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Finding magnetic field from two crossing wires

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Two insulated wires in the x-y plane cross at a 30° angle but do not make electrical contact. Each wire carries 3.0 A current (the angle between current directions is 30°). Find magnetic field in two non-equivalent points located 4.0 cm from the intersection, in the x-y plane, at equal distances from both wires.

    2. Relevant equations
    B = (μ0I)/(2r) For a straight wire

    3. The attempt at a solution
    I'm not sure I understand the problem, or how to tackle it, could you guys help me out analyzing it? I understand the concept if I had only one wire, it would be simple, I know the value of μ0, of I (3.0 A) and of r (0.04 m)

    From what I understood I did this drawing, now my issue is how do angles come into play?

    Attached Files:

  2. jcsd
  3. Nov 1, 2012 #2
    The field from each wire is independent and can be easily calculated for the required positions.
    The two fields add to give a resultant field.

    (In your drawing, the two points are equivalent - you need to find another that is not equivalent)
  4. Nov 1, 2012 #3
    Ooooh I see. So I could do:

    Point 1) Directly perpendicular to the interesction (so at 90°) to the north
    Point 2) In the south-east of the intersection, halfway between the south line and the second wire, so I'd have an angle of 60° from the horizontal, or 30° from the vertical.

    I guess if I do that for point #2, I would simply do:

    BX = (μ0I)/(2r) X Cos 60
    BY = (μ0I)/(2r) X Sin 60

    and then for the magnitude, √(BX2 + BY2)

    Does that work?
  5. Nov 1, 2012 #4
    4 cm from the intersection (draw a circle 4cm radius round the intersection)

    Equidistant from both wires (draw a line that bisects the angle between the wires - there are two such lines)

    Where the bisectors cross the circle you will find treasure.
  6. Nov 1, 2012 #5
    If I followed you correctly, it would look like this?

    But if I find B at both points, won't it give me the same answer?

    R is the same in both cases, they are simply at opposite sides of the circle

    Attached Files:

  7. Nov 2, 2012 #6
    There are TWO bisectors - you only have one.
  8. Nov 2, 2012 #7
    Ok I don't follow anymore, I just don't understand the concept of the problem when I'm reading it. Could you help me figure out what I'm looking for?
  9. Nov 2, 2012 #8
    Yes you do. You just have hit a blind-spot. It's staring you in the face.

    When two lines cross they make two angles, not just one: α and 180-α. You can draw a bisector through α and another through 180-α. Look again at your drawing.
  10. Nov 2, 2012 #9
    Ok well this part I understood (I hope), I redrew my figure... It's just from here I don't know how to approach the problem. I understand that I have to add the points to get the total field, but it seems like I will get the same results for the 4 points since I am using

    B = (μ0I)/(2r)

    I'm thinking maybe the angle of the point crossing the circle from the interesection will affect my result for each individual points, but if it's the case, I wouldn't be sure how to take it into account

    Attached Files:

  11. Nov 2, 2012 #10
    Ah yes! there's another little trap in the question.

    The field has the same magnitude that's true (r is the same) - but has it the same direction?
  12. Nov 2, 2012 #11
    ugh I was afraid it was going there, I could never understand how to figure the direction of the magnetic field

    I know I have to use the right hand rule, but my question is two-fold

    1) How do I know whether the current in the wire goes towards the extremity, or away from the extremity? Is it simply a convention?

    2) I just don't know how to analyze the right hand rule, I know the way my fingers curl is the way of the magnetic field, but this becomes too abstract. Do you have sort of a rule of thumb to help me in deciding the direction?
  13. Nov 2, 2012 #12
    Oh, I agree with you on the right-hand rule - it's cr@p!
    I never use it.

    It's very simple - the concept is based on a screw-motion. If you turn a screw clockwise it moves forward. Anti-clockwise is backwards.
    Similarly, (given no friction) if you push a screw forward it turns clockwise and vice-versa.

    The screw is going down the direction of current. It rotates in the direction of the magnetic field. How easy is that?
  14. Nov 2, 2012 #13
    Heh actually that's quite easy to use, more straight-forward than the hand, but my problem remains the same. Once I find the direction of the magnetic field I just don't know how to apply it - and this isn't only for this problem, it's for any problem in general.

    I'm sorry, I know I'm very slow understanding this problem, but the whole concept is very alien to me unfortunately. At least I got the drawing part, but it's the whole concept of figuring out the direction that has me completely lost
  15. Nov 2, 2012 #14
    It takes a while to get used to it but it's quite easy really.
    You can figure out the direction of the field from the corkscrew rule I just gave you. The magnitude is simply the formula H = etc.

    In this problem it's been made a bit easier because the wires are in the plane of the paper and that means the field (in the plane) can only be straight up towards you or down through the table.
    All you have to figure out (for each wire) is where the field is coming up and where it's going down.

    If the field from both wires is in the same direction, they add. If not - they oppose.
  16. Nov 2, 2012 #15
    Ok I'm starting to figure it out piece by piece. My question is, to find the magnetic field direction, I need to know the direction of the current. Maybe it's because my brain is fried right now, but how do I determine the direction of the current in a wire when I have no positive or negative pole?

    Or... Would I be wrong in saying that both points to the right of the intersection go inside the paper, while both points to the left of the intersection go towards me? If that's the case, wouldn't the answer be 0?
  17. Nov 2, 2012 #16
    It's stated in the question: the angle between current directions is 30°. (not 150° notice)

    You'll find that specifies absolutely the relationship of the current to the two points of interest (the diagram could be mirror imaged - but that makes no difference because of symmetry)
  18. Nov 2, 2012 #17
    Ok so basically one current will point to the right, and one to the left?
  19. Nov 2, 2012 #18
    Is that a guess?
    Two roads cross at an angle. Traffic on one road is travelling at 30° to the traffic on the other. That's sort of like a slip road. They are travelling almost in the same direction.
  20. Nov 2, 2012 #19
    Ok you're right, they are both travelling in the same direction, I re-read correctly the part about the 30 degree angle, it gives us the direction of the current relative one wire to another

    So if both currents flow in the same direction (let's just say left to right), I push a screw, it goes clock-wise, so inside the paper. But since both wire have the same current direction, wouldn't all four points be in the same direction?

    So my magnetic field would be 4x each individual points
  21. Nov 2, 2012 #20
    That's the trap.
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