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Magnetic field. I need help on understanding this problem.

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Two long parallel conductors carry currents I_1 = 3.0A and I_2 = 3.0A, both directed into the page. Find the magnitude and direction of the resultant magnitude field at point p?

    2. Relevant equations

    3. The attempt at a solution

    I need help with this problem. The professor only provided the answer but not the steps. I'm having a hard time understanding what to do. I can do all the other problems, but when it comes to a problem like this (with having to apply vectors) I do not know what to do. Help me.

    First I tried drawing the magnetic fields (figure 2). Then I found the angles - 23 degrees and 67 degrees.

    I am stuck after this. How do I solve using vectors?

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  3. Apr 10, 2013 #2

    Simon Bridge

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  4. Apr 11, 2013 #3
    what do you mean by tangential?
  5. Apr 11, 2013 #4

    Simon Bridge

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    The B field around a wire follows circles - at distance r, the B vector will be a tangent to the circle, centered on the wire, with radius r.

    Is ##\vec{B}_1## the field due to ##I_1## and ##\vec{B}_2## the field due to ##I_2##?
    If so, they are pointing in the wrong directions. Please see the accompanying link.
  6. Apr 11, 2013 #5
    oh yes, im sorry. your correct. i miswrote the B1 and B2.
  7. Apr 11, 2013 #6

    Simon Bridge

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    OK - so... do you know how to add vectors?
  8. Apr 11, 2013 #7
    hmm is the special right triangle the triangle with the 90 degrees or is it the one with the added vectors?
  9. Apr 11, 2013 #8
    i know how to add vectors like moving head to tail, but not very well when it comes to the math.
  10. Apr 11, 2013 #9
    so at this point i would get a resultant vector that goes straight down. but im confuse. do i do the x and y component for B1 and B2 separately or do i do the x and y component for the resultant component?
  11. Apr 11, 2013 #10

    Simon Bridge

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    To use x-y components you need to define an x and y axis - which is not provided for you.
    Instead you should use the head-to-tail triangle of vectors and your knowledge of trigonometry.

    One special triangle you have is the pythagorean triplet 5-12-13 ... point P is at the right angle.
    It is probable that the B vectors also make two sides of a special triangle. They do conveniently make a right-angle with each other. If the resultant vector does go straight down, then it must be 5-12-13 as well.

    There are lots of ways of specifying the direction of a vector.
    If you label the position of I1 as Q and I2 and R, then the straight down direction is the Q-R direction ... or "parallel to ##\overrightarrow{QR}##"
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