Proton travelling through a magnetic field

[JonathanH13]

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.

 

[ehild]

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.

It is no cheating if you calculate the angular position as
alpha = 2pi*(t/T). And you know that the proton moves along a circle with constant speed, so the velocity vector is tangent to the circle. Knowing alpha, you also get the direction of the velocity.

ehild

 

[JonathanH13]

Thanks for your reply. You describe angular position, but I require cartesian coordinates in order to simulate this:

I calculate alpha = 2pi*(t/T), using t = 1e-9 (1 nanosecond), and T (Period of one revolution) = 131.19e-9 seconds

This gives me an alpha of 0.0478 (is that in Radians?)

To attempt to convert that into an instantaneous position (my original problem),

I use Y = Sin(alpha)*Radius and X = Cos(alpha)*Radius

So, in in one nanosecond the proton is now at:

X = 0.1258 meters
Y = 0.00599 meters

Is that correct?

 

[ehild]

it is correct, but I got x=0.1248 m.

ehild

 

[JonathanH13]

Ah, I think I see the error with my calculation.

I am now using X = Cos(alpha)*(Velocity*1e-9)

That is, if velocity is distance per second, and I require distance per nanosecond, hence(Velocity*1e-9). This delivers x and y values that describe a circle with the correct radius!

Thanks!