Proton and magnetic field question to see if i got the answers right?

[taylor.simon]

hey i just need to know if these answers are right i think i got it hopefully


A proton of mass 1.67 x 10-27 kg is fired with a velocity of 1000 ms-1, into a uniform magnetic field of 2.00 T where the velocity and magnetic field are at right angles.

(a) What is the magnitude and direction of the force acting on the proton?

ok so the
charge of a proton is +1.6x10^-19
velocity = 1000
the magnetic field (B) = 2

so it would be f=QVxB = 1.6x10^-19x1000x2= 3.2x10^-16
F=3.2x10^-16
still don't know how to find the direction of the force any help would be appreciated

(b) What is the acceleration of the proton?
A=f/m
a=3.2x10^-16/1.67x10^-27=1.916167665x10^-43
A=1.916167665x10^-43

 

[rl.bhat]

To find the direction of the force, use Flemming's left hand rule.

 

[tomcue]

(c) Based on the direction of the force, what is the direction of the proton's motion?

The direction of the force on the proton would be perpendicular to both the magnetic field and the velocity of the proton. This means that the proton would experience a centripetal force, causing it to move in a circular path with the magnetic field acting as the center of the circle. Therefore, the direction of the proton's motion would be perpendicular to the magnetic field and the velocity, in a clockwise or counterclockwise direction depending on the orientation of the magnetic field.