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Homework Help: Biot-Savart Law- magnetic field produced by single wire.

  1. Dec 10, 2012 #1
    Biot-Savart Law-- magnetic field produced by single wire.

    1. The problem statement, all variables and given/known data

    A wire of length L, carrying current I lies along the x-axis as shown in the picture. A
    point P is located a distance 2L below the right end of the wire, as shown.
    a) Determine the direction of the contribution dB to the magnetic field at P due to the
    element dx of the wire.
    b) Write an expression for the contribution, dB, to the magnetic field due to element dx.
    c) Calculate the magnetic field B at point P due to the entire wire.

    2. Relevant equations


    3. The attempt at a solution

    This is study for an exam and as such my prof refuses to tell us whether or not our answers are wrong, since it doesn't help us learn or something, I DUNNO. I've done the problem as best I could but didn't finish because I'm almost 100% positive I'm doing something wrong but am not sure how to rectify the problem.

    So far I have that looking at the image provided B would be facing into the page.


    with r2=((2L)2+x2) and |d[itex]\vec{s}[/itex]x[itex]\hat{r}[/itex]=sinθ=x/((2L)2+x2)1/2

    I worked that down to:


    This is where I stopped. It seems to me that what I have so far doesn't seem to incorporate the y component at all except where (2L)2 is mentioned, but this isn't really like other problems I've done where dBy is negligible due to symmetry so I wondered if I was doing it wrong. Does this look correct so far?

    EDIT: Solved on my own.

    with r2=((2L)2+x2) and |d[itex]\vec{s}[/itex]x[itex]\hat{r}[/itex]=sinθ=2L/((2L)2+x2)1/2


    integrate from 0 to L...

    dB=μ0I2L/4π * L/[2L2((2L)2+L2)1/2]
    0IL/[(2L)2+L2] where 2L=R
    and if L<<R
    then B= μ0I/4πR

    Attached Files:

    Last edited: Dec 10, 2012
  2. jcsd
  3. Dec 10, 2012 #2
    Re: Biot-Savart Law-- magnetic field produced by single wire.

    You're doing it correctly. Keep in mind that each infinitesimal element of the wire produces a magnetic field in the same direction as every other differential element - all into the page or in the positive [itex]\hat{z}[/itex] direction. So not only is there NOT a chance for magnetic field contributions to be in the oppositie direction and thus potentially cancel the field in the 'z' direction, but you also have the luck of having the magnetic field all pointed in an easy direction (in the direction of one of the coordinate axes.) Thus you do not have to worry about projecting dB into the direction of the various axes and then doing 3 separate integrals.

    The calculation shouldn't depend on 'y' because all of the differential wire elements are all the same 'y' distance away from the point who's magnetic field is in question. And you know that 'y' value, it's just '2L'. Every distance between differential wire elements and the point 'P' then is just dependent on 'x' with the help of the constant '2L' and the pythagorean theorem.

    Also I notice that you end up with an integral that involves a slight trick to evaluate. You have to split up the 3/2 power in the denominator to
    [tex]\frac{x}{\sqrt{a^2 + x^2}^3}[/tex]
    and then use trig substitution where [itex]x = a \tan{u}[/itex]. Just trust me the dust will settle with the algebra and the third power won't be an issue. 'a' is just your '2L' constant.

    Edit** Actually yeah, just noticed your original sine term was wrong, but looks like you corrected it already and solved the problem.
    Last edited: Dec 10, 2012
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