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

spsch
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?

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
spsch
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.