Square copper wire loop within a magnetic field

[cookiemnstr510510]

Homework Statement

In the figures below, a copper wire of circular cross section, A, has been bent into a square loop of side length, c, and arc welded at the seam for electrical continuity. Assume that the resulting square loop has a resistance, R, and a mass, M. The loop is originally held between the poles of a large magnet such that the entire loop is within the B-field except for the very bottom wore segment (Magfield2.jpg) When the loop is released, it begins to fall under the influence of gravity.

a) what is the magnitude and orientation of the current generated in the wire loop when it has non zero speed, V?

b) What is the magnitude and direction of the resulting magnetic force on the wire loop when it has non-zero speed, v?

c) use Newtons second law to describe the resulting acceleration of the wire loop

d) use your prior answers to describe the terminal speed in terms of the given quantities.

Homework Equations

ε=vlB
I=ε/R

The Attempt at a Solution

A) I=$\frac{ε}{R}$=$\frac{vlB}{R}$=$\frac{vcB}{R}$
I see that the magnetic flux is decreasing, but not sure how to figure out the direction.
B)So when it has nonzero speed this means it is dropping. Less and less of the wire loop is within the magnetic field.
Without knowing the direction of the current I cannot calculate the direction of force, however let's put that aside for a second.
I can still say:
F=ILxB=ILBsin(90)=ILB=$\frac{vcB}{R}$(cB)=$\frac{B^2vc^2}{R}$
The RHR will prescribe the direction of force once I know which direction the current is flowing
C)F=ma→a=$\frac{F}{m}$=$\frac{B^2vc^2}{Rm}$
D)Just solve the above for v?

 

[gneill]

Check out Lenz's law, Faraday's Law.

 

[cookiemnstr510510]

Check out Lenz's law, Faraday's Law.

Ahh okay, I haven't got there yet. Ill check it out

 

[rude man]

F=ILxB=ILBsin(90)=ILB=$\frac{vcB}{R}$(cB)=$\frac{B^2vc^2}{R}$

So far, so good. (Of course you need to find the direction of current).

The RHR will prescribe the direction of force once I know which direction the current is flowing
C)F=ma→a=$\frac{F}{m}$=$\frac{B^2vc^2}{Rm}$
D)Just solve the above for v?

Well, what is F? Do you see forces balancing so there even IS a terminal velocity?