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

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

enter image description here 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.

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

1 Answer 1

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Your final expression for B is correct and could have been found using Ampere's law. To get the electric field you need the gradient of the charge density in the wire. Without the electric field the speed of the charged object would be constant and the magnetic force would be centripetal, changing from point to point as the B field and the velocity direction both change. This motion (in three dimensions) might be approximated by a numeric simulation.

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