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Effect of Hall effect on resistance

  1. Jun 29, 2011 #1
    The charge carriers in a current carrying wire subjected to a magnetic field will move to a side due to the Hall effect. But doesn't that also decrease the effective cross section area through which the charge carriers are moving? Does the resistance increase? If so, how significant is it in electric machines like motors?
     
  2. jcsd
  3. Jun 30, 2011 #2
    74 views, but no replys. But, I need this answer:(
     
  4. Jun 30, 2011 #3

    berkeman

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    Staff: Mentor

    What is the context of the question? Is it for schoolwork?
     
  5. Jun 30, 2011 #4
    No, not for school work. It's a burning curiosity. School is closed for the summer, so it's not easy to ask the professors. That's why I am here. I can't seem to wrap up the theories to actually carry out the calculations (I am a physics major and have done just one ENM course). I have a coil with high enough resistance to run the test from a home power supply and multimeters, but don't have a magnet to run the test. I will appreciate any sort of answer and an explanation- with or without the math.

    If the effect was significant, people could have made more efficient motors with coils of rectangular cross section. I have never seen such a motor, so I presume the effect is not significant. But, I am still don't feel satiated without a proper explanation.
     
  6. Jun 30, 2011 #5

    berkeman

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    I'm no expert on the Hall Effect, but I believe that it is pretty low in metals because the density of conduction electrons is high. Note how the Hall Voltage ratios inversely with the density of the carriers:

    http://en.wikipedia.org/wiki/Hall_effect

    BTW, square cross-section wires are sometimes use in coils (I don't know about motors), because the coil packing is more efficient.
     
  7. Jun 30, 2011 #6
    Thanks. I still have a lot to understand. What I am suggesting is, due to the Hall effect, the current in the coil will decrease. The same number of charge carriers have to pass through a smaller area (the carriers are getting deflected to a side). I=neA(Vd). "A-Area" will decrease, but (Vd-drift velocity) will not increase proportionally as the carriers have more obstructions to their motion.

    I cannot relate it to the Hall voltage formula. Perhaps I need to start at a more basic level, and carry out some weird integrations. I don't know. I will keep on reading. Thanks.
     
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