Magnetic Field Induced By A moving Charge Which formulas and/or rules I need to use to calculate the magnetic field induced by an electron moving at constant velocity. Can I use biot-savart law or do I need to use maxwell equations instead?
 A: I believe the Biot-Savart Law is non-relativistic (it is only applicable when $v\ll c$) so if that's all you want, go for it! Of course, that will not give you the whole picture. To completely calculate the magnetic field of a moving charge, it's best to use Maxwell's Equations, or -- even better -- just use Coulomb's law and the Special Relativity.
The method I'll outline below is from The Feynman Lectures on Physics: Vol. II Chapter 26. The basic idea is that the four-potential is a four-vector, and so should transform according to the Lorentz Transformations. In other words, if you had two frames $S$ and $S'$, with $S'$ moving rightwards with a velocity $v$ with respect to $S$, then the components of a four-vector $\mathcal{A}^\mu$ transform between them as:
\begin{aligned}
\mathcal{A}'^{\,t} = \gamma \left(\mathcal{A}^t - \frac{v}{c} \mathcal{A}^x\right) \quad \quad \quad \quad \quad \quad \mathcal{A}^{\,t} = \gamma \left(\mathcal{A}'^{\,t} + \frac{v}{c} \mathcal{A}'^{\, x}\right)\\ 
\mathcal{A}'^{\,x} = \gamma \left(\mathcal{A}^x - \frac{v}{c} \mathcal{A}^t\right)\quad \quad \quad \quad \quad \quad \mathcal{A}^{\,x} = \gamma \left(\mathcal{A}'^{\,x} + \frac{v}{c} \mathcal{A}'^{\, t}\right)\\ \\ 
\end{aligned}
where the two equations on the left are the "forward" Lorentz Transformations (going from $S\rightarrow S'$), and the two equations on the right are the transformations that take you from $S'$ to $S$. Also, by the superscripts "$t$" and "$x$" I mean the "time" and "space" components of the four-vector $\mathcal{A}^\mu$, and I have assumed that $S'$ is moving along the common $x-x'$ axis.
So the procedure is simple:

*

*Consider the charge at rest in the frame $S'$ (so that it is moving rightwards according to someone sitting in the lab frame $S$). Write out the four-vector $\mathcal{A}'^\mu$ in this simple case, which is just the four-vector for a stationary charge: $$\mathcal{A}'^\mu = \begin{pmatrix}\dfrac{\phi'}{c}\\ 0\end{pmatrix},$$ where $\phi'$ is just the Coulomb potential. (There is no magnetic field in this frame, and so the vector potential $A=0$.)


*Do a Lorentz Transformation from $S'$ to $S$, so that  $\mathcal{A}'\rightarrow \mathcal{A}$. $$\mathcal{A}^\mu = \begin{pmatrix}\dfrac{\gamma \phi'}{c}\\ \dfrac{\gamma v}{c^2}\phi'\end{pmatrix}$$


*Calculate the magnetic field in the new frame using $\mathbf{B} = \nabla \times \mathbf{A}$. If you do the above calculations carefully, you should be able to show that $$\mathbf{A} = \frac{\gamma}{4\pi\epsilon_0 c^2} \frac{qv}{\sqrt{\left( \gamma^2(x-vt)^2 + y^2 + z^2\right)}}\hat{\mathbf{x}}$$ From which you can calculate the magnetic field in $S$.
I particularly like this method since it only requires you to know (i) the Coulomb potential and (ii) the Lorentz Transformations, both of which are rather fundamental.
