Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.14 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1560 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 4.35 V/m, (b) in the negative z direction and has a magnitude of 4.35 V/m, and (c) in the positive x direction and has a magnitude of 4.35 V/m?

Answer 1

here,

the charge on proton , q = 1.6 * 10^-19 C

the magnetic feild , B = 2.14 mT (-i) = -0.00214 i mT

the velocity of proton , v = 1560 j m/s

a)

the electric feild , E = 4.35 k V/m

the net force on the proton , F = q * ( E + v X B)

F = 1.6 * 10^-19 * ( 4.35 k + (1560 j) X ( - 0.00214 i))

F = 1.23 * 10^-18 K N

the magnitude of the net force is 1.23 * 10^-18 N

b)

the electric feild , E = - 4.35 k V/m

the net force on the proton , F = q * ( E + v X B)

F = 1.6 * 10^-19 * ( - 4.35 k + (1560 j) X ( - 0.00214 i))

F = - 1.62 * 10^-19 K N

the magnitude of the net force is 1.62 * 10^-19 N

c)

the electric feild , E = 4.35 i V/m

the net force on the proton , F = q * ( E + v X B)

F = 1.6 * 10^-19 * ( 4.35 i + (1560 j) X ( - 0.00214 i))

F = 1.6 * 10^-19 * ( 4.35 i + 3.34 k) K N

the magnitude of the net force , |F| = sqrt(4.35^2 + 3.34^2) * 1.6 * 10^-19 N

|F| = 8.77 * 10^-19 N