What Happens to a Proton in a Magnetic Field?

[shanie]

Homework Statement

A proton with the speed 3.7*10^6 m/s is moving in a magnetic field, parallel to the field.
The magnetic B-force is 0.45T.

a)Draw a figure that shows the proton in the magnetic field. Indicate the direction of the proton's velocity, the direction of the magnetic field and the direction of the magnetic force.
b) Determine the magnetic force.
c) Determine the radius of the circular motion that the proton will make.

Homework Equations

a) Using the right hand rule
b) F= qvB
c) r= mv/qB

The Attempt at a Solution

a) The picture I drew looked like this
http://img134.imageshack.us/img134/3895/namnls1hx8.png

where v is the direction of velocity, Fm the magnetic force and the B-force is the cross going into the paper.
b) F = (1.602*10^-19)*3.7*10^6*0.45=2.67*10^-13 N
c) r=mv/qB=[(1.673*10^-27)*(3.7*10^6)]/[(1.602*10^-19)*0.45]
r=0.086 m

Could anyone tell me if this is correct? I would really appreciate some help!

 

[rl.bhat]

A proton with the speed 3.7*10^6 m/s is moving in a magnetic field, parallel to the field.The magnetic B-force is 0.45T.

It should be perpendicular to the field.
Rest of the calculation is correct