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 in BIOT-SAVART law

  1. Dec 27, 2006 #1
    Dear sir/lady

    I have a question about the magnrtic force of steady curent .
    In Biot - Savart law to evalute of B (magnetic Field ) , below the Integral we have to do a cross product Idl'*(r-r')/|r-r'|^3 that r and r' are the vector position of the field and source . How we can evaluate this Integral if the vector r-r' is zero . ?
    for example if we have a U shape Incomplete circuit and put a metal bar as the fourth side to complete it , when a stationary current I circualtes in square , a force will exert on the bar . if we would like to evaluate the force exerted by three sides on the fourth one , we will encounter this problem for the toppest and the lowest point of fourth one , on corners .Because vector r-r' =0 and B approches to infinity . Please help me .
    Last edited: Dec 28, 2006
  2. jcsd
  3. Dec 27, 2006 #2
    If you have a square shape current loop you can use superposition (i.e. the additive linearity operator).
  4. Dec 27, 2006 #3
    If you have a totally localized current distribution, you'll get divergences, same as if you have point charges in electrostatics, i.e. in Coulomb's Law. Try solving the magnetic field for a cylindrical charge distribution and see what you get.
  5. Dec 28, 2006 #4
    please explain me more !!
  6. Dec 28, 2006 #5
    Well, just work it out. The Biot-Savart law is pretty much equivalent to Ampere's law, so I'll work out the most trivial example of Ampere's law.

    If you have an infinitely long wire with a current I running through it, and it has no spacial extent, then if I make a circle centered on the wire, I find that
    [tex] \oint \mathbf{B} \cdot d \mathbf{l} = \mu_0 I[/tex]
    which leads to
    [tex] B (2 \pi r) = \mu_0 I[/tex]
    from which we see that the magnetic field is infinite at r = 0.

    Now let's work the same problem, only this time with a wire of radius a and constant current density J. Outside the wire, we get the same result from Ampere's law:
    [tex]B = \frac{m_0}{2 \pi} \frac{\pi a^2 J}{r}[/tex]
    But inside, the magnetic field is different. Inside, the current enclosed in the loop is given by [tex]I = \pi r^2 J[/tex], which means that
    [tex]B = \frac{\mu_0}{2 \pi} \pi r J[/tex]
    The magnetic field is nice and well-behaved inside the wire! But what's the difference?

    Well, in the first case, the current density can be described as a delta-function, so that [tex]\nabla \times \mathbf{B} = \mu_0 I \delta(x)\delta(y)[/tex] if, say, the wire is resting on the z-axis. These delta-functions are wildly singular, and in fact this differential equation looks very similar to that of a point electrostatic charge [tex]\nabla \cdot \mathbf{E} = q \delta(\mathbf{r}) / \epsilon_0[/tex]

    In the second case, the current distribution can be described by step-functions, which although their derivatives are discontinuous, the actual functions are finite and well-behaved everywhere. This is the origin of the difference, and why the magnetic field would diverge for a wire of zero physical extent.
    Last edited: Dec 28, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook