4\left[ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 \right] . \end{align*} (I note that $q > 0$); and by $k$, I denote a parameter defined as $$k = \frac{4c}{q} = \frac{ 1}{ \left[ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 \right]}. $$ From [2], since $c = 1 >0$, $$\int \frac{dx}{\sqrt{X}} = \dfrac{1}{\sqrt{c} } \sinh^{-1}\frac{2cx+b}{\sqrt{q}}.$$ So. \begin{align} \vec{B}\left({\vec{r}}\right) &= I\frac{\mu_o}{4\pi}\int_{-\infty}^{\infty} \frac{ dr_{l,z} \left[ (r_{x} - r_{l,x}) \hat{e}_y - (r_{y} - r_{l,y} )\hat{e}_x \right]}{ \sqrt{\left[ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + r_{z}^2\right] -2r_{z}r_{l,z} + r_{l,z}^2 }} \\ &= I\frac{\mu_o}{4\pi} \left[ (r_{x} - r_{l,x}) \hat{e}_y - (r_{y} - r_{l,y} )\hat{e}_x \right] \left[\frac{1 } { \sqrt{c}} \sinh^{-1}\frac{2c r_{l,z}+b}{\sqrt{q}} \right]_{-\infty}^{\infty} \\ &= I\frac{\mu_o}{2\pi} \left[ (r_{x} - r_{l,x}) \hat{e}_y - (r_{y} - r_{l,y} )\hat{e}_x \right] \left[ \lim_{r_{l,z} \rightarrow \infty} \sinh^{-1}\frac{ r_{l,z}- r_z}{\sqrt{ \left[ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 \right]}} \right] \end{align} Since $ \lim_{x \rightarrow \infty} \sinh^{-1} x = \infty $, then $\vec{B}\left({\vec{r}}\right) $ does not converge.

DISCUSSION

I interpret my result as being incorrect. I think that $\vec{B}\left({\vec{r}}\right) $ should converge to $$ \vec{B}\left({\vec{r}}\right) = \frac{\mu_o I \left[ (r_{x} - r_{l,x}) \hat{e}_y - (r_{y} - r_{l,y} )\hat{e}_x \right]}{2\pi \left[(r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2\right]} $$

QUESTION

Where have I gone wrong?

BIBLIOGRAPHY

[1] https://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law

[2] CRC Standard Mathematical Tables and Formulae, 30nd Edition

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You solved the wrong integral:

(i) $\vec{r'}$ is an unit vector

(ii) you forgot the power in the denominator

Without loss of generality, allow that a constant current moves along a contour whose differential length is $d\vec{l} = dr_{l,z}\vec{e}_z$. Plugging in to the Biot-Savart equation, I find \begin{equation} \vec{B}\left({\vec{r}}\right) = I\frac{\mu_o}{4\pi}\int_{-\infty}^{\infty} \frac{ dr_{l,z}\vec{e}_z \times \left[ (r_{x} - r_{l,x}) \hat{e}_x + (r_{y} - r_{l,y} )\hat{e}_y + (r_{z} -r_{l,z} )\hat{e}_z\right]}{ \sqrt{ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + (r_{z} -r_{l,z} )^2 }} \end{equation}

At this point, your integral should be

\begin{equation} \vec{B}\left({\vec{r}}\right) = I\frac{\mu_o}{4\pi}\int_{-\infty}^{\infty} \frac{ dr_{l,z}\vec{e}_z \times \left[ (r_{x} - r_{l,x}) \hat{e}_x + (r_{y} - r_{l,y} )\hat{e}_y + (r_{z} -r_{l,z} )\hat{e}_z\right]}{ \left[ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + (r_{z} -r_{l,z} )^2 \right]^{3/2} } \end{equation}

From the cross product \begin{equation} \vec{B}\left({\vec{r}}\right) = I\frac{\mu_o}{4\pi}\int_{-\infty}^{+\infty} \frac{ dr_{l,z} \left[ (r_{x} - r_{l,x}) \hat{e}_y - (r_{y} - r_{l,y} )\hat{e}_x \right]}{ \sqrt{ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + (r_{z} -r_{l,z} )^2 }} \end{equation}

\begin{equation} \vec{B}\left({\vec{r}}\right) = I\frac{\mu_o}{4\pi}\int_{-\infty}^{+\infty} \frac{ dr_{l,z} \left[ (r_{x} - r_{l,x}) \hat{e}_y - (r_{y} - r_{l,y} )\hat{e}_x \right]}{ \left[ (r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2 + (r_{z} -r_{l,z} )^2 \right]^{3/2}} \end{equation}

From here, you can define $a^2=(r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2$ and $u=(r_{z} - r_{l,z} )$. So

\begin{equation} \vec{B}\left({\vec{r}}\right) = I\frac{\mu_o}{4\pi} \left[ (r_{x} - r_{l,x}) \hat{e}_y - (r_{y} - r_{l,y} )\hat{e}_x \right] \int_{-\infty}^{+\infty} \frac{ du }{ \left[ a^2 + u^2 \right]^{3/2}} \end{equation}

Trigonometric substitution, $u = a \tan{(\theta)}$, takes you to

\begin{equation} \int \frac{ du }{ \left[ a^2 + u^2 \right]^{3/2} } = \int \frac{ a \sec^2{(\theta)} d\theta }{ \left[ \sqrt{ a^2 \left[ 1+\tan^2{(\theta)} \right] } \right]^{3} } = \frac{1}{a^2}\int \cos{(\theta)} d\theta = \frac{\sin{(\theta)}}{a^2} = \frac{u}{a^2\sqrt{a^2+u^2}} \end{equation}

\begin{equation} \lim_{ u \rightarrow \infty} \left[ \frac{u}{a^2\sqrt{a^2+u^2}} - \frac{(-u)}{a^2\sqrt{a^2+(-u)^2}} \right] = \lim_{ u \rightarrow \infty} \left[ \frac{2u}{a^2\sqrt{a^2+u^2}} \right] = \frac{2}{a^2} = \frac{2}{(r_{x} - r_{l,x})^2 + (r_{y} - r_{l,y} )^2} \end{equation}

Putting it back you have the answer you were looking for.

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