Aproton travels through uniform magnetic and electric fields. the magnetic field is in the negative x direction and has a magnitude of 3.21 mt. at one instant the velocity of the proton is in the positive y direction and has a magnitude of 2790 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 5.97 v/m, (b) in the negative z direction and has a magnitude of 5.97 v/m, and (c) in the positive x direction and has a magnitude of 5.97 v/m?

The magnetic force acting on the proton is given by the following:

F = qv×B

F is the magnetic force vector

q is the charge

v is the velocity vector

B is the magnetic field vector

F is the cross product of qv and B, and the right-hand rule can be used to easily determine the direction of F. One way to utilize the right-hand rule is to:

v points in the +y direction and B points in the -x direction, therefore F points in the +z direction. Since v is perpendicular to B, the magnitude of F is given by:

|F| = q(|v|)(|B|)

Given values:

q = 1.60×10⁻¹⁹C

|v| = 2790m/s

|B| = 3.21×10⁻³T

|F| = 1.60×10⁻¹⁹(2790)(3.21×10⁻³)

|F| = 1.43×10⁻¹⁸N

We also have an E field acting on the proton. The magnitude of the electric force on the proton is:

|F| = q(|E|)

|F| is the magnitude of the electric force, q is the charge, and |E| is the magnitude of the E field.

For each case use q = 1.60×10⁻¹⁹C to calculate |F|

a) E is directed in +z and |E| = 5.97V/m, therefore F will point in +z with magnitude |F| = 9.55×10⁻¹⁹N. The electric force points in the same direction as the magnetic force, therefore the magnitude of the net force is just the sum of their magnitudes, 2.39×10⁻¹⁸N.

b) E is directed in -z and |E| = 5.97V/m, therefore F will point in -z with magnitude |F| = 9.55×10⁻¹⁹N. The electric force points in the opposite direction to the magnetic force, therefore the magnitude of the net force is the difference of their magnitudes, 4.75×10⁻¹⁹N.

c) E is directed in +x and |E| = 5.97V/m, therefore F will point in +x with magnitude |F| = 9.55×10⁻¹⁹N. The electric force is perpendicular to the magnetic force, therefore we will use the Pythagorean theorem to calculate the magnitude of the net force...

√((9.55×10⁻¹⁹)^{2} + (1.43×10⁻¹⁸)^{2})

= 1.72×10⁻¹⁸N

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

probably not if net force is zero