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Homework Help: Magnetic Field from Current Segments

  1. Feb 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Find Bz(0,0,z1), the z component of the magnetic field at the point P located at x=y=0,z=z1 from the current I flowing over a short distance dl = |dl|*j located at the point rc = x1*i.


    2. Relevant equations

    Biot-Savart law:
    (μo/4∏)*[I*(dL x r)/(|r|^2)]

    3. The attempt at a solution

    I concluded that the radius would be √(( x1 )2 + (z1 )2) and submitted:

    (μo/4∏)*[I*dL/√(( x1 )2 + (z1 )2)]

    which was wrong. Then I remember that the equation called for r^2 not r, so I submitted:

    (μo/4∏)*[I*dL/(( x1 )2 + (z1 )2)]

    which was also wrong. I thought that the cross product would be even but just in case I tried:

    (μo/4∏)*[I*-dL/(( x1 )2 + (z1 )2)]

    which was still wrong.

    Am I getting the radius wrong? I'm not quite sure what's going on.
  2. jcsd
  3. Feb 21, 2014 #2


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    Staff: Mentor

    Start by writing out the vectors involved. Your radius magnitude squared is okay as ##x_1^2 + z_1^2##, but you'll need the radius vector and dL vector components to perform the cross product. Do the cross product!
  4. Feb 22, 2014 #3
    So the cross product I want is [itex]\vec{dl}[/itex] x [itex]\hat{r}[/itex] and the homework tells me that [itex]\vec{dl}[/itex] x [itex]\vec{r_{x}}[/itex] = [itex] dlx_{1}\hat{k}[/itex]. Shouldn't [itex]\vec{dl}[/itex] x [itex]\vec{r_{x}}[/itex] be negative since [itex]\hat{j} \times \hat{i} = -\hat{k}[/itex]?

    So assuming [itex]\vec{dl}[/itex] x [itex]\vec{r_{x}}[/itex] = [itex] dlx_{1}\hat{k}[/itex], we know [itex]\hat{r} = \frac{\vec{r}}{r}[/itex]. Then would the answer be (μo/4∏)*[(I*dL)/(√(( x1 )2 + (z1 )2)^3)] since there's now an extra r on the bottom?
    Last edited: Feb 22, 2014
  5. Feb 22, 2014 #4


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    Staff: Mentor

    There shouldn't be an "extra" r on the bottom. You've extracted the z-component of the cross product but then ignored it? The z-component of the cross product contains an x1, so why does it disappear when you write the result? And I'm not sure why you've introduced a unit vector in the r direction; they're asking for the z-component only, not the whole vector (and it wouldn't be in the same direction as the r vector anyways).

    So the z-component of the cross product, as you've found, is ##dl x_1## . The square of the magnitude of r is ##x_1^2 + z_1^2##. That, along with the current magnitude and constants of the Biot-Savart law should be sufficient to write the result.
  6. Feb 22, 2014 #5
    I forgot to mention that I tried (μo/4∏)*[(I*dL*x1)/(( x1 )2 + (z1 )2)] which wasn't correct.

    From what I understand the Biot-Savart law is (μo/4∏)*[I*(dL x r hat)/(|r|^2)]. Am I not suppose to be finding the cross product dl x r hat?

    Sorry I'm having so much trouble understanding and visualizing this. Thank you for your patience.
  7. Feb 22, 2014 #6


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    Staff: Mentor

    Okay, I think I owe you an apology there. The cross product does indeed involve the unit vector in the direction of r. That puts another ##\sqrt{x_1^2 + z_1^2}## in the denominator along with the ##x_1^2 + z_1^2## that's already there. That yields ##(x_1^2 + z_1^2)^{3/2}## net. With that and the ##x_1## in the numerator I think you'll be alright.
  8. Feb 22, 2014 #7
    Not at all, you helped me get the correct answer and more importantly I get it now. Thanks so much!
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