Force on a Rectangular Current Loop [Magnetism]

[FermiParadox]

Homework Statement

A force acts on a rectangular loop of wire, pulling it out of a magnetic field. Calculate the force required to pull the loop to the right at a constant speed of 4.42 m/s.

B = 0.44 T
l = 0.350 m
v = 4.42 m/s
R = 0.356 ohms

Homework Equations

F_magnetic=I*l*B
I=(emf)/R
emf=B*l*v

The Attempt at a Solution

1. As we pull the loop, the area exposed to the magnetic field will decrease.
2. Per Lenz's law, this means that the induced magnetic field will 'strengthen' the magnetic field and will act in the same direction, into the page.
3. So current will flow clockwise.
5. The 'right' way to do the problem seems to be fairly simple - just plug in the formulas I've already given, assume that F_magnetic will oppose F_applied and solve for F_applied.

This gave me the right answer, but I'm now looking for a reason why the magnetic force will oppose the applied force. How do we find the direction of magnetic force? It's IL_vector x B_vector, where L_vector is in the same direction as I is - the problem is that the vector will change directions four times as it moves through the loop. How do we know which one of these to cross with B? Wouldn't the magnetic force act on all of the segments of wire - i.e. there'd be a force pushing up on the rectangle, down on the rectangle, one pushing the rectangle to the left, and one pushing the rectangle to the right.

I seek understanding! Help me out!

 

[ideasrule]

Wouldn't the magnetic force act on all of the segments of wire - i.e. there'd be a force pushing up on the rectangle, down on the rectangle, one pushing the rectangle to the left, and one pushing the rectangle to the right.

Your understanding is almost completely correct. There's a force pushing up on the rectangle, down on the rectangle, and pushing the rectangle to the left. There's no force pushing it to the right because the right side of the rectangle is outside of the field.

 

[FermiParadox]

I can't believe I was missing that the entire time - thanks so much!