# Force on a copper loop entering into a magnetic Field B with speed v

Homework Statement:
A rectangular copper loop is entering a magnetic field B with speed v. What is the Force against the loop's motion?
B = 0.03 T
diameter of the cooper string is 0.4 mm
and v = 5 m/s

Loops dimensions are length 10cm, width 5cm.
Relevant Equations:
## V= (change in magnetic flux) / (change in time) ## (I'm not sure about the greek letter, is it phi?)
F = ILB
V = IR
Hi, second problem in one evening, I'm sorry!

But i'm also not quite sure if I did this one right.

I had thought I need lenz's law but there is no current before entering the field so I just use the induced Voltage?
My approach:
## V = \frac {B*A}{t} ##
## IR = \frac {B*A}{t} ## and ## A = v*t (1s) * width (0.05m) ##
so ## I = \frac{B*v*width}{R} ## and ## R = rho* \frac {2v+w}{pi*(0.0004)^2} ##
then ## I = \frac{B*v*width*(pi*(0.0002)^2)}{2v+w} ##
Because ## F = ILB ## I have after canceling some terms:
## F = \frac {B^2*pi*(\frac {d}{2})*width*v}{rho*(2*v+width)} ##

It seems overly complicated? Could someone maybe point to where I went wrong?

kuruman
Homework Helper
Gold Member
It looks good except for the wire resistance. The resistance of a wire is given by ##R=\dfrac{\rho L}{\pi r^2}##, where ##L## is the length of the wire and ##r## is its radius. What are these two quantities in this case? Specifically, why is the length ##2v+w##? Does the loop perimeter get to be longer when it moves faster? Also, in the last equation for the force you forgot to square ##(\frac{d}{2}).##

• spsch, Michael Price and berkeman
Hi Kuruman! Thank you for helping me on this post as well.

Originally I only had w as L when I first worked on the problem. Because only this section of the wire experiences a net force.
But since the current is induced through all the wire I thought I should use the length that is exposed to the magnetic field.

Should I use the full length of the loop instead? (That kind of makes sense now that I think about it because the current should go through the whole loop, right?)

So then L would be ## 2*width + 2*length ## or 0.3 meters.
R is correct I believe, d/2. And yes, I missed to square it in my post here, thanks for pointing this out too. I wanted to correct but it doesn't let me anymore.

kuruman
Homework Helper
Gold Member
Should I use the full length of the loop instead? (That kind of makes sense now that I think about it because the current should go through the whole loop, right?)
Right. The length of wire has the resistance it has even when no current is running through it.

• spsch
Right. The length of wire has the resistance it has even when no current is running through it.
Thank you!