A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.66 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2930 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 5.51 V/m, (b) in the negative z direction and has a magnitude of 5.51 V/m, and (c) in the positive x direction and has a magnitude of 5.51 V/m?

a)

b)

c)  

Explanation:

We have:                    

B: is the magnetic field = (2.66x10⁻³ T)(-)      

v: is the velocity of the proton = (2930 m/s)    

a) To find the magnitude of the net force (F) we need to use the Lorentz force equation:

 (1)

E: is the electric field = (5.51 V/m)            

q: is the proton charge = 1.6x10⁻¹⁹ C

Hence, the magnitude of the net force is:

b) When E = (5.51 V/m)(), the net force is (equation 1):

Then, the magnitude of the net force is:

c) When E = (5.51 V/m)(), the net force is:  

Hence, the magnitude of the net force is:

I hope it helps you!                                                                                                

answer: a ball

explanation: because it will bounce