EMF when a loop is pulled away from a current-carrying wire

[guyvsdcsniper]

Homework Statement

On a horizontal table, there lies a very long straight wire (current I), and a square conduct-ing loop (side L). The loop is oriented at 45 degrees, with its closest corner a
distance L/2 from the wire. If the loop is pulleddirectly away from the wire with constant speed v,
what emf is generated?

Relevant Equations

Emf

Setting up the integral to find the flux that is giving me trouble.
I know that I will have to break up this integral into 2 parts, the first part account for when the start of the loop is increasing in area, and another right when I pass the halfway mark of the loop and the area begins to decrease again.

I have denoted the area as wds, where w is the the width and ds is it the infinitesimal distance from the wire.
I believe it is a correct approach to have w for the lower half of the loop to be $(s-\frac{L}{2})$
I was told by a friend I should multiply w by 2 to account for both halves of the loop.

Where I feel uneasy is my w for the area of the loop decreasing in size. I have it as $(s+\frac{L}{2})$. Does this choice make sense?
I also feel uneasy about picking my bounds. I know for sure that my first integral will start at L/2. Would the upper bound be $2\sqrt(L)$?
My friend told me the first upper limit should be $2\frac{\sqrt(L)}{2}$.

I would appreciate any help I can get with this problem.

 

[kuruman]

Doing two separate integrals is the clean approach. For the lower triangle, I would integrate with respect to variable $y$ which is measured from the lower corner to the strip of width $dy$ and length $w$. Then the area of the strip is $dA=2wdy$ and its distance from the wire is $s=L/2+y.$ Put the flux together and integrate. Note that $0\leq y \leq \frac{L}{\sqrt{2}}.$ Treat the top triangle in a similar manner starting from the diagonal of the square using appropriate variable $y'$ and limits of integration.

Your last equation for $\Phi$ in your notes is incorrect. There is no distance to the strip in the denominator of the integrand.

 

[guyvsdcsniper]

Doing two separate integrals is the clean approach. For the lower triangle, I would integrate with respect to variable $y$ which is measured from the lower corner to the strip of width $dy$ and length $w$. Then the area of the strip is $dA=2wdy$ and its distance from the wire is $s=L/2+y.$ Put the flux together and integrate. Note that $0\leq y \leq \frac{L}{\sqrt{2}}.$ Treat the top triangle in a similar manner starting from the diagonal of the square using appropriate variable $y'$ and limits of integration.

Your last equation for $\Phi$ in your notes is incorrect. There is no distance to the strip in the denominator of the integrand.

Ah I think your distance from the wiire $s=L/2+y.$ is a lot easier to visualize forme.

So if I integrate wrt y, then: $\vec B=\frac{μI}{2piy}$.
If I try to evaluate the lower corner my integral should look like this (ignoring constants for now):
$\int_0^\frac{L}{\sqrt(2)} 2(\frac{L}{2}+y)dy$ ?

 

[kuruman]

First off in LaTeX you enter pi as \pi to render it properly.

Secondly, the field due to an infinite wire at distance $s$ from the wire is $B=\dfrac{\mu_0I}{2\pi r}.$ Make a fresh drawing showing $y$ as previously defined. Look at the drawing. What is $r$ in terms of $y$ and $L$?

Thirdly, the element of flux that you have to integrate is $d\Phi=BdA$. Write separate expressions for $B$ and $dA$. Multiply them. Show all your work.

 

[guyvsdcsniper]

First off in LaTeX you enter pi as \pi to render it properly.

Secondly, the field due to an infinite wire at distance $s$ from the wire is $B=\dfrac{\mu_0I}{2\pi r}.$ Make a fresh drawing showing $y$ as previously defined. Look at the drawing. What is $r$ in terms of $y$ and $L$?

Thirdly, the element of flux that you have to integrate is $d\Phi=BdA$. Write separate expressions for $B$ and $dA$. Multiply them. Show all your work.

I understand what you are saying. I worked it out with a friend via your method and we got the right answer. Thank you. I will post my work when I get the chance.

 

[kuruman]

I understand what you are saying. I worked it out with a friend via your method and we got the right answer. Thank you. I will post my work when I get the chance.

If you got it right and are satisfied, there is no need for posting the solution unless you wish to practice your LaTeX skills. I asked you to post just in case you got stuck or made a mistake somewhere. Good job!