Consider the magnetic field of a straight wire along the $z$-axis carrying a steady current $\II=I\,\zhat$. The Biot-Savart Law says that \begin{eqnarray*} \BB = - {\mu_0 I\over 4\pi\epsilon_0} \Int_{-L}^{L} {(\rr-\rrp)\times d\rrp\over|\rr-\rrp|^3} \end{eqnarray*} and we have \begin{eqnarray*} \rr &=& r\,\rhat + z\,\zhat \\ \rrp &=& z'\,\zhat \end{eqnarray*} so that \begin{eqnarray*} - (\rr-\rrp)\times d\rr &=& - \bigl(r\,\rhat + (z-z')\,\zhat\bigr) \times dz'\,\zhat \\ &=& r\,dz'\,\phat \end{eqnarray*} which gives the expected right-hand rule behavior for the direction of the magnetic field. We therefore have \begin{eqnarray*} \BB &=& {\mu_0 I\over 4\pi\epsilon_0} \> r\,\phat \Int_{-L}^{L} {dz'\over\bigl(r^2+(z-z'^2)\bigr)^{3/2}} \\ &=& - {\mu_0 I\over 4\pi\epsilon_0} {(z-z')\,\phat\over\sqrt{r^2+(z-z'^2)}} \Bigg|_{-L}^L \end{eqnarray*}

An important special case is an infinite wire — after all, a finite straight wire can not support a steady current. Taking the limit as $L\to\infty$, we obtain \begin{eqnarray*} \BB = {\mu_0 I\over 2\pi r} \> \phat \end{eqnarray*} We will see next week that this result can be derived more simply using symmetry and Ampère's Law, an approach which fails however for the finite wire.