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Magnetic field from a wire at a point

  1. Jun 20, 2017 #1
    1. The problem statement, all variables and given/known data

    Picture of the problem:
    biot%20savart_zpsfbh66mhc.png

    I first need to calculate the magnetic field in point P caused by the piece of wire at z = R (so the top wire parallel to the y-axis).



    2. Relevant equations

    Biot-Savart law for magnetic fields:

    ##\vec {dB} = \frac {μ0*I} {4*π} * \frac {\vec {dl} ×\vec {r}} {r^3}##


    3. The attempt at a solution

    Here is what I did:

    ##\vec {dl} = dy * \vec {ey}## (where ey is the unit vector in the y direction)
    ##\vec r = (P - y)* \vec {ey} - R * \vec {ez}##

    then:

    ##\vec {dl} ×\vec r = -R*dy*\vec {ex}##

    so now I can write:

    ##\vec {dB} = - \frac {μ0*I*R} {4*π} *\frac {1} {(P - y)^2 + R^2)^{3/2}} * dy * \vec {ex}##

    Now I can integrate this using the substitution ##u = P-y## and ##du = -dy##, and then integrate from ##-L## to ##0##, which gives me an expression for the magnetic field:

    ##\vec B =\frac {μ0*I} {4*π*R} *(\frac {P} {\sqrt {P^2+R^2}} -\frac {P+L} {\sqrt {(P+L)^2+R^2}}) * \vec {ex}##

    Now this answer is almost correct except there should be a minus in front. However my minus disappeared when I used the substitution ##du = -dy##. Does anyone know where I went wrong?
     
    Last edited: Jun 20, 2017
  2. jcsd
  3. Jun 20, 2017 #2

    kuruman

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    Check the details of the integration. You probably used a tangent substitution. How is the positive angle defined?
     
  4. Jun 20, 2017 #3
    Integration:

    ##\vec B =- \frac {μ0*I*R} {4*π} * \int \frac {1} {(P - y)^2 + R^2)^{3/2}} * \vec {ex} \, dy##

    ##\vec B = - \frac {μ0*I*R} {4*π} * \int \frac {1} {(u)^2 + R^2)^{3/2}} * \vec {ex} \,* -du##

    I am allowed to use the solution to some standard integrals, like this one, so I can just write out the solution directly:

    ##\vec B = - \frac {μ0*I*R} {4*π} * [\frac {-1} {R^2} * \frac {u} {\sqrt {u^2+R^2}}] \vec {ex}##

    Then I substitute ##u = P-y## back in and evaluate the indefinite integral from ##-L## to ##0## which gets me to my final answer mentioned above:

    ##\vec B =\frac {μ0*I} {4*π*R} *(\frac {P} {\sqrt {P^2+R^2}} -\frac {P+L} {\sqrt {(P+L)^2+R^2}}) * \vec {ex}##

    This is what I did, as you can see my minus sign cancelled out. I am not sure what you mean by the positive angle.
     
  5. Jun 20, 2017 #4

    kuruman

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    Why is the z-component of ##\vec{r}## negative?
     
  6. Jun 20, 2017 #5
    ##\vec r## should point from the line segment ##\vec {dl}## to point ##P## right? So then it would have a negative z-component since it should point from ##z=R## to ##z=0##
     
  7. Jun 20, 2017 #6

    kuruman

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    Yes, you're right. I think you put one minus sign too many with dy and the limits of integration. If you use the the right hand rule and the diagram below, you get
    ##\vec {dB} = - \frac {μ_0*I*R} {4*π} *\frac {1} {[(P + y)^2 + R^2]^{3/2}} * dy * \hat {x}##
    Note the positive sign in the denominator. Also note that the limits of integration should be from zero to L, not negative L because the smallest value for the y-component of ##\vec{r}## is P and the largest value is P + L.
    Wire Segment.png
     
  8. Jun 20, 2017 #7
    But shouldn't ##r = \sqrt {(P-y)^2+P}##? Since y is always negative in this case so the y component of ##\vec r = P-y##
     
  9. Jun 20, 2017 #8

    kuruman

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    I think your problem is with the back substitution from ##u## to ##y##.
    You have the indefinite integral (after cancellation of the two negative signs)
    ##\vec B = \frac {μ_0I} {4πR} \int \frac {du} {(u^2+R^2)^{3/2}} \hat {x} = \frac{u}{R^2\sqrt{R^2+u^2}}##
    Since ##u=P-y##, when ##y=0##, ##u = P## and when ##y=-L##, ##u=P+L##. These are the lower and upper limits that you should put back in the indefinite integral for ##u##.

    Edited to clarify points and fix mistakes.
     
  10. Jun 20, 2017 #9
    Yeah I think I have figured where I am going wrong, though I don't understand why yet. I did perform the back substitution of the limits as you mention in your post, however I am integrating from -L to 0, whilst you are integrating from 0 to -L. I thought you always have to integrate in the direction of the unit vector? Or am I misunderstanding this?
     
  11. Jun 20, 2017 #10

    kuruman

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    Not necessarily. It is element dy that is always assumed to be positive. The limits of integration add a negative sign or not.
     
  12. Jun 20, 2017 #11
    mmm okay, so how do I know what to use as integration limits then? In this case dy points in the same direction as the current, so I figured I should just integrate in that direction too, since its also the direction of the unit vector. How would I know to integrate in the other direction in this case?

    Thank you for your help by the way
     
  13. Jun 20, 2017 #12

    kuruman

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    Ignore the unit vector and its direction. The key things are the integrand, f(q) dq and the limits of integration. Note that I used q, a dummy variable of integration that eventually disappears when you evaluate the integral. You should always bear in mind that when you integrate, the "something" in d(something} can be any symbol. I think your confusion arises because you are treating y as both a dummy variable of integration and a Cartesian coordinate. In post #6 I have y increasing to the left. Strictly speaking, I should have called it something else, say q, because y should be reserved for the Cartesian coordinate that increases to the right.

    Look at the drawing you made when you worked on this problem. Call the distance along the y-axis of element ##d \vec{l}## q, set up the definite integral and see where it takes you.
     
  14. Jun 20, 2017 #13

    TSny

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    Your answer looks correct to me. Note that your answer gives a negative number times the unit vector ex. This is the correct direction for B. I don't think there should be a minus sign in front.
     
  15. Jun 21, 2017 #14
    Alright I found a similar exercise in with more elaborate explanations and they also don't arrive at a minus sign, so I think the answer is just wrong, like TSny said. Thanks a ton for your help though Kuruman since I have a much better understanding of how to perform the integration now, though even with your tips I would still arrive at my original answer.

    Thanks!
     
  16. Jun 21, 2017 #15

    kuruman

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    Right you are. When I first looked at this, I concentrated on the denominator and ignored the ##P+L## in the numerator. :sorry:
     
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