Thin wire alternating exerts force on a nearby charge

[CptXray]

Homework Statement

There's a thin, straight, infinite wire conducting alternating current:
$$I(t) = I_{0}\exp(-\kappa t^2),$$
where $\kappa > 0$.
Calculate the force exerted on point charge $q$ that is located in distance $\rho$ from the wire. Consider relativistic effects.

Homework Equations

Ampere's law:
$$\int\limits_{\partial V} \vec{B} \vec{\mathrm{d}l} = \int \limits_{V} \vec{j} \vec{\mathrm{d} s}$$
$$\nabla \times \vec{E} = - \frac{\partial{\vec{B}}}{\partial{t}}$$

The Attempt at a Solution

First, the magnetic field of a straight infinite wire:

Using Ampere's law, where $\partial{V}$ is a circle, centered on the wire:

$$\int\limits_{\partial V} \vec{B} \vec{\mathrm{d}l} = \int \limits_{V} \vec{j} \vec{\mathrm{d} s}$$
$$B_{\phi} 2 \pi r = \mu_{0} I$$
$$B_{\phi} = \frac{\mu_{0}I}{2\pi} \frac{1}{r} \hat{\phi},$$
where $B_{\phi}$ is a magnetic field in azimuthal direction and $r$ is the distance from the wire.
Now, I know that the magnetic field is changing with time:
$$- \frac{\partial{B_{\phi}}}{\partial{t}} = 2kt\frac{\mu_{0} I}{2\pi}\frac{1}{r} \hat{\phi}$$
I tried doing double curl $\nabla \times (\nabla \times \vec{B}) = \nabla (\nabla \cdot \vec{E}) - \Delta E$, but I have a feeling that it gets me nowhere.

 

[trurle]

If the charge is not initially moving, the only effect would be reaction force due scattering of EM wave emitted by wire antenna (Thomson scattering).

After electron will start moving, it will generally move in cycloidal path (due interaction with variable magnetic and electric field) away from wire.

 

[CptXray]

Ok, if there's anyone else interested in this topic it's really well explained in "Introduction to electrodynamics" third edition by David Griffiths Chapter 10.

 

[TSny]

Hello, @CptXray . Were you able to solve the problem? I tried to approach it by setting up an integral for the potential A and then finding the electric field E from A. However, the integral seems too difficult to evaluate due to the integrand depending on retarded times.

 

[CptXray]

@TSny, apparently the trick is to write the potential $$\vec{A}(\rho) = \frac{\mu_{0}}{4 \pi} \int_{-\infty}^{\infty} \frac{I(t - t_{r}) \hat{e}_{z}}{\sqrt{z^2 + \rho^2}},$$ where $t_{r} = t - \frac{1}{c}\sqrt{z^2 + \rho^2}$. The other step is to see that function under itegral is diverging really quickly so you can take time derivative $\frac{\partial{}}{\partial{t}}$ of a function under integral. Next you calculate the integral in time $t=0$. Result should be:
$$\vec{E}(\rho) = \frac{ \mu_{0} I_{0} \sqrt{k} } { 2 \sqrt{\pi} }\exp\big({-\frac{k}{c^2}\rho^2}\big) \hat{e}_{z}.$$
Sorry it's written really briefly, but I have an exam in few hours.

 

[TSny]

@TSny, apparently the trick is to write the potential $$\vec{A}(\rho) = \frac{\mu_{0}}{4 \pi} \int_{-\infty}^{\infty} \frac{I(t - t_{r}) \hat{e}_{z}}{\sqrt{z^2 + \rho^2}},$$ where $t_{r} = t - \frac{1}{c}\sqrt{z^2 + \rho^2}$. The other step is to see that function under itegral is diverging really quickly so you can take time derivative $\frac{\partial{}}{\partial{t}}$ of a function under integral. Next you calculate the integral in time $t=0$. Result should be:
$$\vec{E}(\rho) = \frac{ \mu_{0} I_{0} \sqrt{k} } { 2 \sqrt{\pi} }\exp\big({-\frac{k}{c^2}\rho^2}\big) \hat{e}_{z}.$$
Sorry it's written really briefly, but I have exam in few hours.

Oh. I didn't know that you only needed the result for the special time t = 0. Thanks.