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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?
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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theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Creating multiple names is also against the Code of Conduct at http://openstudy.com/codeofconduct 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?

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I might be able to help with this one. Do you understand the setup?

qsx
 2 years ago
Best ResponseYou've already chosen the best response.0there are some problems going around, i want to just finish it as soon as possible so that some problems can reduced

qsx
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, i alo attached the picture of the setup

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1So, you understand how the EMF is produced?

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Well, 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.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1http://hyperphysics.phyastr.gsu.edu/hbase/magnetic/magfor.html#c2 Lorentz Force Law, this is!

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I'm trying to find how it's applied to the rod..

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I would invite people to this question, but no one I fanned is on...

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I sent a couple emails 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.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Like, I know what direction the charges move in.. And what force each charge has.. But my mind isn't beyond that.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I have an idea, but I'm not sure it's valid, so I won't say it in case it's wrong.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Oh, thank goodness! Easier than I planned: http://hyperphysics.phyastr.gsu.edu/hbase/magnetic/genwir2.html

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1So 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?

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Cool! So now you can calculate the EMF! That's your part a.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Well, Lorentz Force Law will tell you which way the electrons will be forced! http://hyperphysics.phyastr.gsu.edu/hbase/magnetic/magfor.html#c2 Look to the right hand rule.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Let me know what you think on that.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Alright!dw:1374787554478:dw Use my drawing, and draw the magnetic field and the velocity of the rod.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I have to go. Take care, and good luck!

Festinger
 2 years ago
Best ResponseYou've already chosen the best response.3I 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\[F_{wire}=IL\times B\]

Deba_001
 2 years ago
Best ResponseYou've already chosen the best response.0(a) e= bvl (b) direction is found by using R.H.Thumb rule (c) Heat dissipation = B2*v2* L2/R

qsx2
 2 years ago
Best ResponseYou've already chosen the best response.0This 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

qsx2
 2 years ago
Best ResponseYou've already chosen the best response.0@Festinger are my answers correct?

Festinger
 2 years ago
Best ResponseYou've already chosen the best response.3I 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
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