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Magnetic force on loop

  1. Aug 1, 2017 #1
    Screenshot_2017-08-01-20-46-59-1.png 1. The problem statement, all variables and given/known data
    A rectangular loop of wire, L = 28.8 cm and W = 15.2 cm, carries a I1 = 2.02 A current and lies in a plane (d = 11.0 cm), which also contains a very long straight wire carrying a I2 = 12.7 A current as shown in the figure below.

    2. Relevant equations
    F=uo*i1*i2*l/perpindicular distance

    3. The attempt at a solution
    So if we label the loop as 1 for the top, 2 for the right side, 3 for the bottom and 4 for the left side we see that 1 and 3 cancel due to symmetry.

    Then calculating 2 and 3

    F2=(4pi*10^-7)(2.02)(12.7)(.288)/.11=.000084N

    F4=(4pi*10^-7)(2.02)(12.7)(.288)/(.11+.152)=.0000353N

    The actual answer is 7.79×10-6 N

    Any combination of F2 and F4 that I have calculated does not give the correct answer, I have tried adding them, subtracting one, subtracting the other. So what have I done wrong?
     
  2. jcsd
  3. Aug 1, 2017 #2

    kuruman

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    How sure are you that this is correct?
     
  4. Aug 1, 2017 #3
    Which part are you referring to? The solution 7.79×10-6 N?
     
  5. Aug 1, 2017 #4
    Sorry forget that last post, I thought that was the solution should be it 2pi*distance?
     
  6. Aug 1, 2017 #5
    I think your equation should be divided by the term 2*pi*distance and not just distance.

    The force will be different in sign depending on the direction of the current flow. If they (I1 and I2) are in the same direction, the force will move the wires closer to each other. If they flow in opposite directions, the force will push them apart. You can see this if you look at the right hand rule and examine the magnetic field for both situations.

    With loop currents in both directions compared to wire 2, you will have to subtract the two results.

    If you do these two things, it should match up with the actual answer.
     
  7. Aug 1, 2017 #6
    Okay so,


    F4=(4pi*10^-7)(2.02)(12.7)(.288)/2pi(.11+.152)

    F2=(4pi*10^-7)(2.02)(12.7)(.288)/2pi*.11

    Clearly it matter which one is positive and which one is negative so how does one tell which is positive and which is negative? The only thing I can think of is that it the magnetic field is going into the page so using IxB for f2 we use our right hand point our fingers in the direction of the current so up, then curl them into the page and that gives the thumb going left so f2 should be negative, while f4 we out our fingers down curl them in toward the page and the thumb is right so the force is positive?
     
  8. Aug 1, 2017 #7

    haruspex

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    Assuming you are using positive for forces to the right, yes.
    Rather than using handedness rules, I find it much easier to visualise the field lines.
    If you have currents the same direction along two parallel wires, the field lines run the same way around them, so at a sufficient distance merge to become one system of enveloping lines. The lines like to shrink in length and spread apart from each other, so they pull the wires together.
    With the currents in opposite directions, the two sets of lines run opposite ways around, leading to a bunching up between them. That pushes the wires apart.
     
  9. Aug 1, 2017 #8
    If I was using the right hand rule how would I know that the magnetic field was going into the page? I just kind of guessed and it was correct. That will be my last question
     
  10. Aug 1, 2017 #9

    haruspex

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    It goes into the page on one side and out of the page on the other.
    The right hand rule for this says to point your right thumb in the direction of the current in the wire. Curled fingers then point in the direction of the field.
     
  11. Aug 1, 2017 #10

    So for the right side, I would point my thumb in the direction of the current so upwards then curl my fingers which way? How do I determine whether the magnetic field is coming out or going in
     
  12. Aug 1, 2017 #11

    kuruman

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    Imagine grasping the long wire with your right hand so that your thumb points in the direction of I2. Curl your right hand fingers around the wire. They go into the screen on the right side of the wire and out of the screen on the left side of the wire. So does the field.
     
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