Question

What is the acceleration of a proton moving with a speed of 9.5 m/s at right angles to a magnetic field of 1.8 T

Answer 1

Step-by-step explanation:

F = qvBsin(angle)

F = 1.6 x 10^-19 x 9.5 x 1.8 x Sin (90)

F = 273.6 x 10^-20N

a = F/M

M of a proton = 1.67 x 10^-27kg

a = 273.6 x 10^-20/1.67 x 10^-27

a = 1638323353.29m/s