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Question: Will a stationary charge near a conductor with a steady current experience any force by the conductor?Answer: No; from theory.
I'll like to know if anyone has come across any empirical observation, either direct of indirect, that confirms there cannot be (absolutely sure) that confirms zero force.
But what has never been verified is only assumed. The thing is, probably no one ever tested it and no one is interested. Furthermore, it may not be simple task to isolate a charge and make it steady.
" Will a stationary charge near a conductor with a steady current experience any force by the conductor?"Yes it will.It will be attracted to the conductor- regardless of any current flowing through it.Why wouldn't it?
Quote from: theThinkerQuestion: Will a stationary charge near a conductor with a steady current experience any force by the conductor?Answer: No; from theory. That is correct.Quote from: theThinkerI'll like to know if anyone has come across any empirical observation, either direct of indirect, that confirms there cannot be (absolutely sure) that confirms zero force. To be honest that's like asking for empirical evidence that a wheel will roil. It's something that all physicists know by heart and what their research abd experiments are based on as are the things built by engineers. However you could look into the original publications which led to the laws of electromagnetism. I'm just not sure where you'd find it. Quote from: theThinkerBut what has never been verified is only assumed. The thing is, probably no one ever tested it and no one is interested. Furthermore, it may not be simple task to isolate a charge and make it steady.All of the laws of physics have been carefully tested so don't worry about that. Especially Maxwell's laws of electrodynamics. A huge amount of work has gone into them. And if any part of them were wrong it'd have shown up by now. Especially what you're asking about. If it was wrong then certain experiments and inventions would not work correctly.
I give an example. Aristotle believed that a heavier object falls faster than a lighter one. Everyone just repeated after him for almost 1600 years until Galileo actually demonstrated by letting fall two metal balls, one large and one small. Aristotle was wrong.
Because the force on a charged particle due to a magnetic field is a the cross product of the particles velocity and the magnetic field. In this case the velocity is zero so the force is zero. That's why it wouldn't.
As we see from the figure, Jf, is directed longitudinally along this uniform conductor, as is E since they are parallel. Thus, at the surface, E will be tangential, and since the tangential components of E are continuous by (9-21), there will be a tangential field outside the conductor which is given by E = Jf/σ. This is in marked contrast to the static case where not only was E = 0 inside the conductor, but it was necessarily normal to the surface, as we saw in (6-2) and Figure 6-1b.
I have tried to do the experiment using electrically charged torsion balance. I used ping-pong balls covered with aluminum foil, which then are connected to 4 kV DC source (hence one ball has potential of +2 kV while the other has -2 kV).But it's hard to reach a firm conclusion since either balls are attracted to even neutral objects.
Quote from: PmbPhy on 01/09/2016 02:20:10Because the force on a charged particle due to a magnetic field is a the cross product of the particles velocity and the magnetic field. In this case the velocity is zero so the force is zero. That's why it wouldn't.While this is true it is an oversimplification. For starters the presence of the charge near the conductor repels the like charges and attracts the opposite charges which changes the charge distribution and results in an attractive force between the conductor and the charge. The flow of current doesn't really change this situation. However, beyond even that there does seem to be a case for electric fields outside of a current carrying resistive wire due to the current.Furthermore, in the 2nd edition of Electromagnetic Fields by Wangsness on page 210 it is stated:Quote from: WangsnessAs we see from the figure, Jf, is directed longitudinally along this uniform conductor, as is E since they are parallel. Thus, at the surface, E will be tangential, and since the tangential components of E are continuous by (9-21), there will be a tangential field outside the conductor which is given by E = Jf/σ. This is in marked contrast to the static case where not only was E = 0 inside the conductor, but it was necessarily normal to the surface, as we saw in (6-2) and Figure 6-1b.