- #1
taylor.simon
- 7
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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
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