 #1
spsch
 111
 21
 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?
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?