Magnetic field due to a single moving charge The Biot-Savart law can only be used in the case of magnetostatics (constant current) so how do we calculate the magnetic field of a single charge moving at constant velocity at a distance r. I tried by calculating the displacement current
using but i was not sure wether the biot savart law can be applied to displacement currents.
Please don't use relativity if possible because i have no experience with relativity yet.
 A: 
A point charge $\:q\:$ is moving uniformly on a straight line with velocity  $\:\boldsymbol{\upsilon}\:$ as is the Figure. The electromagnetic field at a point $\:\mathrm{P}\:$ with position vector $\:\mathbf{x}\:$ at time $\:t\:$ is
\begin{align}
\mathbf{E}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right) & \boldsymbol{=}\dfrac{q}{4\pi \epsilon_{\bf 0}}\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}},\quad \beta\boldsymbol{=}\dfrac{\upsilon}{c}
\tag{01a}\\
\mathbf{B}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right) & \boldsymbol{=}\dfrac{1}{c^{ \bf 2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right)\vphantom{\dfrac{a}{\dfrac{}{}b}}\boldsymbol{=}\dfrac{\mu_{0}q}{4\pi }\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}}
\tag{01b}
\end{align}
Equations (01) are relativistic. They come from the Lienard-Wiechert potentials.


Biot-Savart Law


After a quick calculation with Biot-Savart Law (using the Dirac $\:\delta\:$ function) I found the solution
\begin{equation}
\mathbf{B}_{_{\mathbf{BS}}}\left(\mathbf{x},t\right) \boldsymbol{=}\dfrac{\mu_{0}q}{4\pi }\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}}
\tag{02}
\end{equation}
which compared with that from the Lienard-Wiechert potentials, see above equation (01b)
\begin{equation}
\mathbf{B}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right)\boldsymbol{=}\dfrac{\mu_{0}q}{4\pi }\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}}
\tag{03}
\end{equation}
it looks as an approximation for charges whose velocities are small compared to that of light $\:c$
\begin{equation}
\mathbf{B}_{_{\mathbf{BS}}}\left(\mathbf{x},t\right)\boldsymbol{=}
\lim_{\beta \boldsymbol{\rightarrow} 0}\mathbf{B}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right)\boldsymbol{=}
\lim_{\beta\boldsymbol{\rightarrow} 0}\left[\dfrac{\mu_{0}q}{4\pi }\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}}\right]\boldsymbol{=}\dfrac{\mu_{0}q}{4\pi}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{3}}
\tag{04}
\end{equation}


(1)
EDIT
Answer to OP's comment : 

how did you get equation 02 when v << c. – DHYEY Jun 29 '18 at 11:49
  



From Jackson's : Biot and Savart Law
\begin{equation}
\mathrm d\mathbf{B}=\dfrac{\mu_{0}}{4\pi}I\dfrac{\left(\mathrm d\boldsymbol{\ell}\boldsymbol{\times}\mathbf{r'}\right)}{\:\:\Vert\mathbf{r'}\Vert^{3}} 
\tag{BS-01}
\end{equation}
\begin{equation}
I=q\upsilon\delta\left(x'-r\cos\phi\right), \qquad \mathrm d\boldsymbol{\ell}=\mathbf{i}\mathrm dx', \qquad \mathbf{r'}=x'\mathbf{i}\boldsymbol{+}\alpha\mathbf{j}\boldsymbol{+}0\mathbf{k}
\tag{BS-02}
\end{equation}
\begin{equation}
\mathrm d\mathbf{B}=\dfrac{\mu_{0}q}{4\pi}q\upsilon\delta\left(x'\!\boldsymbol{-}\!r\cos\phi\right)\dfrac{\left(\mathbf{i}\boldsymbol{\times}\mathbf{r'}\right)}{\:\:\Vert\mathbf{r'}\Vert^{3}}\mathrm dx'=\dfrac{\mu_{0}q}{4\pi}q\upsilon\delta\left(x'\!\boldsymbol{-}\!r\cos\phi\right)\dfrac{\left(\alpha\mathbf{k}\right)}{\:\:\left(x'^2\!\boldsymbol{+}\!\alpha^2 \right)^{3/2}}\mathrm dx'  
\tag{BS-03}
\end{equation}
\begin{equation}
\mathbf{B}=\dfrac{\mu_{0}}{4\pi}q\upsilon\alpha\mathbf{k}\int\limits_{\boldsymbol{-}\boldsymbol{\infty}}^{\boldsymbol{+}\boldsymbol{\infty}}\dfrac{\delta\left(x'\!\boldsymbol{-}\!r\cos\phi\right)}{\:\:\left(x'^2\!\boldsymbol{+}\!\alpha^2 \right)^{3/2}}\mathrm dx'=\dfrac{\mu_{0}q}{4\pi}\dfrac{\upsilon\alpha\mathbf{k}}{\:\:\left(r^2\cos^2\phi\!\boldsymbol{+}\!\alpha^2 \right)^{3/2}}= \dfrac{\mu_{0}q}{4\pi}\dfrac{\left(\upsilon\mathbf{i}\right)\boldsymbol{\times}\left(\alpha\mathbf{j}\right)}{\:\:\left(r^2\cos^2\phi\!\boldsymbol{+}\!\alpha^2 \right)^{3/2}}
\tag{BS-04}
\end{equation}
\begin{equation}
\mathbf{B} =\dfrac{\mu_{0}q}{4\pi }\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{3}}
\tag{BS-05}
\end{equation}

A: Actually there is only a slight modification in the form of the potential otherwise its the same.
The magnetic vector potential gets modified to $$
\mathbf{A}(\mathbf{r}, t) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}', t_r')}{|\mathbf{r} - \mathbf{r}'|} d^3\mathbf{r}$$
where $$t_r' = t - \frac{1}{c}|\mathbf{r} - \mathbf{r}'|$$
is the retarded time. 
There is no relativity in here but causality is only taken into account.
Calculating the magnetic fields is a bit more difficult you can reffer to
Lienard Wiechert Potential
A: You can use the Biot-Savart formula to find magnetic field with very good approximation, if electric field everywhere is a potential field (it can be expressed as gradient of a potential function). This is because the Biot-Savart field obeys the Ampere-Maxwell law with displacement current due to that potential electric field. In other words, for systems where electric field is a potential field, correction to Biot-Savart magnetic field due to displacement current is zero.
A charged particle moving with constant velocity has electric field that moves in space but if the speed is much lower than speed of light, at any instant electric field can be expressed as gradient of a potential function (giving a $\gamma$ - contracted Coulomb field). So you can use the Biot-Savart formula if the charge speed is low enough.
A: dB = (µ/4π)(i/r²)(dlXr'), dl and r' are vectors, |r'| = 1 (r with ^)
i = dq/dt
dl/dt = v
dB = (µ/4π)(dq/r²)(vXr'), v is vector and |v| = v
B = (µ/4π)(q/r²)(vXr') = (µ/4π)*(q/r²)vsin(θ)
B = µqvsin(θ)/4πr²
comments:
two vectors u and t and θ angle between them, |uXt| = |u|*|t|*sin(θ)
when we have a function the equation dB = Constant*dq when you use integration you can get B=C *integral dq = C *q
