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

  • #1
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 feild (B) = 2

so it would be f=QVxB = 1.6x10^-19x1000x2= 3.2x10^-16
F=3.2x10^-16
still dont 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=1.916167665x1…
A=1.916167665x10^-43

is this right
 

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