# Conceptual question about magnetic fields

1. Mar 8, 2009

### Kruum

I have a physics exam tomorrow and while recapping things for the exam, I was wondering one thing. The magnetic force exerted to a current carrying wire in a uniform magnetic field is $$\vec{F}=I \vec{l} \times \vec{B}$$. But a current carrying wire induces a magnetic field, so the magnetic field wouldn't be uniform anymore. So I guess the question I have is, if the wire isn't fixed would the force exerted be the same or do I some how have to add the effects of the outer magnetic field and the field induced by the wire.

2. Mar 8, 2009

### Redbelly98

Staff Emeritus
You don't need to include the field created by the wire itself. There are a couple of ways to look at this:

An object cannot exert a (net) force on itself, according to Newton's third law .

Also, this is akin to the force on a charge due to an electric field. One does not include the field of the charge itself to calculate the force.

Hope that helps!

EDIT:
... and good luck tomorrow.

Last edited: Mar 8, 2009
3. Mar 8, 2009

### Kruum

Thank you, Redbelly98. So this holds when the wire is moving?

I'd like to still know in detail why this is. Let's assume the wire is perpendicular to the outer field. So the magnetic field exerts a force on the wire, so the wire starts to move, if it's not fixed. And the wire would induce a magnetic field, which then increases or decreases the outer field. So when the wire moves dx meters the magnetic field isn't the same as where the wire started. I've been pondering this and all I've come up with is that the chances in the magnetic field are rapid enough so we can neglegt them.