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Anyone like EM?

  • Thread starter happy thoughts
  • Start date
  • #1
happy thoughts
there is a sraight wire carrying current i. an electron is approaching the wire, in the same plane as the wire, at a 45 degree angle with velocity v0 when it is a distance r0 from the wire. the electron makes it to a distance of rf from the wire before it is repelled away. what is i?

I was thinking that I know that K.E.=Me(v0*sin(45))/2 initially in the y direction (this would be the y-component of K) if y is taken as the coordinate perpendicular to the wire and x the one parallel to the wire. K.E.=0 once the electron is at a distance of rf. I also know that Work=integral Fdr in the y direction.

so here is an equation that should work:

K.E.=Me*v0*sin(45)/2=[inte][e*vx*mu0*i/(2*pi*r)]dr from r=r0 to r=rf

e=charge of electron
mu0=permeability of free space
vx=x-component of velocity
i=current i'm trying to solve for
r=dist. from wire

the problem is that i can't figure out a way to express vx as a function of r so that i can integrate this.

somebody help!!

thanks, rick.
 

Answers and Replies

  • #2
841
1
Originally posted by happy thoughts
K.E.=0 once the electron is at a distance of rf. I also know that Work=integral Fdr in the y direction.
Eh? A magnetic field does no work on a charged particle; the particle's speed is constant.
 
  • #3
happy thoughts
that makes great sense- the force is perpendicular to the direction of motion. im thinking though: the particle does move in the y direction and there is a force component in the y direction- so it seems that there is some work done when considering the y component only even though there is obviously no net work done. could it be that one might imagine some work done in the y direction when doing calculations. this would imply that the work done in the x direction is exactly negative of the work in the y direction. does that make sense at all?
 
  • #4
happy thoughts
just in case anyone is interested, here is a nasty solution that works:

1) find an average B between r0 and rf

(1/ro-rf)*mu0*i*ln(rf/r0)/(2*pi)

2) find an average vx

vx(1+1/sqrt(2))/2

3) use the work idea above, namely

[inte]Fdr where F=q*vx*B

if anyone knows the better and easier approach to this problem please let me know.
 
  • #5
508
0
Why not use Newton'w Law?
(I'm too lazy to fill in all the constants, so it's just a rough sketch)

Lorentz force is: F = v x B
Newton says: v x B = a

Since vz = Bx = By = 0, this boils down to:

I. vy Bz = ax
II. -vx Bz = ay.

Now, let's write Bz = -c/y, where c is a measure of the field's strength that we want to find. Then,

I. cy'/y = x''
This integrates to
c ln(y) = x'
Now, as Ambitwistor pointed out, in the final state, x' = v0. Thus
c ln(rf/r0) = v0 - vx0, or
c = (v0-vx0) / ln(rf/r0).

As I said, just a sketch.
 
  • #6
krab
Science Advisor
896
2
Originally posted by happy thoughts
I was thinking that I know that K.E.=Me(v0*sin(45))/2 initially in the y direction (this would be the y-component of K) if y is taken as the coordinate perpendicular to the wire and x the one parallel to the wire. K.E.=0 once the electron is at a distance of rf.
This approach won't work. KE is not a vector with x and y components. KE is a scalar. As arcnets said, just use F=ma. F definitely can be decomposed into x and y components.

just in case anyone is interested, here is a nasty solution that works: 1) find an average B between r0 and rf (1/ro-rf)*mu0*i*ln(rf/r0)/(2*pi) 2) find an average vx vx(1+1/sqrt(2))/2
This is a naive approach to say the least. One cannot hope to solve a non-linear problem of this nature by taking averages.

Use arcnets approach. Or if you want to be fancier, derive the vector potential and use the fact that the x-component of the canonical momentum is conserved.
 
  • #7
happy thoughts
This is a naive approach to say the least.
thanks for the "friendly encouragement" Bruno... 8=* !

As it turns out, after doing the correct approach, the averaged approach turns out to be exactly what it was for the correct approach, but only in the case of an initial 45 degree angle (lucky me):

if you draw a line through the particle parallel to the wire and let the angle between that line and the velocity vector be [the],

Fy=q*v*B*sin([the])

vy=v*sin([the])

*********************************************************

F=m*a

(q*v*[mu]0*i*cos([the])/(2[pi]*y)=m*d/dt(vy)

solve for i here and you get an equivilent expression in the case of [the]initial=[pi]/4
 
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