TBR physics ch 8, electrostatics

[neelyboy]

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Confused on three things from the chapter.

First, page 142, passage I question 4, "For the circuit in Figure 2, which way does the induced current flow?" Am I supposed to use Lenz's law here? The explanation is very vague. I know the induced current would be counterclockwise( the correct answer) if it were ENTERING the field as the external field is 'in' and thus the system will shift to 'out' and produce a counterclockwise field ...but the rod is already immersed in the field so shouldn't the answer be that no current will flow?

Secondly, p 146, question 23. Why wouldn't it be one-fourth? I thought I was supposed to use coulombs law and do f = kq1q2/r^2. Instead the explanation is to use V= EL (I don't even know where this equation came from). Why not use coulombs here as it's two charged particles???
EDIT: then on page 152, # 3 now they use coulomb's law. How to tell when to use one or the other ?!!!


Finally, p 129 second paragraph. How do you know the torque is into the page and thus clockwise? And p 130, 8.7a why is it a counterclockwise torque? I understand the RHR for magnetic force/v/B and induced current...but I never learned one for torque. Please help.

 

[GyroGyro]

For the first question, I don't think they are using Lenz law. You know that the electrons are moving down the rolling rod thing from using F = qvb. That mean the current is moving up through the rod, making the current counterclockwise. I think that I remember missing this too thinking the same way.

For question two, you cannot use that formula because you are dealing with charged plates.
Since V is constant, and V = ED, when D increases by two, the electric field is halved.

Thus since F = qE, the force is halved. I know EK does a good job of explaining when to use each equation (like only use F = kqq / r^2 when dealing with point charges but F = qE when dealing with things such as capacitors), but I forget if TBR does the same.

For the concern on page 129, i would just look at the picture and the field lines. The - charges will want to go to areas of high potential (i.e. the starting of the field lines or the "+" area) whereas the + charges want to go to the area of low potentital (i.e. the end of the field lines or the "-" area). Thus, they would go clockwise since that would be the fastest way of reaching that. This is kind of a crappy explanation, but if you see it a lot it might make more sense.

For the question on page 130, again the - charge will be attracted to the + end of the field lines (i.e. where they originate which is from the right). The + charges will do the opposite. Thus, rotating clockwise would make sense as they traverse only 90 degrees as opposed to going counterclockwise which would involve going 270 degrees. Again, maybe this is the wrong way to think of it, but its how I have always done it and it has always worked for me.

 

[neelyboy]

Thanks for the response, just a couple more questions. For the induced current flow, I get that it moves up as directed by the RHR-1 bc the palm is facing up and the magnetic force for a positive charge is up. But how do you know if it's clockwise or counterclockwise?? Do you do RHR-2 for induced current? Even then, you can arrange your thumb up but still curl your fingers in OR out, and thus that is where I'm confused. And for the torque I think I see what you're saying, but for p. 130, how is it 270 to go counterclockwise and 90 for clockwise? If you could define that a little more/how it related I think I'd totally get it. Thanks again.

EDIT: I think I get the induced current now...is it just bc the electrOns are forced downwards, that follows a clockwise path? And since current is defined for positive charge flow it is just the reverse?? Hence the answer is counterclockwise??

 

[BerkReviewTeach]

First, page 142, passage I question 4, "For the circuit in Figure 2, which way does the induced current flow?" Am I supposed to use Lenz's law here? The explanation is very vague. I know the induced current would be counterclockwise( the correct answer) if it were ENTERING the field as the external field is 'in' and thus the system will shift to 'out' and produce a counterclockwise field ...but the rod is already immersed in the field so shouldn't the answer be that no current will flow?

• The bar is sliding to the right, so you can determine the direction of the current in one of two ways. The first is to apply the right hand rule to the moving rod. Your thumb points to the right (direction of the rod), your index finger points into the page (direction of the magnetic field), and your thumb ends up pointing up the rod. This means that the induced current will push charge up the rod, resulting in a counter clockwise current that passes through the resistor. You could have also used Lenz's law, noting that the B field pointing into the page within the loop is increasing as the rod moves to the right. A counterclockwise current would be generated to create a B field pointing out of the page within the loop to offset the increasing flux withint the loop. I like the Right-hand rule method better, but both apporaches work.

Secondly, p 146, question 23. Why wouldn't it be one-fourth? I thought I was supposed to use coulombs law and do f = kq1q2/r^2. Instead the explanation is to use V= EL (I don't even know where this equation came from). Why not use coulombs here as it's two charged particles???

• Coulomb's law applies to the force between point charges, which is NOT the case here. In this question, it is a single charge existing between two parallel charged plates (like a capacitor or the plates in gel electrophoresis). The force on the particle in the field is F = qE. It is the plates that are moved apart from one another, not the particle, so the field weakens. The key caveat is that the voltage is held constant, so the field strength decreases solely because of the greatest distance between the plates. The distance is doubled, so the field strength is cut in half, resulting in half the force. Something you might want to take note of, ESPECIALLY on the real MCAT, is that questions with that much word usually have a bunch of stipulations in place so that only two things are changing.

Finally, p 129 second paragraph. How do you know the torque is into the page and thus clockwise? And p 130, 8.7a why is it a counterclockwise torque? I understand the RHR for magnetic force/v/B and induced current...but I never learned one for torque. Please help.

• The key thing is knowing that field lines point the way a (+)-charge would migrate. In Figure 8-10, the (+)-charge is being pulled to the right and the (-)-charge is being pulled to the left by the field. Because the two charges are connected, there is a tension between them. The (+)-charge being above the (-)-charge will be pulled down by the tension force and the (-)-charge being below the (+)-charge will be pulled up by the tension force. The result is that the (-)-charge is pulled left and up while the (+)-charge is pulled right and down, causing a circular path in the clockwise direction (hands on a clock between 12 and 3 travel down and right while hands on a clock between 6 and 9 travel up and left). Example 8.7a is the opposite arrangement of the dipole AND the opposite arrangement of the field lines, so you should get the same type of rotation as you observed in the example in Figure 8-10.

Hopefully these explanations help.

 

[neelyboy]

Wow thank you, that was definitely a great help. I understand everything now except for the first question...I know if the rod were JUST entering the field as in p 136, figure 8-15, t = 1 then it would act to oppose the original field...but if it is already immersed in the field as it looks, then wouldn't it have no current like at t= 2 of figure 8-15. I mean I know the rod is moving through the field and getting more X's but isn't it already immersed??

• Coulomb's law applies to the force between point charges, which is NOT the case here. In this question, it is a single charge existing between two parallel charged plates (like a capacitor or the plates in gel electrophoresis). The force on the particle in the field is F = qE. It is the plates that are moved apart from one another, not the particle, so the field weakens. The key caveat is that the voltage is held constant, so the field strength decreases solely because of the greatest distance between the plates. The distance is doubled, so the field strength is cut in half, resulting in half the force. Something you might want to take note of, ESPECIALLY on the real MCAT, is that questions with that much word usually have a bunch of stipulations in place so that only two things are changing.

• The key thing is knowing that field lines point the way a (+)-charge would migrate. In Figure 8-10, the (+)-charge is being pulled to the right and the (-)-charge is being pulled to the left by the field. Because the two charges are connected, there is a tension between them. The (+)-charge being above the (-)-charge will be pulled down by the tension force and the (-)-charge being below the (+)-charge will be pulled up by the tension force. The result is that the (-)-charge is pulled left and up while the (+)-charge is pulled right and down, causing a circular path in the clockwise direction (hands on a clock between 12 and 3 travel down and right while hands on a clock between 6 and 9 travel up and left). Example 8.7a is the opposite arrangement of the dipole AND the opposite arrangement of the field lines, so you should get the same type of rotation as you observed in the example in Figure 8-10.

Hopefully these explanations help.[/QUOTE]

 

[neelyboy]

Bump. I'm still confused about the p 142 question # 4 question. I understand that the electrons flow down the rod and thus the current flows up. But how do you tell if it's clockwise or counterclockwise? Would I move my hands a certain way? And if you look at it from Lenz's law's perspective...why does the B it encounters increase? Isn't it already immersed as the rod is not leaving or entering and hence shouldn't no flow of current occur?

 

[whiteshadodw]

Bump. I'm still confused about the p 142 question # 4 question. I understand that the electrons flow down the rod and thus the current flows up. But how do you tell if it's clockwise or counterclockwise? Would I move my hands a certain way? And if you look at it from Lenz's law's perspective...why does the B it encounters increase? Isn't it already immersed as the rod is not leaving or entering and hence shouldn't no flow of current occur?

i would completely forget Lenz's law for the MCAT. its not on the AAMC PS topics list. if anything, its just tripping you up. this question was pretty simple i feel, except you had to remember that current by some stupid rule, flows in the direction opposite to electron flow. if you used right hand rule, the electrons would flow clockwise, which means current flows counterclockwise.

 

[hdub]

TBR Chapter 8 Physics, Passage 3 Q # 11 (In Practice Exam Portion)

I need some things set straight.
1) How do I know if the magnetic field points into the page or out of the page when all that I am given is an electric field that goes from + to - (ie point down therefore Electric Force will be down); Am i just supposed to assume that out of the page for a cathode ray tube?! I don't think so.. so fill me on on what I missing. And in general, how do I determine the direction of the magnetic field when I am just given an Electric Field vector..

~The only thing I can think of is that to for a q to be undeflected Felectric=Fmagnetic so if electric force down, then magnetic force should be up because they must be equal & opposite.. then when velocity is too fast Fb>Fe and will be deflected in direction of Fb (obviously opposite for an electron because these are for positive q references) = why answer B not A??