Imagine a wire carrying a DC current. The wire is U-shaped, so the velocity of the current must be changed during the curve of the U. If we imagine a similarly-shaped pipe carrying water, we see that the provider of the force that accelerates the water through the curve is whatever holds the pipe stationary, like being anchored to a wall.
Is the case the same for the electrical current, and mechanically applied forces acting on the conductor are responsible, or is there emf generated in the curve which does the job?
The system has momentum which you can consider to be carried by either the electrons or the electromagnetic field. In either case in order to effect a dp/dt you need an external force. This could be directly supplied by the wall or indirectly via magnets (think particle accelerator) or any number of other ways.
Hmm, now that I think about it in order for current to be flowing you’d have to have a closed loop so the other side of the loop would experience an equal and opposite reaction.
I’ve never seen such a force on a conductor discussed in any text on electrical engineering or physics. Nor is it given any consideration when designing electrical machinery.
There are large forces on current-carrying conductors but these are a result of the associated magnetic fields. These magnetic forces are so strong they will set the rotor of an electric motor going at high speed and pull a pretty big load along with that rotation. The primary and secondary coils of a transformer have to be well constrained to prevent them from being forced apart. Noises from the core of transformers resulting from loose laminations physically vibrating with the magnetic field alternations is not common but it does happen.
Of course I should warn you that there are a number of things that happen that I’ve never heard of.
Electrons in an electrical current don’t actually move that fast - they travel at their “drift speed” which is much lower than the speed of light. So would there really be enough momentum to cause a reaction?
The momentum density equals 1/(2c)epsilonE[sup]2[/sup], and theoretically it doesn’t matter how small it is. But as I said above for every infinitesimal current segment in a current loop there’s an equal and opposite one and integrating the dp/dt over the loop you wind up with zero.
I haven’t either, but the fields are definitely carrying momentum so theoretically there must be some reaction. It must be incredibly tiny….perhaps I’ll try to calculate it tomorrow.
Also, the only reason I tried to answer this question was because no one else did. So being a good samaritan I then get jumped all over. (Sorry, I don’t know how to make the nifty smilies)
You might find it iteresting that crossed static magnetic and electric fields also have a non zero Poynting Vector. I know this was a surprise to me when I first found out.
Let’s say we’ve got a loop of copper wire carrying a current of 1 A. That gives a drift velocity of electrons in the conductor of about 7.4x10[sup]-5[/sup] ms[sup]-1[/sup].
1 A = 1 Cs[sup]-1[/sup].
1 C has a mass of 5.7x10[sup]-12[/sup] kg.
Momentum = mv = 5.7x10[sup]-12[/sup] kg x 7.4x10[sup]-5[/sup] ms[sup]-1[/sup] = 4.2x10[sup]-16[/sup] kgms[sup]-1[/sup]
Each second, you’re reversing the direction of 1 C, so the time rate of change of momentum = 8.4x10[sup]-16[/sup] kgms[sup]-2[/sup] = 8.4x10[sup]-16[/sup] N.
Now, the force exerted on the loop by its own magnetic field is u[sub]o[/sub]I[sup]2[/sup] = 1.3x10[sup]-6[/sup] N.
So the force resulting from the momentum of the electrons is there, but it’s ten orders of magnitude smaller than the magnetic force.
So, in answer to the OP, the current changes direction because the electrical field follows the curve of the wire, and a very tiny reaction force reults on the wire.
Let’s say we’ve got a loop of copper wire with a cross-sectional area of 1 mm[sup]2[/sup] carrying a current of 1 A. That gives a drift velocity of electrons in the conductor of about 7.4x10[sup]-5[/sup] ms[sup]-1[/sup].
1 A = 1 Cs[sup]-1[/sup].
1 C has a mass of 9.11x10[sup]-31[/sup] kg / 1.6x10[sup]-19[/sup] = 5.7x10[sup]-12[/sup] kg.
Momentum = mv = 5.7x10[sup]-12[/sup] kg x 7.4x10[sup]-5[/sup] ms[sup]-1[/sup] = 4.2x10[sup]-16[/sup] kgms[sup]-1[/sup]
Each second, you’re reversing the direction of 1 C, so the time rate of change of momentum = 2 x 4.2x10[sup]-16[/sup] kgms[sup]-1[/sup] x 1 s[sup]-1[/sup] = 8.4x10[sup]-16[/sup] kgms[sup]-2[/sup] = 8.4x10[sup]-16[/sup] N.
Now, the force exerted on the loop by its own magnetic field is u[sub]o[/sub]I[sup]2[/sup] = 1.3x10[sup]-6[/sup] N.
So the force resulting from the momentum of the electrons is there, but it’s ten orders of magnitude smaller than the magnetic force.
So, in answer to the OP, the current changes direction because the electrical field follows the curve of the wire, and a very tiny reaction force results on the wire.
So given an initially stationary circular loop with no current flowing, if you throw a switch and start current flow through the loop, will the loop rotate (very slowly) backwards to compensate for the electrons moving, or is there enough angular momentum stored in the electric and magnetic fields to compensate fior this?
(Sorry, I don’t know how to make the nifty smilies)