Question

A proton moves through a region containing a uniform electric field given by E with arrow = 60.0 ĵ V/m and a uniform magnetic field B with arrow = (0.200 î + 0.300 ĵ + 0.400 k) T. Determine the acceleration of the proton when it has a velocity v with arrow = 210 î m/s. a with arrow = m/s2

Answer 1

Eletric force acting on the proton, Fe = q*E

= 1.6*10^-19*60 j

= 9.6*10^-18 j N

magnetic force acting on the proton, Fb = q*(v cross B)

= 1.6*10^-19*( (210 i) cross (0.2i + 0.3j + 0.4k ) )

= 1.6*10^-19*(210*0.2*(i cross i) + 210*0.3*(i cross j) + 210*0.4*(i cross k) )

= 1.6*10^-19*( 0 + 63 k + 84*(-j))

= -1.34*10^-17 j + 1.008*10^-17 k
net force, Fnet = Fe + FB

= 9.6*10^-18 j -1.34*10^-17 j + 1.008*10^-17 k

= -0.826*10^-18 j + 1.008*10^-17 k

|Fnet| = sqrt(0.826^2 + 1.008^2)*10^-18

= 1.30*10^-18 N

acceleration, a = Fnet/m

= 1.30*10^-18/(1.67*10^-19)

= 7.78 m/s^2 <<<<<<<----------------Answer