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?

Question

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?

anonymous · Answer

attachment

anonymous · Answer

@theEric

theeric · Answer

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?

theeric · Answer

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

anonymous · Answer

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

anonymous · Answer

Yes, i alo attached the picture of the setup

theeric · Answer

So, you understand how the EMF is produced?

anonymous · Answer

nope

theeric · Answer

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.

theeric · Answer

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

anonymous · Answer

ohh, i got the concept

theeric · Answer

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

theeric · Answer

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

anonymous · Answer

please

theeric · Answer

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.

theeric · Answer

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

theeric · Answer

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

theeric · Answer

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

theeric · Answer

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?

anonymous · Answer

yes

theeric · Answer

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

anonymous · Answer

part b?

theeric · Answer

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.

theeric · Answer

Let me know what you think on that.

theeric · Answer

think about* that.

anonymous · Answer

i need more help :(

theeric · Answer

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

theeric · Answer

Hmm?

theeric · Answer

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

festinger · Answer

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$$F_{wire}=IL\times B$$

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

@Fifciol

anonymous · Answer

@Festinger 
are my answers correct?

festinger · Answer

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