Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Biot–Savart Law

  1. Apr 7, 2013 #1
    I'm not sure I get how:[tex]\frac{\mu_{0}}{4\pi}\int_{C}\frac{I(d\vec{l}\times\vec{r})}{r^{3}}[/tex] = [tex]\frac{\mu_{0}}{4\pi}\int_{C}\frac{I(d\vec{l}\times\hat{r})}{r^{2}}[/tex]

    the r is a displacement vector in the first, and in the second it's a unit vector, but why is this so?
     
    Last edited: Apr 7, 2013
  2. jcsd
  3. Apr 7, 2013 #2
    The second expression is correct, the first one is not.
     
  4. Apr 7, 2013 #3
    Sorry, meant to put a cubed there. Fixed
     
  5. Apr 7, 2013 #4
    There's nothing really to get: [itex] \widehat{r} = \frac{ \vec{r} }{r} [/itex].
     
  6. Apr 7, 2013 #5
    I see,another dumb question I asked, sorry for wasting your time......
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Biot–Savart Law
  1. Biot Savart law question (Replies: 10)

  2. Biot-Savart Law (Replies: 2)

  3. Biot-Savarts law (Replies: 1)

  4. Biot-savart law (Replies: 2)

Loading...