- #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?