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

if the forces acting on an object are unbalanced, the object will accelerate in the direction of the net force on the object.

explanation:

this is best explained by having an understanding of newton's second law, which states that f=ma. you see, when the forces are unbalanced, net force (sum of all forces) is not zero. therefore making it necessary for the object to have an acceleration in the direction of the net force.