Calculating Induced EMF in a Moving Square Loop

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A square loop of wire moving towards a long straight wire carrying a steady current is analyzed for induced EMF. The magnetic field from the wire is given by the formula μI/(2πr), and the challenge lies in incorporating the loop's velocity into this non-uniform magnetic field. The magnetic flux for a stationary loop is calculated, but this does not induce current. Substituting x = vt after integration and taking the time derivative of the magnetic flux is suggested, but discrepancies with the book's answer arise. The discussion emphasizes careful handling of derivatives, particularly with the logarithmic function involved.
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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