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Particle in Magnetic Field

by KleZMeR
Tags: field, magnetic, particle
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KleZMeR
#1
Nov2-07, 04:35 PM
P: 59
1. The problem statement, all variables and given/known data

In a purley magnetic field B, motion of particle in x-y plane is circle; use this property, with result from part c to show on average the particle travels with constant velocity U:
(answer given): U=(1/B^2)*B*(B*Vo)+Vdrift
where Vdrift=(1/B^2)(ExB)


2. Relevant equations
"x" implies cross product
E=0i+Ej+0k
B=0i+0j+Bk
F=q(E+VxB)
results from Part c) showed by Galilei transform:
E'=(E+UxB)
V'=(V-U)



3. The attempt at a solution

Well right now i'm solving for U. I also have the answer so i've plugged in vdrift and try to work that out, usually "show" means working backwards in this course. I am not sure if I should use trig functions for the circle, I mean, it says to use that property of pure magnetic feild, I know at one point i have to incorporate the trig rotation because in part f it says my x(t) and y(t) are sin and cos functions.
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KleZMeR
#2
Nov2-07, 07:02 PM
P: 59
well i'm a little farther, the equations of circular motion are what i'm having a little problem with, do they depend on the lorenz force? or is it just:
Wc= Omega Cyclotron
Components:
V'x=WcVy
V'y=-WcVx

With these I can de-couple the equation, but with the equation of motion dependent on the lorenz force i get V'y=(q/m)(E-BVx) and i do not know if i can decouple these because it is two terms...

i think obvious solution to use is u=v-v' so i'm looking for these two i guess? but it says to use result from part c) that E'=(E+UxB) , but i got this from the u=v-v' relationship so i think this is "'the result" i should be using??
KleZMeR
#3
Nov2-07, 08:56 PM
P: 59
ok so pure magnetic field means no E??? there is E in Vdrift so i wonder if E stays in equation of motion

KleZMeR
#4
Nov2-07, 10:08 PM
P: 59
Particle in Magnetic Field

ok, well i found in this forum somewhere that E'(y)=V*B'(z), is that right???
Gokul43201
#5
Nov3-07, 02:33 AM
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I don't know why the given answer has a drift speed coming from an E-field, when the problem states that there is no E-field. Also, what is the "result from part c"?

A particle traveling along a circle clearly can not have a constant velocity. The simplest way to show that it has a constant speed is to show that the net force on the particle is normal to its velocity, and this is trivially true for this problem.
KleZMeR
#6
Nov3-07, 12:21 PM
P: 59
ok, I did state what the result from part c) is in the problem: results from Part c), I showed by Galilei transform:
E'=(E+UxB)
V'=(V-U)

right now I am working in the drift frame, so maybe thats why the extra term is there, but I emailed my professor and asked him the same question Gokul43201 asked (why is there an E in purely magnetic field and where does Vdrift come from), have no response

I thought "on average" might imply some sort of average value function that should be applied? that's what the answer looks like now that i think about it. an integral of (1/B) might return a (1/B^2) function,

the next two steps are integrating equations of motion in lab frame (no drift) and finding V(t), r(t), but these two X(t) and Y(t) also incorporate a Vdrift as coefficient before the trig functions, this Vdrift takes the place of Vo, but Vo is said to be zero in lab frame? this is kindove off topic, but gives some insight into the two different situations.
KleZMeR
#7
Nov3-07, 12:25 PM
P: 59
ok, so i use rule A(dot)B = |A||B| cos(theta) and show theta at 90 always????


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