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Please help me understand movement of electron in a magnetic field

  • Thread starter michaelw
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http://www.vaxop.com/pic.GIF [Broken] is my little picture

what is drawn is a magnetic field pointed N to S (left to right)
the helical path shown is the path of an electron taken in that field

my question is, why is that so?
at the far left, the magnetic field is ~constant, and according to the RHR, the electron (which has a parallel and perpendicular velocity component) will have its perpendicular component moved clockwise

then the magnetic field decreases as u approach the center.. heres where i get confused.
since its decreasing, the electron will move in a way to increase the magnetic field to the right (lenz's law) and thus will continue to move clockwise (is this right?? or is the movement just determined by the magnetic field already present)

then the magnetic field increases towards the right
the electron will move to counter this, and induce a magnetic field of its own pointing to the left.. thus it *should* move counter clockwise (but this isnt whats seen in the diagram)

where is my thinking flawed?

I remember if you have a closed wire, and you move it into a magnetic field, there will be a magnetic field created to oppose the magnetic field, thus current will run clockwise (if the original magnetic field is out of the page, the induced will be into the page) and the opposite is true if the flux is decreasing..
shouldnt the same thing be happening here?

also, as the magetic field increases, what will happen to the radius and the magnitude of the velocity perpendicular to the field? since Fb = qvB = ma, acceleration obviously must change, but a = mv^2/R. so will the velocity change, or the radius? or will changing the velocity induce a change the radius? or will a changing radius induce a change in velocity? (by change in, i mean change in magnitude)

this is quite confusing..
 
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An electron follows a helical path when it enters into the magnetic field at an angle between 0 and 90 degress ,

Lets say the electron enters at an angle Q , now the velocity of the electron will have two components , one of which will be prependicular to the field and another parallel to the field.

The one which is prependicular to the field will help electron acquire a circular path (remember in your previous post,what happened when electron entered prepepndicular to the field??) , and the component of the velcoity [itex]V cosQ[/itex] will suffer no force due to the field because V x B = 0 this time . So as a whole the lectron alose acquires a circular path + travels in staraight line .The resultant path is depicted in the animation given in your post.
 
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then the magnetic field decreases as u approach the center.. heres where i get confused.
since its decreasing, the electron will move in a way to increase the magnetic field to the right (lenz's law) and thus will continue to move clockwise (is this right?? or is the movement just determined by the magnetic field already present)
wHAT HAPPENED IN YOUR PREVIOUS POST , WHEN MAGNETIC FIELD WAS BEING INCREASED?... the radius kept on decreasing but the sense of the motion remained the same.

In this case when magnetic field goes on decreasin , radius goes on increasing , thus the electron forms a bigger loop this time with a bigger area , with a brand new bigger area , the flux through it increases as per Lez Law to counteract the decreasing magnetic field.
 
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ah ha!
thanks that helped a lot :)
so.. please tell me this next statement is right

to compensate for the decreasing magnetic field, the electron increases its radius to keep the flux the same..
since B = mv/qR, and B and R are decreasing, v must also decrease (by more than R)

is that right?
 
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michaelw said:
ah ha!
thanks that helped a lot :)
so.. please tell me this next statement is right

to compensate for the decreasing magnetic field, the electron increases its radius to keep the flux the same..
since B = mv/qR, and B and R are decreasing, v must also decrease (by more than R)

is that right?

Nothing will happen to v, remember magnetic field will never do any work on the electron , the v remains the same . As R=MV/QB , As B decreases , R increases. got it? ...I think we talked a lot about it in your previous post.Please read the last post again.
 

OlderDan

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Dr.Brain said:
Nothing will happen to v, remember magnetic field will never do any work on the electron , the v remains the same . As R=MV/QB , As B decreases , R increases. got it? ...I think we talked a lot about it in your previous post.Please read the last post again.
I think we need to come to some consensus here about what magnetic fields can and cannot do to a moving charge. In the last problem I pointed out that a time varying magnetic field induces an electric field that does work on the charge. In the strictest sense, I agree with you that a magnetic field can do no work on a charge. What I believe was overlooked by everyone in the previous problem was the electric field associated with the explicit time dependence of the magnetic field.

If my view of the previous problem is correct, then the OP needs to understand why this problem is different, not the same. In this problem, there is no time varying magnetic field. Rather, in this case a charge is moving from place to place experiencing a magnetic field that varies with position, but not with time, which is a fundamentally different situation as viewed by a stationary earth obsever. In this problem no work is done on the charge because it experiences different magnetic fields in different places. In the previous problem work was done, in my view, because the magnetic field itself was varying with time.

I don't think anything is gained by viewing this problem in terms of magnetic flux linking the path of the particle, or the field it needs to generate to resist a change in that flux. It should be viewed, as you are suggest, in terms of the magnetic force acting on the particle to change its direction of motion without changing its velocity.
 
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ah

ok i think i understand now
so basically this can all be solved using
R=MV/QB

that if b decreases, r increases, and vice versa, and speed remains the same
 

OlderDan

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michaelw said:
ah

ok i think i understand now
so basically this can all be solved using
R=MV/QB

that if b decreases, r increases, and vice versa, and speed remains the same
In this problem, yes, at least for the circular part of the motion. Depending on how detailed you want to get, you can consider the effect on the pitch of the helix from two things: 1) A larger radius in a weaker field for the particle travelling at constant speed would mean longer time to go around one full "orbit", which would lengthen the pitch if the velocity from left to right in your diagram is constant. 2) The direction of the magenetic field is more or less left to right, but does have a radial component away from or toward the axis of symmetry in your diagram. What direction is the force on the particle from this radial component of the field? You can decide if you need to pursue this.
 

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