Magnetic Force Experienced by a Loop of Wire

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Homework Statement


The problem is shown in the link below:

http://postimage.org/image/do2b7xz0b/ [Broken]

Homework Equations



Biot-Savart Law's and dF=Idl x B

The Attempt at a Solution



My attempt is really just a mess. I essentially want to break it up into 2 segments. (1) Being the straight horizontal segment from -w to w and (2) the semi-oval segment. For the straight segment, I get dl x B to be ydx*z_hat and when I take the integrals in the equation dF=Idl x B, I get a Force of 2wIy*z_hat and since y=0 all along this point I get a Force of 0 due to the line segment at the origin. As for the oval segment I'm probably only correct as far as y=h-x^2 is what needs to be integrated. Any help would be appreciated!
 
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Answers and Replies

  • #2
rude man
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Well, you're right about dividing the problem into two sections, and you figured out the straight (base of the inverted parabola) right.

So the next step is to write the equation for the inverted parabola: y = y(x).

I also suggest "going advanced" and using bold lower-case letters for the unit vectors, thus i_hat = i etc. In fact, use bold for all vectors, e.g dF = i*ds x B etc.
 
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  • #3
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Yeah, isn't that what I have? y=h-x2
 
  • #4
rude man
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That ca'nt be the right equation, can it? What happened to w?
Let x = +w or -w, what do you get for y?
 
  • #5
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well in that case just y=h-w^2

but that doesn't make much sense to me, I was assuming something along the lines:

y=-a(w+x)(w-x) ??
 
  • #6
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Ouf, nevermind, x=w and y=h-w2 make sense to me now. I reviewed a little bit of my highschool quadratic forms! :P
 
  • #7
rude man
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Ouf, nevermind, x=w and y=h-w2 make sense to me now. I reviewed a little bit of my highschool quadratic forms! :P
Well, no, you need y as a function of x.

Hint: what's the standard formula for a parabola with the base at (0,0)?
 
  • #8
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y=x2
 
  • #9
rude man
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Make it y = ax2 and you got a deal.

Now, llok at that graph y = ax2. Picture what you have to do to that graph to make it look like your loop (minus the straight section).
 
  • #10
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Make it y = ax2 and you got a deal.

Now, llok at that graph y = ax2. Picture what you have to do to that graph to make it look like your loop (minus the straight section).
This is what led me to believe that it was y=h-x2. Do you want me to say y=-ax2+h ? or am I completely approaching this incorrectly?
 
  • #11
rude man
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You're getting warm. Of course, you realize that it can't be right:
plug x = +w or x = -w into it, and what do you get for y?

Look at your last formula: y = h - ax2. a is not part of your figure, so how about solving for a by forcing y = 0 when x = + or -w?

I'm wondering if nowadays they don't teach enough analytic geometry before embarking on calculus. In my day, analyt came first, and I'm talking ivy league college, not high school. You have to master analyt first!
 
  • #12
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hmm..

y=h(1-x2/w2) ??
 
  • #13
rude man
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Cigars!

Now, you will need to integrate the magnetic force along the entire parabola. You have already stated that dF = i*ds x B, which is right. You know i and B, what do you do about ds?
 
  • #14
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I'm not sure, I feel as if I need to express ds in such a way that it takes into account both the change in y and the change in x. Possibly a ratio?
 
  • #15
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I take the derivative of y with respect to x and voila?!
 
  • #16
rude man
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You're on the right track.

Draw an element of ds on your parabola. What would ds be in terms of x and y, and, remembering it's a vector, using unit vectors i and j?

Is ds different on the left vs. the right-hand part of the parabola?

BTW let's call the current I instead of i so as not to confuse it with the unit vector.

Also, note that I is defined couterclockwise.
 
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  • #17
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I wanna say di/dj = ds and so:

y=h-hx2/w2

di/dj = dy/dx = -2hx/w2
 
  • #18
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hmm actually,

ds = dxi + dyj

dy=-2hx/w2 * dx

dx/dy=w2/-2hx

dx= -w2/2hx *dy

therefore,

ds = (-w2/2hx*dy)i + (-2hx/w2*dx)j
 
  • #19
rude man
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hmm actually,

ds = dxi + dyj
Almost. Draw a small piece ds along the rigth-hand part of the parabola. Which way does it point? (Remeber the direction of I).

[/QUOTE]

I have to leave for a few hrs. Try to get ds for both sides, remembering the direction of I. Also realize that you will eventually need to do an integration, so you need ds x B to be a function of one variable only.
 
  • #20
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How about. ds x B = (wkydx - 2khx^2/w^2)k

should i sub in my equation for y in for y?
 
  • #21
rude man
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Start with ds, you need to get this first in terms of x and y, then in terms of x. Only then can you take the cross-product with B. B will also first have to be in terms of x before you can do the cross-multiplication to get dFand the eventual integration to get F.
 

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