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:

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answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.