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Basic Magnetic Sources help

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Wire #1 carries a current of 8.0A along the +x-axis. Wire #2 has a current of 6.1A along the +y-axis. What is the magnitude of the magnetic field at the point (1.4,1.3)m

    2. Relevant equations

    I'm not sure if Long straight conductor will be the right formula to solve for magnetic field.

    B = uo*I/(2*pi*r)
    3. The attempt at a solution

    I spend an hour trying to solve this basic problem. I kept reading the textbook, reading my teachers notes, but doesn't seem helpful. They only show an example of one wire, but not two wires.
    So then I found the r by using the Pythagorean theorem.
    I understand that the current isn't the vector. but how am I suppose to solve the magnetic field if there are two currents, but both are perpendicular?


  2. jcsd
  3. Nov 2, 2011 #2


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    First find the magnetic field that you would get from just one of the wires. Then do the same thing for the other wire.
    Paying attention to the direction of each magnetic field you calculate, you then simply add them using vector addition. Remember that the two magnetic fields are vectors; don't simply add their values unless they happen to point in the same direction.

    Hope that helps.
  4. Nov 2, 2011 #3
    Thanks for your reply,

    yes I think I did that, though, but somehow I did

    B= B * i(hat) + B* j(hat)

    Is that right, because I'm getting one wrong. I think I'm not understand this.
  5. Nov 3, 2011 #4


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    We still need to figure out the direction of the magnetic field vectors (it's not in the i-hat or j-hat directions.) Have you been taught about the right hand rule for straight wires? Here is a figure:


    You can read more details at the following link; scroll down to the section titled Right-Hand Rule #2:
  6. Nov 3, 2011 #5
    Thanks for your reply,
    Yes I did learn the right hand rule, and I mostly understood them. But I think I understand how to do it since my friend helped me. But thanks for the link. I'll look into it for future use.
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