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Quick question: Biot-Savart

  1. Nov 24, 2006 #1
    Hey guys I have a wire oriented along the x axis with lengths -L/2 to L/2

    The first part of the Biot Savart formula has a I dl in vector notation crossed with (r-r')/(r-r')^3. I used I(i hat) for the I dl vector and to cross it, i used r as my observation point and r' as source. When I did the integral where the observation point D is located at D j(hat) i got a negative B oriented in the k (hat) axis. Now according to the answer and picture it should be positive.

    Can you please give me tips on how to determine my I dl vector and maybe I flipped the r and r' and it should be source - observation point??
     
  2. jcsd
  3. Nov 24, 2006 #2

    OlderDan

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    If the page is the x-y plane, all the dL x (r-r') in the problem are out of the page (positive z axis). Maybe you need to write the unit vector as (r - r')/|D - r'|. Could be you have a sign problem with D - r'.
     
    Last edited: Nov 24, 2006
  4. Nov 24, 2006 #3

    Doc Al

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    The Biot Savart formula takes [itex]\vec{dl}\times\hat{r}[/itex], where [itex]\hat{r}[/itex] is the unit vector specifying the direction of the observation point with respect to the source (this should be parallel to your (r - r')). So I don't see how you got [itex]\hat{i}\times\hat{j} = -\hat{k}[/itex]. My guess is that you did interchange r and r'.

    Edit: Looks like OlderDan beat me to it.
     
  5. Nov 24, 2006 #4
    The differential magnetic field is:
    [tex]
    d \vec B = \frac{\mu_0 I}{4\pi} \left( \frac{d\vec l' \times \hat R}{R^2} \right)
    [/tex]

    So I believe you are saying that, you have a wire oriented along the [itex] x [/itex] axis. Where the orgin is in the middle of the wire.

    So you need to resolve the following:
    [tex] d\vec l' [/tex]
    [tex] \hat R [/tex]
    [tex] R^2 [/tex]


    Note that [itex] d\vec l' [/itex] is the differential length of the wire pointed in the direction of the current. You know the length of this differential length because you have it on the [itex] x [/itex] axis. Thus you can say that the length is [itex] dx [/itex]. Now which direction is this vector supposed to point? This is the direction of the current, is it positive ([itex] +x [/itex]), or negative ([itex]-x[/itex]).

    [itex] \hat R [/itex] is a unit vector that points from the differential line element ([itex] d\vec l' [/itex]) to the field point (your point of interest). Remember that a vector pointing from one point to another is:
    [tex] \vec V_{AB} = \vec V_B - \vec V_A [/tex]

    So if you want vector pointing from the source point [itex] d\vec l' [/itex] to the point of interest (say P - pointed to by a vector with the un-primed notation ([itex] \vec R [/itex])

    You want a vector that points from the source point (where the charges are located - this is the PRIMED notation), to a point of interest (the UNPRIMED notation). So that would be unprime _vector minus primed_vector.

    EDIT: I guess two people beat me to it :)
     
    Last edited: Nov 24, 2006
  6. Nov 24, 2006 #5
    biotsavart.JPG

    Ok using that information, I am now attempting a problem where a wire forms a circular loop. I have drawn it roughly on paint brush.

    I still get 0. My observation/field point is supposed to be located at the origin which makes the source vector negative. I dont understand what I am doing wrong
     
  7. Nov 24, 2006 #6

    Doc Al

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    Forget the details of the integration for a moment. What must the direction of the magnetic field be at the center of that current loop?
     
  8. Nov 24, 2006 #7
    it has to be pointing in the positive X direction
     
  9. Nov 24, 2006 #8

    Doc Al

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    Right. So check that each element of whatever you are integrating is giving a contribution in the +x direction.
     
  10. Nov 24, 2006 #9
    Ok i think the problem is located at where I am defining my dl. I said cos()k+sin()j . That must mean the current is traversing in a clockwise direction?

    Which means I would have to make both negative? I do not understand how to define a direction for current when it is traveling in a circle. I understand what your saying, to make the numerator's cross product yeild a positive value in the i direction. But I dont understand the concept of defining a vector dl for a circle.

    I re did the first line charge problem and it is now coming out positive and correct. So thanks for the help so far!
     
  11. Nov 24, 2006 #10
    ok I am getting 0, the vector I wrote for the current is wrong. I need help writing that please. I made the vector negative to yeild a postive i direction but the integral of cos is sin. Then 0 to 2 pi is 0.
     
  12. Nov 24, 2006 #11
    What vector would be tangent to the loop at any point?

    EDIT: Remember [itex] d\vec l' [/itex] is a vector with a differential magnitude. What is [itex] d \vec l [/itex] in cylindrical coordinates?
     
    Last edited: Nov 24, 2006
  13. Nov 24, 2006 #12
    Oh i was trying o keep dl in cartesian because my cylidrical/spherical is really weak. Ummm dl in cylindrical should be r i + (theta) j + z k and in this case z would be 0 so only a r and a theta?

    oh and a tangent vector.... not sure in this case. All I can think of is velocity which has to do with current so the current vector should be tangent?
     
  14. Nov 24, 2006 #13
    You can keep it in cartesian, it just makes the problem harder. What are i, j, k in cylindrical coordinates?

    Would you prefer to do this problem in cartesian?
     
  15. Nov 24, 2006 #14
    No I have to learn cylindrical so I can be more proficient. So lets keep it in cylindrical. I thought cylindrical in i j k form would be

    r i + theta j + z k
     
  16. Nov 24, 2006 #15
    Just a little review here.

    Remember that a vector consists of a magnitude and direction. So when we talk about a vector like, [itex] \hat i [/itex] we are really saying that the vector is [itex] (1) \vec x [/itex], where [itex] (1) [/itex] is the magnitude and [itex] \vec x [/itex] is the direction. The reason I bring this up, is when we are writing [itex] d \ vec l [/itex] in the Biot-Savart law we are are asking for a tiny piece of the current with a differential magnitude, and then also a direction.

    So in general in cartesian, [tex] d \vec l = \hat x dx + \hat y dy + \hat z dz [/tex].

    In cylindrical, [tex] d \vec l = \hat r dr + \hat \phi r d\phi + \hat z dz [/tex]

    What does [itex] \hat i [/itex] mean in cylindrical coordinates? It is pointing in the direction of the x-axis right? Well this doesn't mean much to us in cylindrical coordinates (we aren't taking advantage of the coordinate system). For example if we wanted to move in a direction of a circle we would only need to follow the direction of [itex] \phi [/itex].

    Remember that i, j, k are all orthogonal to each other. r, phi, z in cylindrical are also orthogonal.

    EDIT: So if you want to use [itex] \hat i [/itex] in cylindrical coordinates, you would need to express them as unit vectors of r, phi, z
     
    Last edited: Nov 24, 2006
  17. Nov 24, 2006 #16
    [tex] d \vec l = r d\phi \hat k + dr \hat j [/tex]


    Edited, I updated according to the axis the circle loop is located. Sorry for being so slow, but you are helping a lot!
     
    Last edited: Nov 24, 2006
  18. Nov 24, 2006 #17
    Oh wait a second, would my K hat vector in the dl above be negative to indicate the current is traveling counter clockwise?
     
  19. Nov 24, 2006 #18
    dl is supposed to point in the direction of the current.

    Actually if you want to use cylindrical coordinates, a vector pointing in the direction of phi is tangent to the circle. So as you increment in the phi direction you trace the circle (the direction of the current).

    Does that make sense?

    So if you use [tex] d \vec l = \hat \phi r d\phi [/tex] you are tracing the circle in the direction of the current.

    What book are you using?
     
  20. Nov 24, 2006 #19
    Physics for Scientists and Engineers 3ed FISHBANE GASIORWICZ THORNTON
     
  21. Nov 24, 2006 #20
    Ok i see what you mean about the direction of current and phi.

    As for my source point vector, it has a magnitude of lower case r and I will choose the direction to be right on the y axis so r j hat

    Plugging that in gives me this numerator:

    dr j (hat) + r d(phi) k (hat) X -r j(hat)

    and then that gives me r dr in the i hat direction, positive.

    Then i get stuck with an integral that is dr/r^2 integrating from 0 to R and once again, I get negative!

    I am sorry I keep getting mixed up.
     
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