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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
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