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modphysnoob

EMF

  • one year ago
  • one year ago

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  1. modphysnoob
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    |dw:1362247626272:dw|

    • one year ago
  2. modphysnoob
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    so we are supposed to find current in the loop assuming theere is a resistor there so I will start off with using magnetic force forumla q v X B but the book set it equal to Electric force

    • one year ago
  3. prakharluv
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    is there are any dimensions of loop?

    • one year ago
  4. Jemurray3
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    The induced emf in the circuit will be (ignoring the minus sign for the moment) \[ EMF = \frac{d}{dt} \Phi_B = \frac{d}{dt} B\cdot A = B \cdot L \cdot \frac{dx}{dt} = BLv\]

    • one year ago
  5. Jemurray3
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    |dw:1362247929761:dw|

    • one year ago
  6. Jemurray3
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    But \[EMF = BLv = IR \] so the current \[ I = \frac{BLv}{R} \]

    • one year ago
  7. modphysnoob
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    I see, so there is no point to be messing around the force at all?

    • one year ago
  8. Jemurray3
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    No, not really. What book is this?

    • one year ago
  9. modphysnoob
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    Fundamental of Applied Electromagnetics by ulaby

    • one year ago
  10. Jemurray3
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    I am unfamiliar with it, sorry. I don't know, unless there's more to the question.

    • one year ago
  11. modphysnoob
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    I will attach that page

    • one year ago
  12. modphysnoob
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    • one year ago
  13. prakharluv
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    ok for this loop,,, The magnetic flux linked with this loop will be, \[\phi=Blx\] Since x is changing with time, the rate of change of flux \[\phi\] will be induce emf given by : \[e=\frac{ -d \phi }{dt }= \frac{ d }{ dt } Blx\] \[e= -Bl \frac{ dx }{ dt }\] e=Blv

    • one year ago
  14. Jemurray3
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    Oh, I see. What they are saying is the magnetic force on the particles is the same as if there was an electric field \[\vec{E} = (\vec{v}\times \vec{B}) \]

    • one year ago
  15. modphysnoob
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    how did they come to that assumption?

    • one year ago
  16. modphysnoob
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    so magnetic field here would push all positive particle down , negative particle up , between them is potential difference

    • one year ago
  17. Jemurray3
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    yes. And they didn't come to any assumptions, it's fairly clear. There is a force acting on them: \[\vec{F} = q\vec{v}\times \vec{B} \] we know that for electric fields, \[\vec{F} = q\vec{E} \] so our situation is just like what we would find if there happened to be an electric field present \[\vec{E} = \vec{v}\times \vec{B} \]

    • one year ago
  18. modphysnoob
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    I understand the formula , the part that confused me is why these two force must be equal

    • one year ago
  19. modphysnoob
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    oh, I think I understand , it is one force , not two force; we are just looking at it as if it two way

    • one year ago
  20. Jemurray3
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    mhmm

    • one year ago
  21. modphysnoob
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    not quite?

    • one year ago
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