A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.59 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2190 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 3.91 V/m, (b) in the negative z direction and has a magnitude of 3.91 V/m, and (c) in the positive x direction and has a magnitude of 3.91 V/m?

a) 1.88*10^-18 N

b) 6.32*10^-19 N

c) 1.9*10^-18 N

Explanation:

The total force over the electron is given by:

the first term is the electric force and the second one is the magnetic force.

You have that the velocity of the electron and the magnetic field are:

by using the relation  j X (-i) =- j X i = -(-k) = k, you obtain:

a) For an electric field of 3.91V/m in +z direction:

b) E=3.91V/m in -z direction:

c) E=3.91 V/m in +x direction:

co2reflects heat radiation back to earth buning coal and oil emits co2 globe warms

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s