# Forces on a wire moving in a magnetic field?

When a wire is placed inside a magnetic field and current starts to flow in that wire, that wire experiences the Lorentz force = $IL \times B$ and beings to move. Aside from induced motional - $\mathscr E$ = $-vBL$, isn't there another force of repulsion/attraction from the magnetic field produced by the flow of current in the wire?

Assuming that the wire's magnetic field is parallel to the exterior magnetic field, there is no effect?

Also, the magnetic field induced by the flow of current in the wire could cause a change in flux and an induced $\mathscr E$ to oppose the change? Even if the wire's induced field is parallel to the exterior field?

Diagram:

I see three kinds of lines:

On the left I see a vertical line of current, some circular lines ($\vec{B}$ field caused by current?) and some orange lines (rightwards? external $\vec{B}$ field?).

The picture on the right might be the exact same thing, look up along the direction of the current.

If so, then say $\vec{B}_1(x,y,z,t)=B\hat{x}$, and $\vec{I}=I\hat{z}$ with the wire located (at $t=0$) at $x=0$,$y=0$. If the wire weren't moving, there would be a $\vec{B}_2$ field due to it, with $\vec{B}_2=\frac{1}{\mu_02\pi\sqrt{x^2+y^2}}\hat{\phi}$. But there is a force on the wire. If the current everywhere is $I$, then there is a force per length of $IB\hat{y}$.

But I don't see a circuit, so no $\mathscr E$. I don't see a mass density so we can't even figure out how the wire moves. So I don't see a question. But I'll try to address some concerns.

aside from Induced motional $\mathscr E = −vBL$, isn't there another force of repulsion/attraction from the magnetic field produced by the flow of current in the wire?

Concern #1, the $L$, there is a force per unit length, but I didn't see a finite wire. Secondly, no circuit, so no obvious emf $\mathscr E$. Secondly for thinly wired circuits there is an electric-force-related emf due to the flux of $\partial \vec{B} / \partial t$, and there is a magnetic-force-related emf due to the change of flux the instantaneous $\vec{B}$ that changes because of the moving wire. An emf ignores forces not collinear with the circuit (but we don't have a circuit), so for a moving wire the difference between the magnetic force (which takes the total velocity, that due to the wire and that due to the flow along the wire) and the emf (which is force per unit charge, and only the part in the direction of the circuit). For a magnetic force and an emf, that means the velocity along the direction of the wire does not contribute.

That doesn't make it a different force however, because the actual flow of current is caused by the flow of a mobile charge from one part of the wire at one time, and another part of the wire at a different time, so it's exactly but that is needed, the velocity of the wire, plus the flow along the wire). So really it's both that matter for the current.

I'm assuming that since the wire's magnetic field is parallel to the exterior magnetic field, there is no effect?

The field $\frac{1}{\mu_02\pi\sqrt{x^2+y^2}}\hat{\phi}$ caused by the current is orthogonal to the current $I\hat{z}$, but the field is only sometimes parallel to the external field $B\hat{x}$. I hope I didn't misread the entire question.

Also, because of the magnetic field induced by the flow of current in the wire, that would cause a change in flux and an induced ϵ to oppose the change?

Again, I don't see a circuit.

Even though that wire's induced field is parallel to the exterior field?

Again, it seems the external field $B\hat{x}$ is only sometimes parallel to the field $\frac{1}{\mu_02\pi\sqrt{x^2+y^2}}\hat{\phi}$ caused by the current.

• Although my diagram is terrible at illustrating everything, this wire is a part of a circuit(assume so...) and has current flowing from a power supply, has mass m, and resistance R and everything you'd expect from a wire. Commented Feb 15, 2015 at 22:36
• @Key If I bend the wire counterclockwise to make a loop the total force is zero, but if I bend it out of the page towards me to make a loop, then I get a nonzero total force. It's hard to know which to give an answer for. Commented Feb 15, 2015 at 22:44
• That is true, but what I wanted to illustrate was a single wire passing(or in) a magnetic field and the Lorentz force acting on it. I now realize, that the only force that acts on the wire, is the Lorentz force = $IL \times B$ Commented Feb 16, 2015 at 6:08
• However, one concern I do have if you could please address it... when the wire induces it's own magnetic field to oppose the external applied field is there going to be a change in flux? Causing an induced-EMF on the wire? Because in a system that a wire moves due to the Lorentz force I can only account for motional-EMF as the induced-EMF in the system, could it be... that the wire's magnetic field opposing the applied magnetic field cause another induced-EMF? Commented Feb 16, 2015 at 16:30
• @Key The motion of the wire makes an emf due to the magnetic force per charge, a changing field (even due to the wire itself) makes an emf due to the electric force per unit charge. Both are real, and both happen. The emf due to the wires own current is called self inductance, and usually needs to be solved as a differential equation. Commented Feb 17, 2015 at 7:47