GR9677 #86


Problem


This problem is still being typed. 
Electromagnetism}Particle Trajectory
There is a force pointing upwards from the Electric field in the ydirection. Suppose the particle is initially moving upwards. Then, the magnetic field would deflect it towards the right... One can apply the Lorentz Force to solve this problem.
If the particle comes in from the left, then the magnetic force would initially deflect it downwards, while the electric force would always force it upwards. Continue applying this analysis to each diagram. It turns out that one has cycloid motion whenever the electric and magnetic fields are perpendicular.


Alternate Solutions 
flyboy621 20101105 22:24:06  With a Bfield pointing out of the page, the particle must be turning right whenever it is moving. That leaves only (B) and (C), but (C) would be the trajectory if there were no E field. Therefore (B) must be correct.  

Comments 
rav 20140924 11:09:58  I'm almost certain that there is an error in the solutions. In section 5.1.2. of Griffiths EM he goes over the motion of moving charged particles in B fields.
If the direction of the velocity is only in a direction perpendicular to B, then the trajectory of the particle is a circle.
If the velocity has a component parallel to the B as well as perpendicular, then we get helical motion.
If there is an Efield and a Bfield which are perpendicular to each other, then we get cycloid motion. This question addresses this case, so the answer should be the cycloid trajectory, E.
amber 20141021 16:18:57 
I agree. The answer is E
The particle is initially at rest, so no magnetic force acts on the particle. The electric force acts on the particle first and then the magnetic field can act on the moving particle.

amber 20141021 16:25:04 
oh wait never mind. I didn't read the question correctly

dc771957 20161024 22:53:42 
If the particle were initially at the origin, or somewhere else along the xaxis, at rest, it would follow motion similar to E only it would be bouncing along the xaxis, not the yaxis. Initially v=0 so the force due to the magnetic field is zero, as the particle begins to accelerate in the +y direction due to the force from the electric field, it begins to feel a force from the magnetic field, which will initially make the particle accelerate in the +x direction. For this setup, the magnetic force will always be trying to force the particle to make a right turn. As the particle begins to move in the +x direction, the magnetic field will start to pull the particle in the y direction, keep in mind that the electric field is still acting in the +y direction. As the particle begins to return to the xaxis, with a velocity in the y direction, following a similar path shape from answer E, it will slow down because the electric field is pulling it upwards. The particle will have enough kinetic energy to resist the pull of the electric field until it returns to the xaxis, where it will briefly stop, before repeating this process. \r\nIf you take a look at answer E, you will see that the magnetic and electric fields will be pulling the particle in the +y direction when it intersects with the yaxis; the electric field always acts in +y, and the particle will have a velocity in the x direction with magnetic field still in the +z direction. So, no, E is not the correct answer, but that didnt stop me from picking when I first looked at this problem.

  superderec 20101127 12:08:33  it's best to suppose that the particle is fired in the xy plane as a projectile (at an angle with the xaxis and with a low velocity) towards the xaxis .As mentioned it is moving only in the xy plane. Since the Efield is in the +y direction, the particle will be repelled by the positively charged plate near the axis producing the Efield. so it never reaches the xaxis. (there is a minimum x) and the particle is being pushed by the Bfield to perform some circular motion. Only possible answer is B. (The concept of left and right and upwards in Yosun's answer does not make sense)   flyboy621 20101105 22:24:06  With a Bfield pointing out of the page, the particle must be turning right whenever it is moving. That leaves only (B) and (C), but (C) would be the trajectory if there were no E field. Therefore (B) must be correct.   his dudeness 20100730 07:15:50  I found it easiest to just draw a diagram of the forces felt by the particle. The Efield always causes a force in the+y direction. When the particle starts moving in the +y direction, the Bfield causes a force in the +x. When it is moving in the +x direction, the Bfield causes a force in the y, etc. Qualitatively, the motion of the particle is thus described by (B).   gt2009 20090527 01:34:28  The correct answer (B) is a helix. Cycloid motion would be (E), which is wrong.
claire 20091009 14:21:19 
Both B and E are cycloid, e is just a special case.
The problem states "a positively charged particle is moving in the xyplane." The motion is in the xy plane, and thus cannot be a helix (which requires motion in all 3 directions)

nimesh.nic 20111101 06:34:07 
exectaly that's what I thought .......

  PhyAnnie 20081104 06:07:49  I was wondering... since the particle is always under the constant electric force of E*q upward, why the answer B doesn't show a upward trend, instead of being horizontal. That's strange.
f4hy 20090403 19:13:19 
It does seem strange, but it is the behavior when B and E are perpendicular. I don't remember which book but one makes this exact point when talking about cyclotrons.

WallStreet 20140907 20:00:08 
think about this, the B field gives a force of x .
If a positive charge is moving to the right, the B field gives a force in the y direction proportional to the velocity. So a particle moving at just the right velocity would show no upward drift (the force from the magnetic field would counter the force from the electric field.
But this only occurs for a special velocity, if the velocity is different, we will end up with different motion.

  greed 20081015 11:28:00  Doesn't the cycloid motion correspond to a drift velocity that is proportional to ? This would make the drift in the x direction, but choice D shows a drift in y. little confused ...
greed 20081015 11:30:34 
oops, never mind: just checked the answers

  cordercom 20081005 17:58:59  I think we can use the Poynting vector in this case. The net flux of energy density and hence the net trajectory of the particle must be in the direction of E cross B= (y X z)=x. So, the net displacement of the particle must be toward the right. Answer (B) seems most appropriate.
See Griffiths, Intro to Electrodynamics 3rd Edition, pg 206 & 347
physicsisgod 20081030 16:34:35 
The Poynting vector describes the direction of energy propogation for the EM field. It says nothing about the energy of particles moving in the EM field. This is a lucky coincidence.

  buddy.epson 20061014 14:27:13  the +z direction is determined by the cross product of +x and +y (aka the righthandrule).   andreas 20051110 05:00:51  And for those who, like me, are confused by conventions, the magnetic field in the +z direction means it comes out of the page.
(I'm used to defining +z to be away from the observer, i.e. into the page)
shafatmubin 20091105 14:13:55 
Use cross product: i x j = k

 

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