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Homework Help: Current Through Resistor in Rectangular Loop

  1. Apr 7, 2010 #1
    Here is the problem and the equations I feel are relevant. The instructor was not clear on whether or not R2 was moving with velocity v, but I assume so because something must be moving (I think). Can somebody give me some instruction on where to go from what I have?

    [PLAIN]http://people.tamu.edu/~malindenmoyer/tamu/~s2010/phys208/lab/lab04/phys208_problem.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 7, 2010 #2

    ehild

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    R1 is moving with speed v to the right. So the enclosed area increases by the rate dA/dt=l*v. The magnetic field depends on the distance from the upper wire, so you have to integrate to get the flux.

    ehild
     
  4. Apr 7, 2010 #3
    Thanks for the reply, that is what I did and here is the solution I got:

    [tex]d\phi =BdA=\frac{\mu_0 i}{2\pi r}ldr[/tex]

    [tex]\phi =\frac{\mu_0 il}{2\pi}\int_a^b \frac{dr}{r}=\frac{\mu_0 il}{2\pi}\log{\frac{b}{a}}[/tex]

    [tex]\frac{d\phi}{dt}=\frac{dl}{dt}\cdot \frac{\mu_0 i}{2\pi}\ln{\frac{b}{a}}\rightarrow \frac{dl}{dt}=v[/tex]

    [tex]\frac{d\phi}{dt}=\frac{\mu_0 iv}{2\pi}\ln{\frac{b}{a}}[/tex]

    [tex]\frac{d\phi}{dt}=\xi =iR_{eq}=i(R_1+R_2)=\frac{\mu_0 iv}{2\pi}\ln{\frac{b}{a}}[/tex]

    [tex]R_2=\frac{\mu_0 v}{2\pi}\ln{\frac{b}{a}}-R_1[/tex]

    Does this look correct?
     
  5. Apr 8, 2010 #4

    ehild

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    It is not quite correct. The current in the loop is not the same as the current in the upper wire. In the figure, the horizontal side of the rectangle should be l as I show in your figure.

    What is the question in the problem?

    ehild
     

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  6. Apr 8, 2010 #5
    I'm sorry I didn't make it clear, but R2 is sliding with velocity V to the right. l is the length of the sliding rod. The problem was: calculate the current in R2.
     
  7. Apr 8, 2010 #6

    ehild

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    I see. Then your formula for the flux is wrong. dA is not ldr, as they are parallel. If R2 moves to the righ nothing happens with the length of the rod, so dl/dt =0

    ehild
     
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