Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.26 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1980 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.92 V/m, (b) in the negative z direction and has a magnitude of 5.92 V/m, and (c) in the positive x direction and has a magnitude of 5.92 V/m?

Answer 1

a)

F_net = F_e +F_b

= -(qE)k +(qvB)k

= [(-1.6*10^-19*5.92)+(1.6*10^-19*1980*3.26*10^-3)]k

= (8.56*10^-20 N)k

b)

F_net = F_e +F_b

= (qE)k +(qvB)k

= [(1.6*10^-19*5.92)+(1.6*10^-19*1980*3.26*10^-3)]k

= (1.98*10^-18 N)k

b)

F_net = F_e +F_b

= -(qE)i +(qvB)k

= -(1.6*10^-19*5.92)i+(1.6*10^-19*1980*3.26*10^-3)k

= (-9.47*10^-19)I +(1.033*10^-18)k

|F_net |=sqrt[(9.47*10^-18)^2+(1.033*10^-18)^2] = 9.53*10^-18 N