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Using Biot Savart law

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    A short current element dl⃗ =(0.500mm)j^carries a current of 8.90A in the same direction asdl⃗ . Point P is located at r⃗ =(−0.730m)i^+(0.390m)k^.

    Find the magnetic field at P produced by this current element.
    Enter the x, y, and z components of the magnetic field separated by commas.


    2. Relevant equations


    For a wire : dB = (μ naught/ 4π) (I) (dL X r→) / (r^2)


    3. The attempt at a solution

    dL = .0005 meters in the j→ direction

    I = 8.9 amps in j→ direction

    point P is located at r→ = -.730 meters i→ + .390 meters k→

    so r is the magnitude of the distance between the point and the wire, so r = .73^2 + .39^2 = .685

    μ naught / 4 pi = 10^-7 / 4 pi = 7.95 * 10^-9

    Now I'm not sure what to do exactly. I know the cross product of dl and r→ is the determinant, and I know how to do that. Doing that I get a result of .000195 i + .000365 k.

    Now do I take (7.95*10 ^-9) * (8.9 amps) (either .000195 or .000365) / (.685 meter^2) to find the i(x) and k(z) components of dB respectively? I've tried things like that and I couldn't get the right answer.

    Note: I always do my calculations in standard units like meters and tesla first then convert at the end so I don't confuse myself so I don't think that was my mistake. I think I'm just not understanding what term actually means what in the equation, or how to work with vectors correctly. A nudge in the right direction would be appreciated, thanks for any and all advice.


    The correct answer is dBx, dBy, dBz = .306, 0 ,573 nT
     
  2. jcsd
  3. Nov 6, 2014 #2

    BvU

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    Pretty good work!
    I found one little flaw in the relevant equation: it's either ##\displaystyle I\;d{\bf \vec l} \times {\bf \vec r}\over \displaystyle |r|^3## or ##\displaystyle I\; d{\bf \vec l} \times {\bf \hat r}\over \displaystyle |r|^2##

    One little flaw in ##\mu_0##: ##\ \ \mu_0 = 4\pi \; 10^{-7}\ \Rightarrow \ \mu_0/(4\pi) = 10^{-7}##

    And one corny little flaw in r = .73^2 + .39^2 = .685 See it ?

    Fix them and you end up spot on !
     
  4. Nov 6, 2014 #3
    Thank you for responding!

    So for the biot savart law I have a question. The first one you posted has dl→ × r→ whereas the second one uses r with a unit vector hat, so just to clarify, the first equation would be used if I had a vector r from my current element to the point of interest in terms of x, y, and z correct? And the 2nd one would be if I was given a vector in terms of a single unit vector? I'm confused on that part, I'm a little fuzzy on what I remember from vectors. Can you explain the difference between those two and why I would use one or the other?

    The μ naught issue I see I made a mistake in the equation now; I must have misread that constant relationship at first.

    But the last flaw is also confusing. I know the magnitude of a vector is the square root of the sum of the squared components, so I thought : r^2 = x(i) ^2 + y(j) ^2 , in the original post I accidentally said r instead of r^2. But I can see that it would be an error regardless if I originally chose the r^3 biot savart equation. So I don't see where I went wrong in that equation with what I chose originally, regarding the r^2.
     
  5. Nov 6, 2014 #4
    They are one and the same remember that "r-hat" is just R divided by it's own magnitude. So in one of the cases one magnitude has been factored out and canceled and in the other not. They are exactly the same.

    [itex] \vec R = |R| Rhat [/itex]

    You would use the one with Rhat if you didn't know the distance but knew the direction (it might put you a step closer even though you can't get a numerical answer). If you know the just the distance (R) but not the direction there is no sense in calculating Rhat so then you would use the r^3 version.
     
    Last edited: Nov 6, 2014
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