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Current Carrying Wire in Magnetic Field

  • Thread starter DeChance
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  • #1
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Homework Statement


Find the total magnetic force acting on the loop
http://tinypic.com/r/2yvvyuw/5

Homework Equations


f[itex]_{m}[/itex] = IL X B
B =[itex]\frac{μ_{0}*I}{2∏*R}[/itex]


The Attempt at a Solution


Magnetic Field from Wire at Top
B = [itex]\frac{4∏*10^{-7}*10}{2∏*2*10^{-2}}[/itex]
B = .0001

Magnetic Field from Wire at Bottom
B = [itex]\frac{4∏*10^{-7}*10}{2∏*5*10^{-2}}[/itex]
B = .00004

Force on top of wire
F[itex]_{m}[/itex] = 8*5*10[itex]^{-2}[/itex]*.0001
F[itex]_{m}[/itex] = .00004

Force on bottom of wire
F[itex]_{m}[/itex] = 8*5*10[itex]^{-2}[/itex]*.00004
F[itex]_{m}[/itex] = .0000016


Hello everyone! Heard about you awesome guys from a classmate, kinda late in the semester but better late than never :tongue:. At this point I'm not sure what to do. I assume the two sides that are parallel to each other will cancel each other out at this point, but I'm not sure. Also I wanted to be sure the formulas I used were used properly above!
 

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Answers and Replies

  • #2
haruspex
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Your equations look right.
I assume the two sides that are parallel to each other will cancel each other out
If you mean the two of length 3cm, yes. (Which way will the forces act on them?)
You are asked for the net force on the loop, but you have only posted the forces on the upper and lower parts of the loop separately. What is the net force, and which way will it act?
 

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