How come the Biot–Savart law is not applicable to derivation of magnetic field around an infinitely long, straight wire? [closed]

Question

Context

[Edited Nov 20, 2021]

When the present question was initially asked, there was a derivation presented that lead to the wrong result. The Biot–Savart integral that I used to write the magnetic field formed by a infinitely long strait wire appeared to yield a wrong result. Though, of course, the Biot-Savart integral should is correct. There was a reasonable explanation of why I was expected to be able to use the Biot–Savart law to obtain the magnetic field. This is because this is a basic question with a well-known answer. There must have been an error in my derivation. Below, Zalnd pointed out the flaw in my derivation, which was as simple as the incorrect exponent in the denominator of my integrand. So, in hindsight, with the benefit of [1], I would like to have changed my question to "How come the physical principle of the Biot–Savart law is not applicable here?" To which the answer would have been something akin to, "Hey, of course Biot–Savart law is applicable. You simply wrote the wrong denominator in your integral."

Edited Question

How come the physical principle of the Biot–Savart law is not applicable to obtain the magnetic field generated by this infinitely long line charge that is in uniform motion in a direction collinear to the line of charge? [Please note that the Biot-Savart law is definitely applicable. This is demonstrated below by Zalnd below]

Bibliography

[1] Should any check-my-work questions be made on topic?

Answer 1

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.