- #1
JonathanH13
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Homework Statement
I am having trouble with a basic deflection problem - a single proton traveling through a magnetic field:
v = Velocity of the proton = 6 000 000 meters per second (1/50th the speed of light, so no relativistic effects)
B = magnetic field strength = 0.5 Tesla
Charge of proton q = + 1.60217e-19 Coulombs
Mass of proton m = 1.67262e-27 kg
Direction = zero radians
Homework Equations
Force = q*v*B sin(theta)
Radius=(Mass*sqr(Velocity))/Force
Period:=(2*pi*Radius)/Velocity
Force=Mass*Acceleration
The Attempt at a Solution
First I find the maximum force on the proton due to its charge and velocity through the field:
Force = q*v*B sin(theta)
In this case the proton is traveling perpendicular to the field, so sin(theta) = 1
and the magnitude of the force is simply:
Force = q*v*B
= 4.8065e-13 Newtons
Now that is fine, and if I want the radius of the circle described by the proton in the field I can use:
Radius=(Mass*sqr(Velocity))/Force or r=mv/qB
= 0.125 meters or 125mm
Which is particularly satisfying, because it relates such high speed, low mass and charge into a realistic spatial dimension.
I can also find the time that it takes to complete one revolution:
Period:=(2*pi*Radius)/Velocity
=131.19e-9 seconds or 131 nanoseconds.
But what I would like to find is the instantaneous position and direction after a given time, say 1 nanosecond.
I tried dividing the circumference by 131 nanoseconds, but I feel like this is cheating.
I also tried to use
Acceleration = Force/mass, but in this case the acceleration is only a change in direction, not in velocity.