# Magnetic Field and Currents/Moving Charges

1. Nov 13, 2005

### thowonasweaterandgo

Hello, I have a quick question. Why is it that a magnetic field has an effect on a current or moving charge, but not on a stationary charge?

2. Nov 13, 2005

### rbj

because the Lorentz force law says so?

3. Nov 13, 2005

### thowonasweaterandgo

SWEET! :grumpy:

I was hoping for something more along the lines of what's happening at the atomic level. My books provides a ton of equations, but no explanation whatsoever as to why they work.

4. Nov 13, 2005

### hypermorphism

If you're looking for a result derived from existing axioms, relativity shows that electrostatic and magnetic fields are just differing frames of reference of a general electromagnetic field. Without relativity, you just have to rely on the equations derived from empirical observations.
In modern physics, quantum electrodynamics models the electromagnetic field as being mediated by virtual and real photons.

Last edited: Nov 13, 2005
5. Nov 13, 2005

### thowonasweaterandgo

ah okay, so basically I just don't have the right tools at the moment to understand why?

6. Nov 13, 2005

### thowonasweaterandgo

you lost me on the last sentence there.

7. Nov 13, 2005

### hypermorphism

Pretty much. However, note that answers to questions such as "why such and such happen" are purely subjective. Physics can really only model how things happen, and the current model may or may not satisfy your concepts of "why".

8. Nov 13, 2005

### hypermorphism

Instead of the "mysterious action-at-a-distance" of Newtonian fields, modern physics has forces mediated by force particles, gauge bosons; these bosons are interchanged between two particles in order to model forces between those particles. You can read more about these in any popular science book. For an interesting read, see Feynman's QED, a layman's intro to a small part of quantum electrodynamics.

9. Nov 13, 2005

### thowonasweaterandgo

anybody else want to give it a shot?

10. Nov 17, 2005

### Dvarek

Would it not simply have to do with the magnetic field is stationary, because it can not act on its own accord. But as soon as another enerygy field comes along and moves past it, it draws or 'chases' after the electrical field? The magnetic field in itself is stationary and does not need any additional energy to substain itself. That is why it is not affected by a stationary charge. Is that what you are wondering or am I way off base?

11. Nov 17, 2005

### Dvarek

Excuse the spelling. LOL

12. Nov 17, 2005

### Dvarek

If I am way off base, could you maybe elaborate more, please?

13. Nov 17, 2005

### thowonasweaterandgo

thanks for the reply btw, I really appreciate it. I know it's a fact that currents and magnets create and feel forces from magnetic fields, while stationary electric charges do not. What I don't understand is what it is about a moving charge as opposed to a stationary one that makes this true? I guess I'm asking what, on an atomic level, is going on here?

14. Nov 19, 2005

### Dvarek

So you understand the principle, just want the specifics, eh?

15. Nov 19, 2005

### krab

Don't know why you say
A charge in a magnetic field experiences a force . Doesn't matter if you are on a large scale or small. Same thing happens.

16. Nov 20, 2005

### rbj

depends on who the observer is.

the electromagnetic action is nothing other than the sole electrostatic action, but with special relativity taken into consideration. take a look at http://en.wikipedia.org/wiki/Magnetic_field :

Maxwell did much to unify static electricity and magnetism, producing a set of four equations relating the two fields. However, under Maxwell's formulation, there were still two distinct fields describing different phenomena. It was Albert Einstein who showed, using special relativity, that electric and magnetic fields are two aspects of the same thing (a rank-2 tensor), and that one observer may perceive a magnetic force where a moving observer perceives only an electrostatic force. Thus, using special relativity, magnetic forces are a manifestation of electrostatic forces of charges in motion and may be predicted from knowledge of the electrostatic forces and the movement (relative to some observer) of the charges.
A thought experiment one can do to show this is with two identical infinite and parallel lines of charge having no motion relative to each other but moving together relative to an observer. Another observer is moving alongside the two lines of charge (at the same velocity) and observes only electrostatic repulsive force and acceleration. The first or "stationary" observer seeing the two lines (and second observer) moving past with some known velocity also observes that the "moving" observer's clock is ticking more slowly (due to time dilation) and thus observes the repulsive acceleration of the lines more slowly than that which the "moving" observer sees. The reduction of repulsive acceleration can be thought of as an attractive force (in a classical physics context) that reduces the electrostatic repulsive force and also that is increasing with increasing velocity. This pseudo-force is precisely the same as the electromagnetic force in a classical context.

in the discussion page of that wiki article is a quantitative expression of the same thought experiment. i reposted it here at PF at this thread: https://www.physicsforums.com/showthread.php?t=96769 . look it over or try working it out yourself. it's not hard.

(last edit) i decided it was easy to copy that to here:

The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration. The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of $\lambda \$ ) and some non-zero mass per unit length of $\rho \$ separated by some distance $R \$. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static non-relativistic repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance $R \$) for each infinite parallel line of charge would be:

$$a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho}$$

If the lines of charge are moving together past the observer at some velocity, $v \$, the non-relativistic electrostatic force would appear to be unchanged and that would be the acceleration an observer traveling along with the lines of charge would observe.

Now, if special relativity is considered, the in-motion observer's clock
would be ticking at a relative rate (ticks per unit time or 1/time) of $\sqrt{1 - v^2/c^2}$ from the point-of-view of the stationary observer because of time dilation. Since acceleration is proportional to (1/time)2, the at-rest observer would observe an acceleration scaled by the square of that rate, or by ${1 - v^2/c^2} \$, compared to what the moving observer sees. Then the observed outward acceleration of the two infinite lines as viewed by the stationary observer would be:

$$a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho}$$

or

$$a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho}$$

The first term in the numerator, $F_e \$, is the electrostatic force (per unit length) outward and is reduced by the second term, $F_m \$, which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors).

The electric current, $i_0 \$, in each conductor is

$$i_0 = v \lambda \$$

and $$\frac{1}{\epsilon_0 c^2}$$ is the magnetic permeability

$$\mu_0 = \frac{1}{\epsilon_0 c^2}$$

because $$c^2 = \frac{1}{ \mu_0 \epsilon_0 }$$

so you get for the 2nd force term:

$$F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R}$$

which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by $R \$, with identical current $i_0 \$.

Last edited: Nov 20, 2005
17. Nov 20, 2005

### thowonasweaterandgo

wow thanks, that is awesome. I really appreciate it! thanks!