Calculating Magnetic Field for Maximum Electron Deflection Angle

  • Thread starter Thread starter Doodad86
  • Start date Start date
  • Tags Tags
    Motion
Doodad86
Messages
2
Reaction score
0

Homework Statement


Determine the required magnetic field to give an electron deflection angle of θ=90deg. The other variables (such as Rd, Re, and the electron energy in keV) are currently dummy variables such that this problem will initially be solved symbolically.

Homework Equations


(1) F=q(v x B)
(2) F = qvbsinθ
(3) QBR = p

The Attempt at a Solution


Using the following figure (apologies for poor paint skills), I've taken the cross product of (v x B), recognizing that vx = 0, By = Bz = 0. This gives me that v x B = (vzBx)y - (vyBx)z where y and z are unit vectors.

Edit: Apologies; I seem to be awful at remembering to attach photos, files in e-mail, etc. Here's the figure: http://i.imgur.com/V5TN9.png

I am having trouble fitting the cross product to another equation to yield any kind of results. Let me know if I should do more work (I've been reading through Griffith's Intro to Electrodynamics for about two hours now trying to make any headway).
 
Last edited:
Physics news on Phys.org
The figure shows x as + into page. That is not a right-handed coordinate system. Either the B field is directed into the page or into the +x direction.

If you intended a left-hand system we would like to know about it. It's very rare, although my high school textbook of the '50's actually used it.


Is the B field understood to be confined to within the circle of radius RB?
 
rude man said:
The figure shows x as + into page. That is not a right-handed coordinate system. Either the B field is directed into the page or into the +x direction.
...
The one I saw in the figure is right handed.

(+)x is into the page, y is up, z is to the right. --- That's RIGHT handed. Sitting in the 1st octant looking towards the origin, x to y to z is counter-clock-wise.
 
rude man said:
Is the B field understood to be confined to within the circle of radius RB?

Yes, B is to be confined within the circle radius RB (accidentally noted RB as Rd in the description; apologies). As I understand it, we have to take into account the relativistic and classical momentum to find the more realistic values of B (assuming all electrons are uniform, originating from the same source traveling in the same direction). I also may be wrong with taking the cross product of the force, and I may just need (2) as well as the following equations:

(4) pv/r = F
(5) E^2 = m^2 * c^4 + p^2 * c^2.
 
I beg everyone's pardon, it is right-handed. How embarrassing for me.
 
Consider this: since you've already got the directions of all relevant quantities specified in the figure, you can work with equations that only involve magnitudes. So you don't need to calculate out the components of the cross product.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top