Velocity along the cylinder axis

AI Thread Summary
The discussion centers on the behavior of an electron moving along the axis of a cylinder in a magnetic field that is perpendicular to this axis. Participants analyze the implications of the Lorentz force and the resulting motion, concluding that the electron's path is circular or helical, depending on its velocity relative to the magnetic field. There is confusion regarding the equations involved, particularly the relationship between radius, velocity, and magnetic field strength, with participants correcting each other on the dimensional accuracy of these equations. The conversation also touches on the assumptions made about the magnetic field's uniformity and the relevance of gravity in this scenario. Ultimately, the complexity of the problem is acknowledged, with participants seeking further clarification and insights from others.
Istiak
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Homework Statement
A cylinder of length l, has a magnetic field B inside it, which is perpendicular to the cylinder axis. An electron of charge -e and mass m enters the cylinder with velocity v along the cylinder axis. If the radius of the cylinder is r=mveB, which of the following ensures that the electron leaves the cylinder with velocity along the cylinder axis?
Relevant Equations
##\vec F=q\vec v\times \vec B##
##\vec F=m\vec g##?
##\vec B=\frac{\mu_0 nI}{4\pi r}##
Screenshot (111).png

Since the question says that "velocity along the cylinder axis" and "magnetic field perpendicular to the cylinder axis". So cross product of velocity and magnetic field becomes their magnitude.

##\vec v\times \vec B=||v|| \\ ||B||##
So

##\vec F=qvB##
##mg=qv\frac{\mu_0 nI}{4\pi r}##
At first sight my assumption was electron can leave along the cylinder axis. But when I came up with the equation, I just realized I can't find a relation of between those given options and my results so perhaps electron can't leave along cylinder axis. While posting the question I was thinking that given ##n## in option isn't related to mine. So my attempt is completely wrong.
 
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What path does an electron take in a constant magnetic field?
 
PeroK said:
What path does an electron take in a constant magnetic field?
I don't think I understood you properly.
negative charges are attracted toward south pole and positive charges are attracted to north pole.
Field line goes south to north pole..?
 
Istiakshovon said:
I don't think I understood you properly.
You should have learned this on your course. If you search online you'll find it's a circle or helix in general.
Istiakshovon said:
negative charges are attracted toward south pole and positive charges are attracted to north pole.
I don't think so!
 
PeroK said:
You should have learned this. If you search online you'll find it's a circle or helix in general.
I had read about cycloid motion but not helical motion. 🤔
 
Istiakshovon said:
I had read about cycloid motion but not helical motion. 🤔
Cycloid motion requires additionally an electric field.
 
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PeroK said:
Cycloid motion requires additionally an electric field.
But in the question there's no electric field.
 
Istiakshovon said:
But in the question there's no electric field.
That's true! Cycloid motion was your suggestion.
 
PeroK said:
That's true! Cycloid motion was your suggestion.
Nope. I just said that I learned it rather than Helical path.
Helical path is followed when velocity is perpendicular to magnetic field. But charged particle doesn't feel any magnetic force when they are parallel to each other. So velocity of electron is along the cylinder axis. I just found two new equations while watching the video ()

pitch and radius of helix. Is there anything else which I should use to solve the problem? @PeroK
 
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  • #10
More interestingly, I don't know direction of for the given problem. How can I deal with velocity direction?
 
  • #11
I must be honest, I don't understand this question. I don't understand the equation ##r = mveB##. That's not dimensionally correct. Second, I don't where the length of the cylinder comes into the equation.
 
  • #12
Oops yep! The equation came magically :|. It's an Olympiad question. I don't know what to do and I don't think it can be solved any way. Anyway I have asked the same question in PSE (https://physics.stackexchange.com/q/692077/323311). I have also messaged an Olympiad contestant (reached IPhO), waiting for others reply :)
 
  • #13
@kuruman what can you make of this?
 
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  • #14
PeroK said:
@kuruman what can you make of this?
First some general observations on OP's relevant equations. Gravity is not mentioned explicitly so it is safe to ignore it. The magnetic field is perpendicular to the axis of the cylinder and it may be assumed uniform. So ##\mu_0 n I## is irrelevant because, I think, the cylinder is not material like a physical solenoid but intangible like a Gaussian surface.

Secondly, the radius as given may look dimensionally incorrect, but I think ##r=mveB## is to be interpreted as a simplification shortcut in the spirit of ##c=1## which is also dimensionally incorrect.

Thirdly, apparently the Lorentz force ##\vec F=-e~\vec v\times \vec B## is the only force acting on the electron. We all know that the resulting trajectory is a circle in the plane perpendicular to the magnetic field. The electron enters the cylinder "along" but not necessarily on the cylinder axis. Assume that it does so through the "back" flat face of the cylinder. It must exit also "along" which means describing a semicircle of radius ##mveB## without crossing the side of the cylinder or going out the "front" flat face first.

Fourthly, I cannot see how one gets quantization of the cylinder length if the last assumption above is correct. All it does is result in inequalities relating the length to the radius. However, it is easy to see how one gets quantization if there is a repetitive pattern in the trajectory of the kind one gets if the velocity has a component along the field or if there is an electric field perpendicular to the field and the initial velocity.

Fifthly (at this point grasping at straws), given that this is an Olympiad question and these are sometimes tricky in the sense that they are innocent-looking but expect one to think outside the proverbial box, the original statement of the problem in the photo says that the magnetic field is "constant". This means constant in time but not necessarily uniform. However, even if one knew the spatial dependence of the non-uniform field, the problem of quantization would remain.

In short, sorry but I cannot be of much help.
 
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  • #15
PeroK said:
I don't where the length of the cylinder comes into the equation.
Well, if the length were zero...

But seriously, I think the answer is Hotel California (choice b)... :smile:
 
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  • #16
PeroK said:
I must be honest, I don't understand this question. I don't understand the equation r=mveB. That's not dimensionally correct. Second, I don't where the length of the cylinder comes into the equation.
Clearly the equation should read r=##\frac{mv}{eB}##, or some dimensionless multiple of that.
We don't know which way the field goes, so we can't tell whether (initially) the force will oppose or conspire with gravity. Presumably we can therefore ignore gravity.
This leaves us with a circular path in a plane containing the cylinder’s axis.

It worries me that, other than option b, the choices seem to suggest a circular motion plus a continuing forward velocity.
 
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  • #17
haruspex said:
Clearly the equation should read r=##\frac{mv}{eB}##, or some dimensionless multiple of that.
Wait a minute, I suspect that was the case. I had seen Latex problem in that site earlier :). Where they usually wrote ##1021## is mass of a sun but the same value was looking ##10^{21}## in android device. :)
 
  • #18
According to an Olympiad contestant (reached ipho), the answer is ##I=2nr##. (asked for hint)

And he have made sure that the ##r=\frac{mv}{eB}##.
 
  • #19
Istiakshovon said:
According to an Olympiad contestant (reached ipho), the answer is ##I=2nr##. (asked for hint)

And he have made sure that the ##r=\frac{mv}{eB}##.
The assumption might be that the electron bounces elastically off the inner wall of the cylinder. I'm not sure why we would assume that, however.
 
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