Current Through Resistor in Rectangular Loop

AI Thread Summary
The discussion revolves around calculating the current through resistor R2 in a rectangular loop where R1 is moving with velocity v. The magnetic flux is derived from the changing area as R1 moves, but there is confusion regarding the motion of R2 and its impact on the flux calculation. The initial assumption that R2 is also moving is clarified, indicating that it does not contribute to the change in length. The correct approach involves recognizing that the area change is due to R1's motion alone, leading to a reevaluation of the flux formula. Ultimately, the key point is that the current in the loop is not equal to the current in the upper wire, necessitating a careful analysis of the system dynamics.
malindenmoyer
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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
 
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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
 
Thanks for the reply, that is what I did and here is the solution I got:

d\phi =BdA=\frac{\mu_0 i}{2\pi r}ldr

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

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

\frac{d\phi}{dt}=\frac{\mu_0 iv}{2\pi}\ln{\frac{b}{a}}

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

R_2=\frac{\mu_0 v}{2\pi}\ln{\frac{b}{a}}-R_1

Does this look correct?
 
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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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.
 
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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