Force on Current Loop

  • Thread starter james11223
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  • #1
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Homework Statement



Here is the problem and the picture that goes along with it. I am really stuck because our teacher never explained a problem like this to us before. Any help is appreciated!! THank you so much

http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/229022_113377112081455_100002275016913_125327_6984938_n.jpg [Broken]


Homework Equations

 
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Answers and Replies

  • #3
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Here are the equations that I was given

http://a2.sphotos.ak.fbcdn.net/hphotos-ak-snc6/223417_113388105413689_100002275016913_125390_5418139_n.jpg [Broken]
 
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  • #4
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Ok, in your formulas, use

[tex]
F = BI l \sin\theta
[/tex]

[tex]
B_{wire}=\frac{\mu_0I}{2\pi r}
[/tex]
to find out the force per unit length b/w two infinite parallel wires carrying currents [tex]I_1, I_2[/tex]. Can you do that?
 
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  • #5
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THank you so much!!
 
  • #6
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Sorry I just realized... irdg what you meant
@praharmitra...

sorry can you explain a little bit more
 
  • #7
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Ok, what I want you to do is find the force per unit length b/w a pair of infinite wires separated at a distance r. Here's what you do.

You have one wire carrying current [itex]I_1[/itex]. What magnetic field does it produce at a distance r from itself?

Now the second wire carrying current [itex]I_2[/itex], is in the presence of the magnetic field you calculated above. Therefore, what is the force per unit length on this wire?

That is your answer for the force b/w two wires.
 
  • #8
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Would you do it like this:??

Fnet = Fnear - Ffar = (µoI1I2)(ℓnear)/(2πdnear) - (µoI1I2)(ℓfar)/(2πdfar) = (µoI1I2ℓ)/(2π) * (1/dnear - 1/dfar) = [(4π x 10-7T·m/A) (2A)(3A)(6m)]/(2π) * [1/1 - 1/3] = 4.74 x 10-5N away from the wire


doest that make sense? llol

Thank you so much for all the help guys!
 
  • #9
311
1
yes, that is correct yes.
 

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