- #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.