# Proton travelling through a magnetic field

1. Jun 30, 2011

### JonathanH13

1. The problem statement, all variables and given/known data

I am having trouble with a basic deflection problem - a single proton travelling 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

2. Relevant equations

Force = q*v*B sin(theta)

Force=Mass*Acceleration

3. 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 travelling 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:

= 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:

=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.

2. Jun 30, 2011

### ehild

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

3. Jun 30, 2011

### 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),

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

X = 0.1258 meters
Y = 0.00599 meters

Is that correct?

4. Jun 30, 2011

### ehild

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

ehild

5. Jun 30, 2011

### 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!