- #1

- 8

- 0

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