A proton travels through uniform magnetic and electric fields. The magnetic fiel

Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.86 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2480 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 5.04 V/m, (b) in the negative z direction and has a magnitude of 5.04 V/m, and (c) in the positive x direction and has a magnitude of 5.04 V/m? (a) Number (b) Number (c) Number Units Units Units

Explanation / Answer

Magnetic force

F=qvB =(1.6*10-19)(2480)(2.86*10-3)=1.134848*10-18 N

Force due to electric field

F=QE =(1.6*10-19)(5.04) =8.064*10-19 N

a)

Net force

FNet =1.134848*10-18 +8.064*10-19=1.94*10-18 N

b)

Net force

FNet =1.134848*10-18 -8.064*10-19=3.28*10-19 N

c)

Net force

FNet =sqrt[(1.134848*10-18)2+(8.064*10-19)2] =1.392*10-18 N

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