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Biot-Savart law

  1. Aug 4, 2012 #1
    In the expression of Biot-Savart law
    B = (µo/4π) ∫ (I dl x r^)/r2
    why dl does not depend on the coordinate systems ?
    in books they are using del X dl = 0
  2. jcsd
  3. Aug 4, 2012 #2
    It's natural [itex]d\ell[/itex] doesn't depend on the coordinate system; it's a vector.

    Still, [itex]\nabla \times d\ell[/itex] doesn't make a whole lot of sense. Could you give some more context for this question?
  4. Aug 5, 2012 #3
    i am attaching a file in which the derivation of biot-savart law is given. Now after equation 6-29 they used del X dl' = 0. i want to know the reason

    Attached Files:

  5. Aug 5, 2012 #4
    It says the reason right in the screenshot:

    Here, [itex]d\ell[/itex] and [itex]d\ell'[/itex] are two different things. The first depends on "unprimed" coordinates (i.e. [itex]x, y, z[/itex]) and the second depends on "primed" coordinates [itex]x', y', z'[/itex]). The operator [itex]\nabla[/itex] differentiates only with respect to the unprimed coordinates. Hence, [itex]\nabla \times d\ell'[/itex] is zero because [itex]x', y', z'[/itex] are not functions of [itex]x,y, z[/itex]; they can't be differentiated with respect to the unprimed coordinates.

    This use of primed and unprimed variables is pretty common in proofs of integral theorems, particularly when you have a function on the left that is, say, [itex]X(r)[/itex], generally the variable of integration on the right is [itex]r'[/itex]. For example,

    [tex]E(r) = \int \frac{\rho(r')/\epsilon_0}{4\pi |r - r'|^2} \; dV'[/tex]

    Here, [itex]r'[/itex] is the dummy variable of integration, and [itex]dV'[/itex] is the associated volume element for that variable.
  6. Aug 5, 2012 #5
    now i got it. thanks a lot
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