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Is the magnetic force always 0?

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data

    For any loop that may have symmetry, the magnetic force in the center must always be 0?

    Ex. circles, rectangles?
     
  2. jcsd
  3. Aug 16, 2011 #2
    Assume the source of the magnetic field is the loop itself...forgot to mention.
     
  4. Aug 16, 2011 #3
    The first question would have to be "force on what?". If you are really thinking about the strength of the field, start with a circular loop - the field is a maximum at the center.
     
  5. Aug 16, 2011 #4
    Like the net force in the center, is it 0 always for loops in the shape of circles, squares, rectangles?
     
  6. Aug 16, 2011 #5
    Do you mean field?
     
  7. Aug 16, 2011 #6
    Nope, I mean force. I tested out with F = il x B for circles, squares and rectangles and they all seem to have no net force in the center because of the symmetry.
     
  8. Aug 16, 2011 #7
    With a simple circular loop, the field is a maximum, not 0. The force is another matter. To have a force, the field has to act on something - a moving charge, a current, a magnetic pole, etc.
     
  9. Aug 16, 2011 #8
    The current runs in the loop
     
  10. Aug 16, 2011 #9
    OK - that is clearer. If the current is a wire passing perpendicular through a circular loop (through the center), then yes, you are correct - there is no ilB force (the wire and the field are parallel). If the the wire is not perpendicular to the loop, then there will be a force because there is now a cross product.

    I think you can safely extend that line of reasoning to any symmetric shape.
     
  11. Aug 16, 2011 #10
    No I mean the loop is the wire and the current runs in the wire.
     
  12. Aug 16, 2011 #11
    Yes. The current in the loop will produce a magnetic field. If you just have the loop with a current in it, there is a field - but no force - regardless of symmetry. To have a force, there has to be something for the field to interact with - like a second wire carrying a current. If that second wire is straight, and is passing perpendicular to the loop through it's center, then the force on the straight wire will be zero because both the magnetic field and the current in the straight wire point in the same direction.

    Again, with a single, current carrying, circular loop, there is a field - but you cannot talk about force until you bring something else into the picture to interact with the field. A field by itself has no force.
     
  13. Aug 16, 2011 #12
    Let's take a square for instance, why can't yuo take the parallel side's magnetic field as the "external" field?
     
  14. Aug 16, 2011 #13
    Perhaps you are talking about the force on the current in one part of a loop from the magnetic field from the rest of the loop? The currents in segments on opposite sides of the loop will act to repel each other by equal amounts. And then, by symmetry, the net force is 0 for the entire loop.
     
  15. Aug 16, 2011 #14
    Yay as my conclusion
     
  16. Aug 16, 2011 #15
    OK - now we are on the same page. Your example of the square shows the net force is zero. Again, it appears you are safe to extend that reasoning to any symmetry.

    However, symmetry is not really required. Regarless of shape, the net force on the loop must be zero or Newton's 3rd Law would be violated.

    (I'm still not clear about "force at the center")
     
  17. Aug 16, 2011 #16
    Hi - this came to my email.

    "Like the net force in the center, is it 0 always for loops in the shape of circles, squares, rectangles?
    ***************"

    There are no forces in the center - only fields. The force comes about when the field acts on something - such as another current.

    I would distinguish between "no forces at the center" and "the force at the center is 0". The latter implies the existence of a force.
     
  18. Aug 16, 2011 #17

    SammyS

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    Can an object produce a net force on itself?
     
  19. Aug 16, 2011 #18
    We don't live in a world of Star Wars, so no...
     
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