Helical motion of a proton in a magnetic field

[yjk91]

velocity of proton following a helical path?? need a bit of help

Homework Statement

A proton with an initial velocity given by (4.0 + 3.0 + 3.0)(105 m/s) enters a magnetic field given by (0.62 T). Describe the motion of the proton.


The proton will follow a helical path with a velocity of 3e5 m/s along the Z -axis, with the circular motion in the XY -plane having a speed of ?? and radius of ??

The Attempt at a Solution

so i got the velocity of the proton along Z axis but I'm not sure how to get the XY axis

my idea was that i could just square x and y velocity then rad it but it didn't work

and would i be able to get the Radius by using this formula?

 

[ehild]

What is the direction of the magnetic field?
Are you sure you copied the velocity along z correctly? It can be 305 m/s, but 3e5 is impossible if it means 300000 m/s.

Do you know the formula for the Lorentz force, the magnetic field exerts on a moving charged particle?

ehild

 

[yjk91]

yeah it's to the z

no we have not learned the Lorentz force yet
can't you find the velocity the ciruclar motion by doing x^2 + y^2 and rad it
but it's not the answer

 

[yjk91]

oh nvm it is haha

 

[yjk91]

qvB = mv^2/R or R = mv/qB.

i'm using this to find the R

so 1.6*10^-27 * 5 * 10^5 / (1.6*10^-19 * 0.62 * .62T)

and i got 0.008065 m and it's not the answer

5 * 10^5 is the velocity on the xand y axis
i tried using the z axis but that's not even close

help?

 

[ehild]

What is what? does 105 mean 10⁵ m/s? And 3e5 means 3˙10⁵ m/s?
So 5˙10⁵ m/s is the speed of the proton in the x,y plane.
The mass of proton is

$m_{p}\ =\ 1.673\ \times\ 10^{-27}\ kg$

Using 1.6 instead of 1.67 is bad.

The charge is
$e\ =\ 1.602\ \times\ 10^{-19}\ C$

It does not hurt using more then necessary significant digits during calculations.
Do the calculation again, with the more accurate proton mass. It should be all right.

ehild

 

[yjk91]

oh thanks
i'll remember