I brought in relativity for the special case of a uniformly moving charge, because it's much simpler to calculate the electromagnetic field with help of a Lorentz boost. Let's do that. In the lab frame (unprimed coordinates (x^{\mu})=(ct,\vec{x})) the particle moves with a constant velocity \vec{v}=v \vec{e}_x. We want to evaluate the field in this reference frame.
To that end we look at the situation from the reference frame, where the particle is at rest (primed coordinates (x'^{\mu}=(c t',\vec{x}')) at the origin. There the electromagnetic field is simply an electrostatic Coulomb field, which reads (in Heaviside-Lorentz units)
\vec{E}'(t',\vec{x}')=\frac{q}{4 \pi} \frac{\vec{x}'}{r'^3}, \quad \vec{B}'(t',\vec{x}')=0.
Now we use the transformation of the space-time coordinates and the em. field under the Lorentz boost
c t'=\gamma(c t-v x/c), \quad x'=\gamma(x- v t), \quad y'=y, \quad z'=z.
E_x(t,\vec{x})=E_x'(t',\vec{x}')=\frac{q}{4 \pi} \frac{x'}{r'^3}=\frac{q}{4 \pi} \gamma \frac{x-v t}{[\gamma^2(x-v t)^2+y^2 +z^2]^{3/2}},
E_y(t,\vec{x})=\gamma E_y'(t',\vec{x}')=\frac{q}{4 \pi} \gamma \frac{y}{[\gamma^2(x-v t)^2+y^2 +z^2]^{3/2}},
E_z(t,\vec{x})=\gamma E_z'(t',\vec{x}')=\frac{q}{4 \pi} \gamma \frac{z}{[\gamma^2(x-v t)^2+y^2 +z^2]^{3/2}}
or in three-vector notation
\vec{E}(t,\vec{x})=\frac{q}{4 \pi} \frac{\gamma}{[\gamma^2(x-v t)^2+y^2 +z^2]^{3/2}} \begin{pmatrix}<br />
x-v t \\ y \\ z \end{pmatrix}.<br />
The magnetic field is given by
\vec{B}=\vec{\beta} \times \vec{E}=\frac{q}{4 \pi} \frac{\gamma}{[\gamma^2(x-v t)^2+y^2 +z^2]^{3/2}} \begin{pmatrix}<br />
0 \\ -z \\ y \end{pmatrix}.<br />
Of course, you get the same result by using the four-current
\rho(t,\vec{x})=q \delta(x-v t) \delta(y) \delta(z), \quad \vec{j}=q \vec{v} \delta(x-v t) \delta(y) \delta(z)
in the retarded (Lienard-Wiechert) potentials and taking the according derivatives
\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A}<br />
or directly the retarded Jefimenko integrals for the field.<br />
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You <b>must not</b> use the naive Biot-Savart Law for the magnetic field, because the current density is <b>not stationary</b>, because it's obviously time dependent!