Magnetic Force on current-carrying wire

1. Aug 28, 2014

fonz

Not necessarily a homework question but this is pretty fundamental. I can't get a decent derivation online.

If the length of the wire in the uniform magnetic field is $l$ and it moves a distance $\delta s$ in $\delta t$:

From the Lorentz Force Law:

$\vec{F} = q\vec{v}\times\vec{B}$

if flux is cut at $\frac{\pi}{2}$:

$\vec{F} = q\vec{v}\vec{B}$

$\vec{v} = \frac{d\vec{s}}{dt}$, $q = \int I dt$

$\vec{F} = \vec{B}\frac{d\vec{s}}{dt} \int I dt$

That is about as far as I get, not sure how the length of the conductor $l$ comes into it?

Last edited: Aug 28, 2014
2. Aug 28, 2014

BvU

Let us try to get the question right: The force you want to calculate is the force on a current-carrying wire. (This is what I read in the title). But then the wire moves a distance $\delta\,\vec s$, so more is happening ?

"If the flux is cut at $\pi\over 2$" means what ? The wire is perpendicular to $\vec B$, $\delta\,\vec s$ is perpendicular to both ?

Can I see a drawing ?

You are apparently not happy with this derivation ? Isn't it sufficient to adjust the $\vec v$ vector for the case the wire isn't stationary ?

3. Aug 28, 2014

fonz

Unfortunately I don't have a diagram, you are correct in your statement that the wire in this example is perpendicular to both the uniform magnetic field and the velocity vector.

I don't really follow the context of that particular derivation you have referred to. I suppose if I were to describe the problem qualitatively the question I am asking is what is the relation between current and velocity for a wire moving through a uniform magnetic field.

Maybe I am slightly off track. I find the concept of the induced emf due to the flux linkage relatively easy to grasp. I am now asking what is the relation to current?

4. Aug 28, 2014

BvU

Yes, but everything can be brought back to the Lorentz force: Induced emf is a $\vec v$ of the charge carriers in a conductor, a wire in this case, due to the wire itself moving. Force on a current-carrying wire is a $\vec v$ of the charge carriers moving within the wire.

My impression was you do not necessarily want to dig into the case of a combination of the two, but I could be wrong, however:
tells me otherwise.

Are you comfortable with the q v = q L/t = q/t L = I L steps in the derivation ?

The effect of the v from moving the wire simply adds (vector wise) to the effect of the movement
within the wire.

Suppose the wire is moving downward in their picture, then all I can think of is that the red velocity vector rotates a bit clockwise (with the component along the wire remaining the same magnitude). And with that rotation, the F vector also rotates by the same amount, thus pushing the charge carriers a bit towards the right end of the wire. I can't do much in the sense of physics with that, however: bringing about a current I in the wire has become a little easier. But we don't look at external stuff like a current source or so.

I think it would be wise to move on to further exercises with wire frames moving through zones with a magnetic field and such. Perhaps the links in the link also provide some insight.