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Biot-savart law

  1. Oct 25, 2008 #1
    Refer to the attachment below,

    anyone can explain the formula highlighted by red box? how the | dl x R | become dl only?
     

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  3. Oct 25, 2008 #2

    Doc Al

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    Note that [tex]\hat{R}[/tex] is a unit vector that is perpendicular to [tex]\vec{d\ell}[/tex].
     
  4. Oct 25, 2008 #3
    yes, I'm understand that dl is perpendicular to unit vector of R, but how the magnitude of them become dl only?
     
  5. Oct 25, 2008 #4

    Doc Al

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    The magnitude of any cross product is given by
    [tex]\vec{A}\times\vec{B} = AB\sin\theta[/tex]

    Applying this to your question, B is a unit vector and theta = 90, so
    [tex]AB\sin\theta = A[/tex]

    Make sense?
     
  6. Oct 25, 2008 #5
    I not really understand about the 2nd formula, could explain more?
     
  7. Oct 25, 2008 #6

    Doc Al

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    What aspect do you not understand? A = dl; B = 1; theta = 90.

    Have you worked with vector cross products before?
     
  8. Oct 25, 2008 #7
    why the B=1 ?

    yes, I know that
    [tex]
    \vec{A}\times\vec{B} = AB\sin\theta
    [/tex]
     
    Last edited by a moderator: Oct 25, 2008
  9. Oct 25, 2008 #8

    Doc Al

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    It represents the magnitude of the unit vector [tex]\hat{R}[/tex], which is 1.

    Perhaps you are confusing [tex]\vec{R}[/tex], which has magnitude of R, with [tex]\hat{R}[/tex], which has magnitude of 1?
     
  10. Oct 25, 2008 #9
    how do u know the [tex]\hat{R}[/tex] has magnitude of 1 ?
     
  11. Oct 25, 2008 #10

    Doc Al

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    It's a unit vector! The purpose of a unit vector is to define a direction; unit vectors--by definition--have magnitude of 1.

    The little "hat" symbol ^ on top of the vector tells you that it's a unit vector. Its direction is the direction of the vector R, but its magnitude is 1.
     
  12. Oct 25, 2008 #11
    ok, I understand why unit vector of R has magnitude of 1.

    now my another question is, why we can cross product of a vector with a unit vector?
    eg. vector of dl x unit vector of R
     
  13. Oct 25, 2008 #12

    Doc Al

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    Why not? A unit vector is a perfectly good vector, just like any other.
     
  14. Oct 25, 2008 #13
    let say vector of dl = hat{x} dl and vector of R = hat{x}3 + hat{y}2 + hat{z} 3

    the cross product of vector of dl x vector of R should be

    |hat{x} hat{y} hat{z}|
    | dl 0 0 |
    | 3 2 3 |

    or

    |hat{x} hat{y} hat{z}|
    | 1 0 0 |
    | 3 2 3 |
     
  15. Oct 25, 2008 #14

    Doc Al

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    True. But realize that here you need the cross product of vector dl and unit vector R hat, not vector R. The magnitude of R hat must equal 1.

    Given your definition of vector R, what would be unit vector R hat? (Hint: Find the magnitude of vector R.)
     
  16. Oct 25, 2008 #15
    vector R = (unit vector R)(magnitude of R)

    and now my question is

    let say vector of dl = hat{x} dx and vector of R = hat{x}3 + hat{y}2 + hat{z} 3

    the cross product of vector of dl and vector of R should be

    |hat{x} hat{y} hat{z}|
    | dx 0 0 |
    | 3 2 3 |

    or

    |hat{x} hat{y} hat{z}|
    | 1 0 0 |
    | 3 2 3 |

    which 1 should be correct? 1st or the 2nd?
     
    Last edited: Oct 25, 2008
  17. Oct 25, 2008 #16

    Doc Al

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    Assuming that the vector dl is purely in the x direction (for some reason), then version 1 is correct. Version 2 treats dl as a unit vector, which it is not.

    In general, dl will be in some arbitrary direction.
     
  18. Oct 25, 2008 #17
    sorry, my mistake, let say the cable carry current lies along the x-axis,

    so vector of dl = hat{x} dx and vector of R = hat{x}3 + hat{y}2 + hat{z} 3

    the cross product of vector of dl and vector of R

    |hat{x} hat{y} hat{z}|
    | 1 0 0 |
    | 3 2 3 |

    [ - hat{y}3 + hat{z}2 ]dx

    is my working correct?
     
  19. Oct 25, 2008 #18

    Doc Al

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    Yes, but the vector hat{x} dx should have components dx, 0, 0, not 1, 0, 0.
     
  20. Oct 25, 2008 #19
    An infinite current filament carries a current of 3A and lies along the x-axis. Using Biot-Savart Law, find the magnectic field intensity in cartesian coordinates at a point P(-1,3,2).

    dH = I (vet)dl x hat{R} / 4piR^2

    let say substitude hat{R} with (vet)R / R

    then dH = I (vet)dl x (vet)R / 4piR^3

    R= hat{x} (-1-x) + hat{y} 3 + hat{z}2

    is my working correct? and what should i do for the next step?
     
  21. Oct 25, 2008 #20

    Doc Al

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    Unless you are being asked to integrate to find the field at that point, you are doing this the hard way. What's the field from an infinitely long current-carrying wire?
     
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