Find the EMF

  • #1
A conducting rod moves with a constant velocity v in a direction perpendicular to a long, straight wire carrying a current I as shown in the figure below.

p31-58.gif


Find the EMF

It should equal U_o*V*L*I/2*pi*r, but how do I get there?
 

Answers and Replies

  • #2
fluidistic
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Hi and welcome to PF!
I'd use Faraday's law of induction.
 
  • #3
Alright, how does this look:

E = -d/dt*(Flux B)

E = -d/dt * BLR

E = dr/dt * -BL

E = -BLV

E = -u_o*I*V*L / 2*pi*r


Why does the negative go away for the correct answer?
 
  • #4
Also, did I skip any steps????
 
  • #5
fluidistic
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I'm not an expert, but just a student like you. What you did looks right to me. I'll wait someone more experienced to help us. I think that the negative sign shows that the emf opposes to the direction of the current I in the wire.
 
  • #6
Well fluid, I appreciate you looking it over! I know for a fact this will be on my final tomorrow and I'm currently studying. This problem seems easy enough using Faraday's law but I hope they way I did it was correct. I wish I was just better at physics than I am. :cry:
 
  • #7
fluidistic
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Well fluid, I appreciate you looking it over! I know for a fact this will be on my final tomorrow and I'm currently studying. This problem seems easy enough using Faraday's law but I hope they way I did it was correct. I wish I was just better at physics than I am. :cry:
I wish you the best of the luck for your exam. I'm taking the final exam on Friday :biggrin:.
I'd appreciate if someone could tell us if you did it properly and why is there this minus sign...
Let's wait.
 
  • #8
Can any experts confirm my answer as correct?
 
  • #9
fluidistic
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When I was in my bed at 3:08 am, I realized that [tex]\frac{d \Phi _B}{dt}[/tex] was negative because the rod is going out from the wire. Hence [tex]-\frac{d \Phi _B}{dt}[/tex] must be positive as the result. I'd love someone to confirm me on this.
 
  • #10
You are correct, fluid. By Lenz's Law, the current would be pointing down and thus negative due to the fact the conducting bar is moving away from the wire. If the opposite were the case, the final result would be positive.

As for my final, the teacher didn't put anything like this on it. :( sigh.... hope I did alright enough to get a C+ or a B- for the class.

Thanks again for your help fluid, hope you do well on your final on friday!
 
  • #11
fluidistic
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112
You are correct, fluid. By Lenz's Law, the current would be pointing down and thus negative due to the fact the conducting bar is moving away from the wire. If the opposite were the case, the final result would be positive.

As for my final, the teacher didn't put anything like this on it. :( sigh.... hope I did alright enough to get a C+ or a B- for the class.

Thanks again for your help fluid, hope you do well on your final on friday!
Ok.
Thank you very much. This is a hard course indeed. Good luck with your result, hopefully you'll earn a better than expected grade.
 

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