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Current Carry Wire With A Loop

  • Thread starter Charanjit
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  • #1
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1. Homework Statement :
A. Determine the magnetic flux through a square loop of side a (see the figure) if one side is parallel to, and a distance b from, a straight wire that carries a current I.

Express your answer in terms of the variables I, a, b, and appropriate constants.

Answer: [(uIa)/(2pi)]ln((b+a)/b))

B. If the loop is pulled away from the wire at speed v, what emf is induced in it?
Express your answer in terms of the variables I, a, b, v, and appropriate constants.

Answer: Help




2. Homework Equations :
Flux=integral(B*da)
E=BLv


I think those are the two equations we need, but not 100% sure.

3. The Attempt at a Solution :
I have already figured out part A, and the solution is posted above, just having issue with part B.

So all I did was replaced L=length with "a", one side of the area. I am totally lost here. So please help me.
 

Answers and Replies

  • #2
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Have you tried using Faraday's law?
 
  • #3
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Well yes, but how to relate velocity?
 
  • #4
collinsmark
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Well yes, but how to relate velocity?
The rate of change per unit time of [tex] \Phi_B [/tex] is a function of the loop's velocity.

Hint: Distance = Velocity times Time. Distance here, as your problem describes it, is your variable 'b'. Next find the negative of the time derivative of [tex] \Phi_B [/tex].
 
  • #5
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Which b are you talking about? The equation E=BLv? That "B" is the magnetic field.
 
  • #6
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Which b are you talking about? The equation E=BLv? That "B" is the magnetic field.
In the problem statement, "...and a distance b from ..."
 
  • #7
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Ok... so how do I import that into the equation? We don't have the "time" variable.
 
  • #8
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Ok... so how do I import that into the equation? We don't have the "time" variable.
Distance is equal to Time times Velocity, or b = vt.

Make the substitution as an interim step. You can always make the reverse substitution (t = b/v) later to get rid of the 't' variable if you need to, once you have your final answer.
 
  • #9
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Ok, but in which equation? E=BLv?
 
  • #10
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Ok, but in which equation? E=BLv?
try the answer to the first part:
Answer: [(uIa)/(2pi)]ln((b+a)/b))
After you make the substitution, you'll still need to use Faraday's law as a step to the final answer.
 
  • #11
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Ahh ok. I got it. Thanks.
 

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