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qsx1 Group Title

2. Two horizontal conducting rails are connected by a resistor R = 0.750 , as illustrated below. There is a uniform magnetic field B = 0.120 T pointing vertically downward. A conducting rod of length l = 2.20 m moves to the left along the two rails at a constant speed v = 3.90 m/s. Assume that resistance between the rod and rails is negligible. (a) What is the induced emf in the circuit? (b) What is the magnitude and direction of the current flow? (c) What is the rate of heat dissipation in the resistor? (d) What force is required to keep the rod moving at its constant speed?

  • one year ago
  • one year ago

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  1. qsx1 Group Title
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    • one year ago
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  2. qsx2 Group Title
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    @theEric

    • one year ago
  3. theEric Group Title
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    Creating multiple names is also against the Code of Conduct at http://openstudy.com/code-of-conduct I bet you want to be efficient, but this is not the way to do it. You can only learn one problem at a time. Why complicate it by trying to learn two?

    • one year ago
  4. theEric Group Title
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    I might be able to help with this one. Do you understand the setup?

    • one year ago
  5. qsx Group Title
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    there are some problems going around, i want to just finish it as soon as possible so that some problems can reduced

    • one year ago
  6. qsx Group Title
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    Yes, i alo attached the picture of the setup

    • one year ago
  7. theEric Group Title
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    So, you understand how the EMF is produced?

    • one year ago
  8. qsx Group Title
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    nope

    • one year ago
  9. theEric Group Title
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    Well, when charges,\(\qquad\) like those in the rod move with respect to an electric field,\(\qquad\) like the rod is in the uniform magnetic field then the charges will experience a force, and thus there is a current. The EMF is the voltage of the electrons, I believe. I'm finding equations now.

    • one year ago
  10. theEric Group Title
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    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html#c2 Lorentz Force Law, this is!

    • one year ago
  11. qsx Group Title
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    ohh, i got the concept

    • one year ago
  12. theEric Group Title
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    I'm trying to find how it's applied to the rod..

    • one year ago
  13. theEric Group Title
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    I would invite people to this question, but no one I fanned is on...

    • one year ago
  14. qsx Group Title
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    please

    • one year ago
  15. theEric Group Title
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    I sent a couple e-mails to people who are on and have answered many physics questions. I'll go back to seeing if I can learn how to solve this, now.

    • one year ago
  16. theEric Group Title
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    Like, I know what direction the charges move in.. And what force each charge has.. But my mind isn't beyond that.

    • one year ago
  17. theEric Group Title
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    I have an idea, but I'm not sure it's valid, so I won't say it in case it's wrong.

    • one year ago
  18. theEric Group Title
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    Oh, thank goodness! Easier than I planned: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir2.html

    • one year ago
  19. theEric Group Title
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    So the EMF is\[EMF=v\ B\ l\ sin(\theta)\]where \(\theta\) is the angle between the velocity and magnetic field directions. Now we can move on, thank goodness. You there?

    • one year ago
  20. qsx Group Title
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    yes

    • one year ago
  21. theEric Group Title
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    Cool! So now you can calculate the EMF! That's your part a.

    • one year ago
  22. qsx Group Title
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    part b?

    • one year ago
  23. theEric Group Title
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    Well, Lorentz Force Law will tell you which way the electrons will be forced! http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html#c2 Look to the right hand rule.

    • one year ago
  24. theEric Group Title
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    Let me know what you think on that.

    • one year ago
  25. theEric Group Title
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    think about* that.

    • one year ago
  26. qsx Group Title
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    i need more help :(

    • one year ago
  27. theEric Group Title
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    Alright!|dw:1374787554478:dw| Use my drawing, and draw the magnetic field and the velocity of the rod.

    • one year ago
  28. theEric Group Title
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    Hmm?

    • one year ago
  29. theEric Group Title
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    I have to go. Take care, and good luck!

    • one year ago
  30. Festinger Group Title
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    I did a derivation for the formula to use here: http://openstudy.com/study#/updates/51f0facfe4b00daf471b20fe For force it is an experimental fact that \[F=qv\times B\] But the force exerted on the electrons is proportional to the speed the electrons move in the conductor, not the velocity the rod moves at. Current is \[I=nqAv_{d}\] where vd here is the drift velocity of the electrons (speed they move in current), n is number of free electrons per unit volume, A is cross sectional area of the wire and q is the charge of each electron. Now, the force on the wire is the sum of all the electrons in the conductor: \[F_{wire}=nALqv_{d}\times B\] Which reduces to\[F_{wire}=IL\times B\]

    • one year ago
  31. Deba_001 Group Title
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    (a) e= bvl (b) direction is found by using R.H.Thumb rule (c) Heat dissipation = B2*v2* L2/R

    • one year ago
  32. qsx2 Group Title
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    This is what i got so far a) = vBl = 3.9*0.12*2.2 = 1.03V b)Direction of the current flow is from left to right. I = /R = 1.03/0.75 = 1.37A c) Rate of heat dissipation, P = I2R = 1.41W d) Force = IBl = 0.36N

    • one year ago
  33. qsx2 Group Title
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    @Fifciol

    • one year ago
  34. qsx2 Group Title
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    @Festinger are my answers correct?

    • one year ago
  35. Festinger Group Title
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    I am not you sure what you mean by left to right, but the direction of the current is anti clockwise when viewed from the top. otherwise everything else looks fine. There are a few ways to get force for this question. 1: F=ILB 2: Power = Fv

    • one year ago
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