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Question about the right hand rule

  1. May 3, 2005 #1
    Hello.

    I have a question about RHR #2 (Force in a conductor).

    If IL X B = F and
    If the fingers represent the direction of the current, the thumb represents the direction of the force and the palm is the direction of the magnetic field, how come the magnetic field can be at an angle other than 90 degrees from IL? In such a case, does the RHR still apply? For example, how would I find the direction of the force on side A (Bottom wire ) in this case:

    http://img.photobucket.com/albums/v228/LianaBlank/untitled.bmp

    B is on the positve X direction (palm)... but IL is at 60 degrees from B.. How do I use my Right Hand rule to find F?
     
  2. jcsd
  3. May 3, 2005 #2

    OlderDan

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    "palm" is the problem. B is not in the direction of the palm. The palm is merely the direction toward which you must curl your fingers to get from the direction of L to the direction of B through the smaller angle. That angle can be anywhere between 0 and 180 degrees
     
    Last edited: May 3, 2005
  4. May 3, 2005 #3

    Doc Al

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    The magnetic field can be anywhere it wants to be with respect to IL, of course. But F will always turn out to be 90 degrees from both IL and B.
    Sure the right hand rule applies. It may be easiest to just consider the component of IL perpendicular to B (and ignore the component parallel to B, since it will create no force!).
     
  5. May 3, 2005 #4

    Doc Al

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    I'm glad you mentioned that Dan (I was just going to myself).

    The "right hand rule" that I use is the one that Dan describes. To find [itex]\vec{A}\times\vec{B}[/itex], I "curl" my fingers from A to B... my thumb indicates the direction of the cross-product.

    (There are several versions of the right hand rule; some are better than others. :smile: )
     
  6. May 3, 2005 #5

    Doc Al

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    Staff: Mentor

  7. May 3, 2005 #6
    Thanks Doc... I'll check it out
     
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