Calculating Induced EMF in a Moving Square Loop

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Homework Help Overview

The problem involves calculating the induced electromotive force (emf) in a square loop of wire moving with a constant velocity near a long straight wire carrying a steady current. The original poster expresses difficulty in treating the non-uniform magnetic field due to the wire and incorporating the loop's velocity into the calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration of the magnetic field expression to find magnetic flux and the challenges of incorporating velocity into the calculations. There are attempts to substitute variables and differentiate to find induced emf, with some questioning the correctness of the derivative taken.

Discussion Status

The discussion is ongoing, with participants providing suggestions for refining calculations and clarifying steps. There is a recognition of the complexity of the problem, and some participants are exploring different interpretations of the integration and differentiation process.

Contextual Notes

Participants note that the original poster's calculations did not align with the book's answer, indicating potential misunderstandings in the treatment of variables and derivatives. The discussion reflects the constraints of homework rules that may limit the depth of guidance provided.

discoverer02
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One more problem that's causing me grief:

A square loop of wire, b meters on a side, moves with constant velocity v (m/sec) toward the right in the plane of a long straight wire carrying a steady current I amperes. Calculate the emf induced in the loop when the side of the loop nearest the wire is at distance x meters from the wire.

The treatment of the non-uniform magnetic field with the velocity is what troubles me here.

I know that the magnetic field coming from the wire = uI/(2pir) and that if I integrate [uI/(2pir)]b](dr) from x to x + b then I get the magnetic flux for the stationary loop but this, of course, induces no current. I'm having trouble seeing how to incorporate the velocity into the non-uniform magnetic field.

I tried to substitute r = vt into the equation, and then take the derivative of the magnetic flux with respect to time to get the induced emf, but that didn't agree with the answer in the book..

Any suggestions are be greatly appreciated.

Thanks.
 
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Originally posted by discoverer02

The treatment of the non-uniform magnetic field with the velocity is what troubles me here.
yes, as i recall, this calculation is a little messy

I know that the magnetic field coming from the wire = uI/(2pir) and that if I integrate [uI/(2pir)]b](dr) from x to x + b then I get the magnetic flux for the stationary loop but this, of course, induces no current. I'm having trouble seeing how to incorporate the velocity into the non-uniform magnetic field.

I tried to substitute r = vt into the equation
i think you should substitute x=vt, after you have finished integrating. remember, r is just a dummy variable of integration, so it shouldn t really appear in your final answer

, and then take the derivative of the magnetic flux with respect to time to get the induced emf, but that didn't agree with the answer in the book..
that is essentially correct. if you show your calculation in a little more detail, i can be a little more specific in where you went wrong.
 
The magnetic flux for the loop with no velocity is:

[buI/(2pi)]ln[(x+b)/x]

If I plug x = vt here and take the derivative with respect to time, I end up with:

emf = [buI/(2pi)][vt/(vt+b)? This doesn't agree with the answer in the book.
 
Originally posted by discoverer02
The magnetic flux for the loop with no velocity is:

[buI/(2pi)]ln[(x+b)/x]

If I plug x = vt here and take the derivative with respect to time, I end up with:

emf = [buI/(2pi)][vt/(vt+b)? This doesn't agree with the answer in the book.

i don t think you took the derivative correctly. remember the t appears twice, once in the numerator, and once in the denominator, and both are inside a logarithm.
 
You're right. What was I thinking? I'll try it again and see what comes out.

Thanks.
 

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