# Thin wire alternating exerts force on a nearby charge

• CptXray
In summary, the force exerted on a point charge located in a distance from a wire is proportional to the magnetic field and inversely proportional to the square of the distance.
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.

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.

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.

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.

@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.

CptXray said:
@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.

## 1. What is thin wire alternating and how does it exert force on a nearby charge?

Thin wire alternating refers to a type of electrical current that flows through a thin wire in an alternating direction. This alternating current creates a magnetic field around the wire, which can then exert a force on a nearby charged object.

## 2. What is the difference between thin wire alternating and direct current?

The main difference between thin wire alternating and direct current is the direction of the flow of electrons. In thin wire alternating, the direction of the current changes periodically, while in direct current, the electrons flow in one direction only.

## 3. How does the strength of the alternating current affect the force exerted on a nearby charge?

The strength of the alternating current affects the force exerted on a nearby charge because the stronger the current, the stronger the magnetic field it creates. This results in a stronger force being exerted on the nearby charge.

## 4. Can thin wire alternating only exert force on charged objects?

No, thin wire alternating can also exert force on other objects that are capable of experiencing a magnetic force, such as other wires with electric currents or magnets.

## 5. How is thin wire alternating used in everyday life?

Thin wire alternating is used in a variety of everyday devices and appliances, such as computers, televisions, and refrigerators. It is also commonly used in power transmission and distribution systems to transport electricity over long distances.

• Introductory Physics Homework Help
Replies
17
Views
469
• Introductory Physics Homework Help
Replies
1
Views
195
• Introductory Physics Homework Help
Replies
3
Views
307
• Introductory Physics Homework Help
Replies
5
Views
397
• Introductory Physics Homework Help
Replies
2
Views
345
• Introductory Physics Homework Help
Replies
4
Views
949
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
281
• Introductory Physics Homework Help
Replies
64
Views
2K