# I Force from "like charges" with and without movement

1. Mar 7, 2017

### Don Bones

I apologize ahead of time if this thread was made in the incorrect post. I have only taken Physics III (Mechanics) and am now taking Physics IV (Electricity/Magnetism). I have yet to find a description of every type of forum here.

In class, I was understanding everything that was mentioned and the professor decided to ask us if we understand. We all nodded and then he decided to mention to us that physics isn't as simple as it seems, and that we are only scratching the surface. Then he proceeded with the following situation.

Q: What happens when I have two positively charged objects next to one another?
A: The like charges repel. Sum of all forces are outward, away from one another.

Q: What happens when I push the two positively charged objects forward?
A: If they move fast enough, the like charges will attract to one another until the net force is directed between the two objects. This would be due to magnetic forces resulting from a current.

This was easy to understand after we looked at magnetic fields from wires of current, But then he asked another question that puzzled everybody.

Q: What happens if I have the charged objects near one another, hold them there with an external force, and then run away from the positively charged objects while observing them (external force removed as he runs and observes)? What does a bystander see if he stays still?
A: The charges now have velocity relative to the person running away. Because of this velocity, the positive charges become currents and the running man observes the charges attracting to one another. The bystander, however, observes the positive charges remaining where they are.

I am assuming this is just a "physics" thing rather than a realistic situation. Still I am confused and it would be nice if this was clarified.

2. Mar 7, 2017

### Khashishi

You still have an electric force pushing the electrons apart. Try to calculate how fast is "fast enough" for the magnetic force to overcome the electric force.
You can find expressions and approximations for the fields here:
https://en.wikipedia.org/wiki/Biot–Savart_law#Point_charge_at_constant_velocity

3. Mar 7, 2017

### Don Bones

I don't see how to use the quote system so sorry if this looks terrible. Also, thanks for the quick response
1. "You still have an electric force pushing the electrons apart." They aren't electrons
2. "Try to calculate how fast is "fast enough" for the magnetic force to overcome the electric force." That isn't the question at hand. No numbers are needed, but the idea here is that if the objects move fast enough the magnetic force will cause them to move towards one another. Yes the electric field is still there but at the "fast enough speed" the magnetic force is strong enough to pull them together for a moment (observable).
3. "https://en.wikipedia.org/wiki/Biot–Savart_law#Point_charge_at_constant_velocity" This doesn't solve my confusion

I'm sorry I wasn't clear in my OP. My confusion lies with the 3rd question. Usually when I think of an object moving, I think of an object going from point A to point B on earth. Then the teacher brought up a scenario where the positively charged objects are remaining still with respect to earth, but have a velocity with respect to the man. Then he stated that the running man will observe the objects moving towards one another IF he is fast enough, and an observer remaining still would experience the exact same thing as the first scenario. I guess my question is this instead.

When considering the magnetic field produced by a charge, is it problematic to have the velocity relative to a moving object? Ex: The magnetic field of a proton at rest relative to a drag racer moving 100km/h. Again, I know how to solve it. My question lies with what my velocity should be for accurate information

4. Mar 7, 2017

### Philip Wood

If I've understood the thought-experiments you're describing, then in all three cases the charges repel. This is because each particle is in the electric field due to the other. Cases (2) and (3) are identical in the frames of reference of the person moving the charges forward in (2) and the running man in (3), assuming that the charges have the same velocity in both these frames. In these frames the charges still repel, but with less force than in the frame in which the charges are stationary or almost stationary (case (1)). The reason the force is less is indeed that in the frames in which the particles are moving parallel to each order, each sits in the magnetic field due to the other, so in addition to the repulsion due to the electric fields, there is an attractive force due to the magnetic fields. But this attractive force is always weaker than the repulsive force, so the resultant force is still repulsive, but weaker than in case (1).

Why do we need all this business of two different sorts of fields? Can't we simply say that the charges repel each other, but that the force is less when the charges are observed in a frame of reference in which they are moving parallel to each other? The answer is that we can, and that quite simple application of Special Relativity theory yields exactly the force diminution that we talk of as due to the effects of a magnetic field.

5. Mar 7, 2017

### Khashishi

Actually, your confusion is such that you don't know where you are confused. If you had gone and done the calculation I suggested, you would have realized this.
How fast is fast enough? Don't just guess. Calculate what the force between the objects is as a function of velocity.

6. Mar 7, 2017

### Staff: Mentor

I second the recommendation to calculate the speed that is "fast enough". It turns out to be fairly important in resolving the problem you are addressing.

No, it is not problematic at all. In fact, it is a pillar of modern physics that you can always do so. If the object is moving inertially then it is particularly simple.

Last edited: Mar 7, 2017
7. Mar 7, 2017

### Don Bones

Okay so:
Scenario (1)
Fe=(kq1q2)/(r2)
That is the force experienced by object 1. q1 is the charge of object 1, and q2 is the charge for object 2. r is the distance between them.I am going to assume that the force due to gravity is negligible.
Scenario (2)
Two forces are being acted on this object.
Fe=(kq1q2)/(r2) and FB=q1vBsinΘ (degree is 90, so sin90=1)
Fe is the same as last time. FB is the force due to a magnetic field created by both positively charged objects in motion. q1 is the charge of the first object, and v is the magnitude of the velocity object 1 and 2 are moving in. According to Biot-Savart's law, the magnitude of a magnetic field B created by q for this equation would be..
B=(μ0q2v)/(4πr2) assuming that the angle between the velocity vector and the vector for the distance from object 1 to 2 is 90, sin90=1.
μ0 is the permeability of free space, q2 was listed before, v is the velocity of object 2, and r is the distance from object 1 to object 2. The goal is to have magnetic force overcome electric force correct? then we are aiming to say
FB > Fe
q1vB> (kq1q2)/(r2)
Substitute B with (μ0q2v)/(4πr2)
0q2q1v2)/(4πr2)> (kq1q2)/(r2)
and then solve for v
v > √(ke4π/μ0)

In order to check my work I turned all the value into their units to analyse them.
√[(Nm2)/(C2) * (C/s)/(Tm)] Convert Tesla into simpler units
√[(Nm2)/(C2) * (C/s * C * m/s)/(Nm)] Simplify
√(m2/s2)
and we receive m/s, the other side of the inequality.

This, however, looks odd. This states the velocity needed to have the two positive objects move towards each other doesn't depend on the charge, the distance between them, or anything about the objects. Rather, the nature our universe. That and when I plug in the value for the constants, I get 8.99*1016 m/s. So what did I miss?

8. Mar 7, 2017

### Staff: Mentor

Yes, that is correct.

It looks like you forgot a square root somewhere.

9. Mar 7, 2017

### Don Bones

Oops, thank you. I see the velocity is now 3.00*108m/s which happens to be the speed of light. So the event where the objects move toward one another couldn't happen because they would have to be travelling at the speed of light. Good to know but that didn't answer the question I meant to ask. I will do some more math before I ask again though.