# Charge $q$ near Current-Carrying Wire

A charge $$q$$ nears a current-carrying wire. How does $$q$$ move? Specifically, what is $$\vec{r}(t)$$ for $$q$$?

I've found the direction of some of the forces acting on the charge $$q$$:

Using the Biot-Savart Law and Coulumb's Law, I can also find the Magnitude of the Magnetic and Electric Fields. I've used a suitable Amperian Loop for Ampere's Law, and computed the Magnetic Field to be as follows:

$$d \vec{B}=\frac{\mu_{0}}{4 \pi} \frac{i d \vec{s} \times \vec{r}}{r^{3}} \\ d B=\frac{\mu_{0}}{4 \pi} \frac{i d s(r \sin \theta)}{r^{3}} \\ B=\frac{\mu_{0} i}{4 \pi} \int_{-\infty}^{\infty} \frac{d s(r \sin \theta)}{r^{3}}$$

\begin{aligned} B=\frac{\mu_{0} i}{4 \pi} \int_{-\infty}^{\infty} \frac{d s(r \sin \theta)}{r^{3}} &=\frac{\mu_{0} i}{4 \pi} \int_{-\infty}^{\infty} \frac{R d s}{\left(s^{2}+R^{2}\right)^{3 / 2}} \\ &=\frac{\mu_{0} i}{2 \pi} \int_{0}^{\infty} \frac{R d s}{\left(s^{2}+R^{2}\right)^{3 / 2}} \\ =& \frac{\mu_{0} i R}{2 \pi}\left[\frac{s}{R^{2}\left(s^{2}+R^{2}\right)^{12}}\right]_{0}^{\infty}=\frac{\mu_{0} i}{2 \pi R}. \end{aligned}

Using Gauss' Law and a Cylinder as a Gaussian Surface, we have

$$E(R)=\frac{\lambda}{2 \pi \epsilon_{0} R}$$

And by the Lorentz Force formula, I have $$\vec{F}=q \vec{v} \times \vec{B}$$. Nevertheless, I'm still struggling to understand how the charge $$q$$ would move to these Electric and Magnetic Fields. Please advise.

• You haven’t computed the wire’s magnetic field. The charge doesn’t feel its own electric field. Sep 12, 2020 at 16:19
• @G.Smith Thanks. I've added my computation of the Wire's Magnetic Field, which I found to be $\frac{{\mu}_{0}I}{2\pi R}$ Sep 12, 2020 at 17:31
• Your first two equations have different powers of $r$ in the denominator and for some reason change notation from $\vec L$ to $\vec s$. Sep 12, 2020 at 17:40
• You need to compute the magnetic field as a vector. Sep 12, 2020 at 17:41
• What is $\lambda$? Wires aren’t charged. They have stationary protons and moving electrons. The net charge is zero. Sep 12, 2020 at 17:43