Magnetic Fields proton beam accelerated

[Smugdruggler]

Homework Statement

A beam of photons, originally at rest, is accelerated horizontally between the plates of a parallel-plate capacitor, the potential difference of which is 1000 V and has a separation of 1.0 cm.
The particle enters a region wher a uniform magnetic field, B=0.913T (directed out of the page. To detect the particle a sensor is located directly below the point where theparticle enters the field.

a) What is the velocity vector for the particle just as it enters the field?
b) What is the magnetic force vector (in component form) exerted on the particle immediately after it enters the field?
c) In order to detect the exiting particle, at what distance below the particle's entrance point does the sensor need to be?

d) Finally, a beam of alpha particles is the accelerated same /\V and fired into the mag. field. To what magnitude should the field be adjusted to so that the particles will be detected by the sensor? Mass of alpha particles is 6.64E-27 kg and charge is +2e

Homework Equations

V=sqrt((2q*V)/m)

The Attempt at a Solution

a) v=E/B I'm not sure if this is the correct equation to use. my velocity vector is heading horizontally into a magnetic field pointed out of the page, right now I'm picturing a birds eye view of a ball flying over an air hockey table (many holes with air pushing up). If x is going out of the page y is going horizontally to the right, then z is going to up, the velocity vector JUST as I enter should be in the x direction with magnitude v=V/B*d

1000V/(.913T*.010m)=1.1E5 m/s

b)The magnetic force vector in component form exerted on the particle immediately after it enters the magnetic field is F=qvBsin(theta) except in this case because it's moving perpendicular I've got sin(90) or 1 so its just F=qvB
I'm assuming from the instructions to use q=+1.6E-19 because I had a beam of protonS and now I have only "THE PARTICLE" which I'm guessing is one proton so:
F=1.6E-19C*1.1E5m/s*.913T=1.607E-14 N in the + or - z direction. and since q is positive then it must be in the +z direction.

c) Already I'm confused if the particle is to travel in the +z direction which I think it should then the sensor should be placed ABOVE where the particle will enter. The magnitude of the distance however is still possible to find...

Since F=|q|vB in this example because of the perpendicular vectors, I use F=(mv^2)/r
(mv^2)/r = 1.607E-14N in the +z direction=(1.67E-27Kg*[1.1E5m/s]^2)/r. Then r=.0013m and diameter = 2r so d=.0025m

d) So using v=sqrt([2q*V]/m) =sqrt([2*2e*1000]/6.64E-27)=3.093E5 m/s
Falpha=(6.64E-27Kg*[3.093E5m/s]^2)/.0013m=4.886E-13N also in the +z direction
oops that's force

should have used m1/m2=(Bo^2)/(B^2) that makes for 1.6E-19/6.64E-27=.913^2/x^2=.002 T That just seems wrong...
How'd I do? Am I doing this right?

 

[wbandersonjr]

a) v=E/B I'm not sure if this is the correct equation to use.

No. The correct equation is the one you put under relevant equations. That will give you the velocity as it enters the magnetic field.

b) your equation is correct, just use the correct velocity. Also, your imagining of a ball above an air hockey table is useful. However the magnetic field does not operate like air streams. Yes, the magnetic field is directed out of the page, like the air is blowing up out of the table. But the force on the particle is determined using a cross product F=qv X B, meaning you need to use the right hand rule to determine the direction of the force on the particle, it is not in the z direction.

c) Looks good, just recalculate with the new F value.

d) You need to determine the velocity when the alpha particles enter the magnetic field. Then determine the magnetic force, F, needed to cause them to have the same radius of curvature. Last, calculate the magnetic field, B, that would cause that force needed.

 

[Smugdruggler]

Thank you very much!

I got to run to class Ill be sure to come back and post a nicely formatted write up for the next person to seek these answers after me!