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Polarity of a horseshoe shaped solenoid

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data

    polarity of a horseshoe shaped solenoid given symbol for power source large line small line.
    Current flows through the right end of the horsehoe and out of the left end.

    The wire is wrapped in a CW direction if looking at the left end of the magnet.
    The wire is wrapped in a ACW direction if looking at the right end of the magnet.

    Im confused that the current direction using right hand grip rule and wrap direction of the coil viewed from the ends are giving the opposite conflicting result.

    2. Relevant equations

    none

    3. The attempt at a solution

    Assuming conventional current from positive (large line) to negative (small line).

    Viewing the right end current direction is ACW so this should be North
    Viewing the left end current direction is CW so this should be South

    Is this correct or should I consider the rules as if the horseshoe was broken in half like a wishbone?

    Thanks..
     
  2. jcsd
  3. Apr 10, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Abigaile! Welcome to PF! :smile:
    No. :confused:

    Wrap your right hand around the left end of the solenoid, so that your fingers are curling the same way as the current …

    your thumb then points into the solenoid.

    Now slide your hand all the way round the solenoid to the other end …

    your thumb now points out of the solenoid …

    isn't that what you expect, one end north and the other end south? :smile:
     
  4. Apr 10, 2012 #3
    Re: Welcome to PF!

    Hi Tim :smile:

    does that mean left end is south and right end is north?:wink:
     
  5. Apr 10, 2012 #4

    tiny-tim

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    Yes! :smile:

    (or the other way round … i can't remember o:))
     
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