1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    19590_e.jpg

    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

    gneill

    User Avatar

    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

    gneill

    User Avatar

    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

    gneill

    User Avatar

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted