(Non-current) conductors in a magnetic field

A quick question.

Let's say you had an infinitely long wire carrying a constant current, and a straight piece of wire parallel to it a certain distance away. Obviously, the current, being a motion of charges, will generate a magnetic field, with its curl being equal to the electrical permittivity times the charge density.

As well, this magnetic field will induce an electromotive force in the piece of wire.

However, the question I have is this: it's obvious to me that there is an electric potential produced by the current-carrying wire on the piece of wire, but is there a potential DIFFERENCE? I'm fairly sure that the induced electric field won't have a curl--the magnetic field is basically constant--but I'm not sure about the mere presence of an electric field at all.

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Danger
Gold Member
If the wire is infinitely long, how can if form a circuit?

I just mean theoretically speaking. "Infinitely long". Long enough that its ends don't distort the magnetic field around the other, shorter wire. Obviously there will be a potential difference applied at the FAR AWAY left end and the FAR AWAY right end. I'm talking purely theory here, in the case of a uniform magnetic field applied to a finite, non-looping wire, by a different, current-carrying wire.

SammyS
Staff Emeritus
Homework Helper
Gold Member
A quick question.

Let's say you had an infinitely long wire carrying a constant current, and a straight piece of wire parallel to it a certain distance away. Obviously, the current, being a motion of charges, will generate a magnetic field, with its curl being equal to the electrical permittivity times the charge density.

As well, this magnetic field will induce an electromotive force in the piece of wire.

However, the question I have is this: it's obvious to me that there is an electric potential produced by the current-carrying wire on the piece of wire, but is there a potential DIFFERENCE? I'm fairly sure that the induced electric field won't have a curl--the magnetic field is basically constant--but I'm not sure about the mere presence of an electric field at all.
Whatever law is there which would lead you to believe " there is an electric potential produced by the current-carrying wire on the piece of wire" ?

You're right--that was my mistake.

But what if the piece of wire was moving perpendicularly away from the current? The magnetic field in the piece of wire would be changing, implying an EMF--right?

And if so, is the potential equal across the entire wire?

BruceW
Homework Helper
a current would be induced if the wire was moving perpendicularly away from the current. and if they were both very long, then the magnetic field is approximately uniform (as we look in the direction of the axis). So the induced current is approximately the same along the wire. In this case, the magnetic field is causing the EMF right? So there is no difference in potential causing the current.

How can a current be generated in the piece of wire (for more than a few moments) if it's not a circuit? A current flowing in a straight, isolated piece of wire implies charges falling out of one end or another, which obviously doesn't actually happen.

To clarify my meaning, this is a diagram of what's happening (I left off the magnetic field lines as they're implied by the current): http://prntscr.com/15imtk

sophiecentaur
Gold Member
As well, this magnetic field will induce an electromotive force in the piece of wire.
Think back on what you've been taught. Don't you remember something about 'Rate of change of magnetic flux'???

BruceW
Homework Helper
How can a current be generated in the piece of wire (for more than a few moments) if it's not a circuit? A current flowing in a straight, isolated piece of wire implies charges falling out of one end or another, which obviously doesn't actually happen.

To clarify my meaning, this is a diagram of what's happening (I left off the magnetic field lines as they're implied by the current): http://prntscr.com/15imtk
Oh right, I assumed you were talking about a circuit. OK, the charge carriers will 'bunch up' at the ends, due to the magnetic field pushing them that way (and there is nowhere for them to go). But then, because there is a build-up of charge, there will be an electric field created, which will counteract the force on the charges due to the magnetic field, so a static charge distribution will be reached.

Often, the assumption is made that the electric force between charge carriers is so much stronger than the magnetic force, that there is negligible 'bunching up', so that approximately, nothing will happen, and the charges do not move.

So, because it reaches that charge distribution, ΔV ≈ 0 (or = 0, in a perfect situation)?

sophiecentaur
Gold Member
If del (/ delta) V is zero then there will be uniform charge distribution??

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BruceW
Homework Helper
So, because it reaches that charge distribution, ΔV ≈ 0 (or = 0, in a perfect situation)?
no, when it reaches that charge distribution, ΔV will be just enough to counteract the magnetic force on each of the charge carriers. In other words, the rate of change of the charge distribution eventually ends up being zero, therefore the EMF is zero. The EMF has 2 terms: first term due to the electric field, and second term due to the magnetic field. These two terms exactly cancel out, so there is no more forces on the charges (in the macroscopic sense). Also, In the limit that the magnetic force is much less than the electric force, then ΔV will still be non-zero. But it will be small. (maybe this is what you meant?)

p.s. the word 'EMF' may be defined differently by different people. Maybe the definition I am using here is not the same as the one you are used to. I think that there are several possible definitions, so it is important to know which one you are using.