qsx1 2 years ago 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?

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1. qsx1

2. qsx2

@theEric

3. theEric

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?

4. theEric

I might be able to help with this one. Do you understand the setup?

5. qsx

there are some problems going around, i want to just finish it as soon as possible so that some problems can reduced

6. qsx

Yes, i alo attached the picture of the setup

7. theEric

So, you understand how the EMF is produced?

8. qsx

nope

9. theEric

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.

10. theEric

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html#c2 Lorentz Force Law, this is!

11. qsx

ohh, i got the concept

12. theEric

I'm trying to find how it's applied to the rod..

13. theEric

I would invite people to this question, but no one I fanned is on...

14. qsx

15. theEric

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.

16. theEric

Like, I know what direction the charges move in.. And what force each charge has.. But my mind isn't beyond that.

17. theEric

I have an idea, but I'm not sure it's valid, so I won't say it in case it's wrong.

18. theEric

Oh, thank goodness! Easier than I planned: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir2.html

19. theEric

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?

20. qsx

yes

21. theEric

Cool! So now you can calculate the EMF! That's your part a.

22. qsx

part b?

23. theEric

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.

24. theEric

Let me know what you think on that.

25. theEric

26. qsx

i need more help :(

27. theEric

Alright!|dw:1374787554478:dw| Use my drawing, and draw the magnetic field and the velocity of the rod.

28. theEric

Hmm?

29. theEric

I have to go. Take care, and good luck!

30. Festinger

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$F_{wire}=IL\times B$

31. Deba_001

(a) e= bvl (b) direction is found by using R.H.Thumb rule (c) Heat dissipation = B2*v2* L2/R

32. qsx2

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

33. qsx2

@Fifciol

34. qsx2