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Homework Help: Magnetic Field Problem

  1. May 7, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the magnitude and direction of magnetic field at point P shown in the figure attached if i = 10 A and a = 8.0 cm? Express the answer in vector notation.

    2. Relevant equations

    [tex]B = \frac{\mu_0I}{2\pi d}[/tex]

    3. The attempt at a solution

    I'm really not sure how to approach this one... I considered using Biot-Savart for each segment of wire... but the problem is the distance r varies... furthermore, keeping the vectors separate will be pretty hairy.

    Am I approaching this problem wrong? Or is it just a ludicrous amount of work?

    Attached Files:

  2. jcsd
  3. May 7, 2010 #2


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    The equation that you quoted applies only to a very long (infinite) wire. You have to use Biot-Savart, there is no other way. The four contributions all point in the same direction so the net field is the sum of the four - no need to worry about components here.

    The amount of work is not ludicrous if you are systematic about it. Use Biot-Savart to calculate the magnetic field due to a wire segment of length a at arbitrary point (x, y).

    For the first segment plug in x = a/4, y = 3a/4. Now imagine the loop rotated 90o clockwise. The second segment then contributes a field with x = 3a/4, y = 3a/4. Rotate two more times to get the third and fourth contributions. Add the four contributions and you are done. :wink:
  4. May 7, 2010 #3
    I not sure what value to use for r in Biot-savart. Could I use a double integral and integrate over r as well considering that it is a changing value?
  5. May 7, 2010 #4


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    There is only one variable to integrate over. No double integral is required. Draw yourself a drawing. Put the origin at the leftmost end of the segment. Point P is at (x, y). Consider a current-carrying element ds at distance s from the origin. What is the distance from ds to P? That is your r and the only variable to integrate over is s because x and y are fixed.
  6. May 8, 2010 #5
    If I'm understanding correctly,

    [tex]B(\hat r) = \int \frac{\mu_0 I}{4 \pi} \frac{dl \times \hat r}{|r|^2}[/tex]

    dl = ds
    [tex] \hat r = <x,y> [/tex]


    [tex]B(<x,y>) = \frac{\mu_0 I}{4 \pi} \int \frac{ds \times \hat r}{|<x,y>|^2}[/tex]

    [tex]ds \times \hat r = ds[/tex]

    Integrate from 0 to a,

    [tex]B(x,y) = \frac{\mu_0 I}{4 \pi} \int^{a}_{0} \frac{ds}{|<x,y>|^2}[/tex]

    [tex]= \frac{\mu_0 I}{4 \pi} \frac{a}{|<x,y>|^2}[/tex]

    Plug in x, y and thus the contribution to B from the first line segment? Rinse and repeat for each other segment?
  7. May 8, 2010 #6


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    This is not correct. The unit vector is in the direction of r which is a vector from ds to point P. As ds moves along the line segment this unit vector changes direction. First you need to express r in terms of x, y and s. Then you need to find unit vector r which is vector r divided by its magnitude.
  8. May 8, 2010 #7
    Does [tex]r = \sqrt{s^2 + \frac{a^2}{16}}[/tex]?


    Correction: [tex]r = \sqrt{s^2 + \frac{x^2}{16}}[/tex]
  9. May 8, 2010 #8
    Your first equation is correct, the correction is wrong.

    This in only one of four equations for r. Each wire segment will be done seperately. You need to add an origin to your diagram and draw in the R vector to prevent confusion.

    Also, express x and y as a fiction of s (parametric integral).
  10. May 8, 2010 #9
    So, generically speaking,

    [tex]r = \sqrt{s^2 + <x,y>^2}[/tex]

    where <x,y> will change depending on which segment I'm considering.

    I could integrate,

    [tex]B(<x,y>) = \frac{\mu_0 I}{4 \pi} \int^{a}_{0} \frac{ds}{|r|^2}[/tex]

    using the generic value for r which will produce an equation that gives B as a function of <x,y>.

    Because each segment has the same current and length, and only the position of P relative to the segment changes (in terms of <x,y>), carefully plugging in values for <x,y> will give the the B field at P as a result of each segment. Correct?
  11. May 8, 2010 #10


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    Actually, neither equation is correct. Look at the drawing. Vector r has magnitude
    [tex]r=\left[(s-x)^2+y^2 \right]^{1/2}[/tex]

    No reason for that. Please read my initial posting to this thread.

    This is incorrect because x and y are fixed. They represent the coordinates of point P. They cnnot depend on where one places element ds.

    Attached Files:

    Last edited: May 8, 2010
  12. May 8, 2010 #11
    Almost. S is your variable of integration and P is fixed. You need to express x(s) and y(s) under the root (with the coordinates of P fixed). There will be a different equation for each line segment.

    Edit: it's correct. I just saw your new drawing.
  13. May 8, 2010 #12
    Taking into account that my value for r is incorrect, is this the correct approach?

    If I plugged in the correct value for r, this would work?

    Thanks for all the assistance so far!
  14. May 8, 2010 #13


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    Follow through my postings, look at my drawing and you will see your way through.
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