How to approach work energy problems (electromagnetics problem)?

[ilovemedi]

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This is passage VII, TBR, page 160 Question 44.

I'm confused as to how to approach energy problems. I always use Einitial + Work = Efinal. To me Einitial = Electric Fiel (Eq) + Magnetic force (qvB) = Centripital force (mv^2/R) - which is the kinetic energy but in motion. I'm confused why the answer says to first solve for v from: qvb=mv^2/R and then plug that "v" to KE = 1/2mv^2. Where's the electric field in all of this? why isn't it included in the equation? basically, how do you approach these type of problems?

 

[Cawolf]

For 44 is the answer D? That's what I got when I solved it. I approached it as a uniform circular motion problem.

angular velocity = w
frequency = f

v = rw

w = 2pi*f

v = R(2pi*f) = 2*pi*R(qB/2*pi*m) = qRB/m

K = 0.5mv^2 = 0.5m(qRB/m)^2 = q^2R^2B^2/2m

For the second part I think it is B? I just eliminated the wrong answers.

C is obviously true - we see the voltage is not in the frequency formula.
D is obviously true as the voltage doesn't effect the B field.
B is false because we just saw KE depends on frequency.

It must be choice B.

 

[ilovemedi]

For 44 is the answer D? That's what I got when I solved it. I approached it as a uniform circular motion problem.

angular velocity = w
frequency = f

v = rw

w = 2pi*f

v = R(2pi*f) = 2*pi*R(qB/2*pi*m) = qRB/m

K = 0.5mv^2 = 0.5m(qRB/m)^2 = q^2R^2B^2/2m

For the second part I think it is B? I just eliminated the wrong answers.

C is obviously true - we see the voltage is not in the frequency formula.
D is obviously true as the voltage doesn't effect the B field.
B is false because we just saw KE depends on frequency.

It must be choice B.

Yes, 44 is D. But my question is, how did you know to approach as uniform circ motion? because I thought you'd need to include at least Electric Force = q/E somewhere? And for the second part, yes, it is B. But how is A false?

 

[Cawolf]

If something is moving in a circle and the frequency is given it is the simplest route to take. That problem is too complicated for me to even try to evaluate - especially because the voltage is alternating.

I just picked the easiest route - they give you frequency, so you should try to use it. I doubt we are expected to know how to calculate the velocity in a cyclotron. It just seems to complicated to look at any other way - for me at least. I try to use what is given first. Sorry I don't have better reasoning.

Also for #45 it is clear that B is false and that is enough to answer the question. A is slightly tricky but you can reason that it must be true. The voltage supplies energy to the particle and therefore increases the kinetic energy - so it would move faster and leave sooner. Since V = Potential Energy/Charge, it is the amount of energy that some charge gains as it passes through the potential.

 

[tshank]

Hi, I'm sorry I don't see how B is obviously false for question 45. If the particle has more velocity, shouldn't it also have more KE? To me, it makes B obviously correct. Please help me see the obvious. Thanks,

 

[tshank]

Or is B false because kinetic energy is the .5 mass x speed (not vector velocity)^2 ??

Or maybe because since the radius is the release point, every particle at that point will have the same Velocity since we determined that KE is proportional to the radius. Therefore since the radius is constant, the KE must also be constant.

 

[Cawolf]

redact

 

[Cawolf]

redact

 

[tshank]

Thanks for the response. So, since the voltage would only change the direction aspect of the particle, not speed, KE can't change because KE is based on speed, not directional change. Basically the voltage would change time it takes to reach that ejection velocity, but not actually change the speed itself (ie. voltage would change its acceleration). My problem was in not realizing that the KE is based on speed not direction. If KE was directional and magnitudinal velocity changes, then B would not be the correct answer - but since it is only speed, B is not true, and is the correct answer (false according to reality).

Another way to look at it, because this is angular KE, it is angular velocity - which is not the same as linear velocity.

 

[Cawolf]

redact

 

[tshank]

Question then: How does this relate to pendulums, or something where there is angular force? How do you judge KE? It can't be done in linear fashion in that case, and has to be done in torque? Is this correct?

 

[tshank]

I must be missing something then. If the voltage across the gap were doubled, the particle would move faster, therefore it would be higher KE, and be is actually TRUE, rather than NOT true. If its true, its not the right answer to the question. What did I miss here?

 

[Cawolf]

Never mind, I was looking at the wrong problem.

You can simply determine that B is false because is question 44 we determined that KE is determined by the frequency - which has no voltage term.

 

[tshank]

Okay. Tricky, tricky - would the MCAT do this? Make one answer inherently connected to a correct answer on another?

 

[Cawolf]

No idea never taken it!

 

[tshank]

Good luck when you do, you seem very bright from your posts on SDN.