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Homework Help: Finding the magnetic field

  1. Oct 28, 2013 #1
    1. Problem statement.
    6h0u8m.jpg

    2. Known equations
    Biot-savart law

    3. Attempt
    I believe since I was given a current value already, I can just use the equation

    B = IA

    with A = 2l + 2w, but I'm having difficulty understanding the "10cm from the center" part, I'm just thinking to pick a spot 10cm from the center because the problem is not telling me any extra directions, if it does need to be at a certain spot, how does that change my answer??
     
  2. jcsd
  3. Oct 29, 2013 #2

    Zondrina

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  4. Oct 29, 2013 #3

    rude man

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    Yes, use the B-S law. Pick any point 10 cm away from the center along a line parallel to two of the sides. Remember, it's a square loop. I'd put the coordinate origin at that point.

    Then you have to integrate dB = (μ0i/4π) dl x r/r3 for the 4 segments. Take advantage of various symmetries to simplify your calculations (you need only 3 integrations as I see it).
     
    Last edited: Oct 29, 2013
  5. Oct 29, 2013 #4
    Ok I am kind of seeing what you mean by 3 integrals, so using the diagram Zondrina created and also labeling the middle as the coordinate origin, from that I can see 3 integrals, the integral from the bottom wire, the top wire and the side wires are symmetrical so they count as one.

    This is my attempt for just the integral from the bottom wire. I want to make sure that is correct first before I continue,

    Basically I let dl = dx, I see that for the r-hat part, only the sin part matters since the cos's cancel each other out, set the bounds from -.25m to 25m and y is a constant at .60m and then solve the integral.

    of1onn.jpg
     
  6. Oct 29, 2013 #5

    rude man

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    You took quite a different approach from what I had in mind, but that's not saying it couldn't be done that way. I put the origin at the "fetus" and did not use an angle in the math.

    But your statement of the B/S law is in error. If your Ř is a unit vector (no dimension) as I suspect, the denominator needs to be x2 + y2. If R is a position vector (dimension = L) then the denominator needs to be (x2 + y2)3/2.

    I would also leave the i in the coefficient even though i = 1A. This is so you can do dimensional checks later on - a very powerful checking tool.

    Your math seems mysterious and unfinished. You can't have any y's in your answer.
     
  7. Oct 29, 2013 #6
    Hmm I see what I did wrong, I modified it and got a different integral. Once I substitution in all the values and such, is the method I am using would give me the correct answer? I am thinking I just repeat the integral for each of the wire and sum them all up would give me the magnetic field at that point?

    But the way u are suggesting to do it, with setting the origin at the fetus, I can't see how you don't need an angle to complete the problem..

    Bounds are from -.25m to .25m, y is .35m

    2znwxmc.jpg
     
  8. Oct 29, 2013 #7
    No Calculus is needed for this problem.
    Since the point where you are calculating the magnetic field is at equal distances from the two ends of each wire, the magnetic field from each wire at that point must be perpendicular to the wire. This means you can use the equation for the magnetic field of an infinite wire, although these wires are finite.

    Since the point is 0.01m from the plane and 0.025m from each edge, the distance to each wire is [itex]r=\sqrt{0.01^{2}+0.025^{2}}=0.0269m[/itex].

    The field from each wire is simply [itex]\textbf{B}_{w}= \frac{\mu_{o} I}{2 \pi r}[/itex]

    The field from each wire makes an angle with the vertical axis of [itex]\Theta = arctan \left(\frac{0.01}{0.025}\right)[/itex]

    The horizontal components of the field will cancel but the vertical components will add.

    The vertical component of each field is [itex]\textbf{B}_{w} cos \left(\Theta \right)[/itex]

    The total field is then [itex]\textbf{B}=4 \left( \textbf{B}_{w} cos \left(\Theta \right) \right)[/itex]
     
  9. Oct 29, 2013 #8
    I am still having trouble how the point is equidistant from the wires, since the point we are looking at is above the center, wouldnt it be closer to one side than the other?
     
  10. Oct 29, 2013 #9
    I believe the problem wants us to assume the point is 10cm away from the plane of the blanket, not 10cm from the center within the plane of the blanket. If the latter were the case we would need to know more information about where the point is, since there is an entire circle of points 10cm from the center within the plane of the blanket and the field would be different at different points on such a circle. Plus the problem mentions a fetus, I assume if the expectant mother is using the blanket, the fetus would be something like 10cm below the center of the blanket.
     
    Last edited: Oct 29, 2013
  11. Oct 29, 2013 #10

    mfb

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    I interpret the problem statement in the same way.

    I'm not sure how you plan to avoid calculus, however, unless you use the result of (non-trivial) calculus in some formula.
    " the distance to each wire" that is the distance to the closest point. And your formula does not give the correct scaling for large distances, so something is wrong.
     
  12. Oct 29, 2013 #11
    As a more rigorous explanation, we first use symmetry to realize the field must be perpendicular to the wire at any point along a line drawn through the center of the wire and perpendicular to it.

    We now observe the dot product in Amperes Law now simplifies, since B is parallel to dl.

    [itex]\oint \textbf{B} \cdot dl = \textbf{B}2 \pi r = \mu_{o} I [/itex]
     
    Last edited: Oct 30, 2013
  13. Oct 30, 2013 #12

    mfb

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    Amperes Law requires an infinite length of the wire. We do not have an infinite length.
     
  14. Oct 30, 2013 #13
    So I used your method but just adjusted the values, you were just off by a factor of 10, you made an error in your conversion. But my total magnetic field I came up with is

    2.759 x 10^-7 T

    Is this the correct total?
     
  15. Oct 30, 2013 #14
    Amperes Law is always valid, it is the integral form of one of Maxwell's equations. The trick is whether or not the integral can be evaluated. If the direction of B is known along a closed path around the wire, then the integral can be evaluated. This usually only occurs for problems with unique symmetry were B is parallel to dl. This problem along with the infinite wire problem has such symmetry.
     
  16. Oct 30, 2013 #15
    Whoops! Sorry about that...

    With the correct lengths of 0.1m and 0.25m I get a field of 2.756x10-6 T.
     
  17. Oct 30, 2013 #16

    mfb

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    I'm not really convinced. Replace the square with an octagon, do you get (roughly!) twice the magnetic field? Certainly not.

    You are considering Ampere's law for a single wire section, but this setup violates charge conservation and cannot satisfy the Maxwell equations without a return wire somewhere - and this return wire does influence the field.
     
  18. Oct 30, 2013 #17
    I am still confused, is the solution suggested by NFuller correct?
     
  19. Oct 30, 2013 #18

    rude man

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    EDIT: I wonder, can superposition be argued? Take 1 segment at a time and the B field magnitude is indeed B = μ0 i/2πR around the integration path if the center of the wire is picked for the integration path, which it is. Multiply by 4.

    Anyway, this should be easily resolvable once we perform the integration.
     
    Last edited: Oct 30, 2013
  20. Oct 30, 2013 #19

    rude man

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    The jury is out! We should attack the problem both ways & compare results. I personally tend to favor NFuller, not necessarily from his argumentative point but from a superposition argument.
     
    Last edited: Oct 30, 2013
  21. Oct 30, 2013 #20

    rude man

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    Why? It's sitting 10 cm from the center of the coil, above the coil, so one side looks just like any other from that point.
     
    Last edited: Oct 30, 2013
  22. Oct 30, 2013 #21
    Ok so you agree with NFuller's method? but I don't understand why a pi is involved, the object is a square so it does not need to involve a pi right?

    From what I see it right now, I think it would just be

    B = (μ0) i / 4 (pi) ∫ dx * sinθ / r^2

    convert the sin in terms of x and y, and then integrate from -.25m to .25m with y = .10m
     
  23. Oct 30, 2013 #22

    rude man

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    @ NFuller and mfb, are you duking this issue out via personal messages? If so, too bad, I would love to see this issue resolved, and I imagine so would the OP.

    Speaking of octagons, a circle is an "infinit-agon". For a circular current loop of equal perimeter 2πR = 2m with 1A of current, the axial field at d = 10 cm from the loop's center is by Biot-Savart
    B = μ0iR2/2(R2 + d2))3/2

    2πR = 2m → R = 0.318m, R2 = 0.1m2, d2 = 0.01m2 so

    B = (1.26e-6 1 0.1 )/2(0.1 + 0.01)3/2
    = 1.73e-6 T
    (compared to 2.76e-6 T computed for the square loop by NField's method. Suspiciously higher?).

    Now, consider a polygon with 100 sides, thus approximating a circle. Each segment if isolated from the rest produces a B field given by Ampere's law (mid-point of each segment assumed). Then if we apply superposition, the B field should be 100 times the 1-segment value! Which I guess is what mfb was getting at with his octagon.

    I at least am faced with a paradox. Is superposition invalid? Why? It's a linear system ... or is the concept of an isolated current-carrying segment simply impermissible?

    How about others joining in? TSny, vela et al?
     
  24. Oct 30, 2013 #23

    mfb

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    No PMs.
    Superposition is valid, but the formula for the individual segments is not right - this would need lines of infinite length.

    I get 1.877 µT with a formula for line segments instead of a manual integration.

    Well, the conclusion is independent of those small prefactors.

    Right, as that conclusion (B-field increases without limit for more edges) cannot be true.
     
  25. Oct 30, 2013 #24

    rude man

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    I will work the integration presently. So far I've been thinking about Fuller vs. mfb. Stand by and don't form any conclusions yet.
     
  26. Oct 30, 2013 #25

    rude man

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    It certainly looks like you have to be right, but I still don't know why Ampere's law doesn't work. An isolated straight segment of wire carrying current i, with a circular path taken about its mid-point: ampere's law is just ∫H dl around the mid-point = i. H must be symmetrical around that circle - how could it not be? So H must = i/2πR. On the other hand, this says the length of the segment is immaterial, which is I think your point is that that can't be true. However, note that the H field is calculable by ampere's law only at the mid-point of the finite segment whereas the infinitely long wire produces the same H field everywhere along the wire. This could resolve the paradox of a isolated segent vs. an infinite wire.

    By Biot-Savart, I also see that an infinite length of wire has to produce an H field greater than that of a finite length, even at the mid-point. The remaining lengths must contribute to H, albeit in diminishing strength, but again the wire is of infinite length. All this can be shown by Biot-Savart.

    I am going to go with Biot-Savart and resolve the paradox by simply stating that an isolated, current-carrying wire cannot exist! I will give my calculation of the OP's problem presently.
     
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