Square copper wire loop within a magnetic field

AI Thread Summary
The discussion revolves around the behavior of a square copper wire loop in a magnetic field when released to fall under gravity. The generated current in the loop, when it has a non-zero speed, is calculated using the formula I = (vcB)/R, where v is the speed, c is the side length, B is the magnetic field strength, and R is the resistance. The magnetic force acting on the loop is expressed as F = (B^2vc^2)/R, and the resulting acceleration is derived from Newton's second law as a = F/m. The conversation also touches on the need to determine the direction of the current to fully understand the magnetic force's direction. Ultimately, the participants discuss the concept of terminal velocity and whether forces balance to establish it.
cookiemnstr510510
Messages
162
Reaction score
14

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?
 

Attachments

  • Magfield1.jpg
    Magfield1.jpg
    54.7 KB · Views: 374
  • Magfield2.jpg
    Magfield2.jpg
    44.3 KB · Views: 367
Physics news on Phys.org
Check out Lenz's law, Faraday's Law.
 
gneill said:
Check out Lenz's law, Faraday's Law.
Ahh okay, I haven't got there yet. Ill check it out
 
cookiemnstr510510 said:
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?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top