Question

When at rest, a proton experiences a net electromagnetic force of magnitude 8.6×10−13 N pointing in the positive x direction. When the proton moves with a speed of 1.6×106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.3×10−13 N , still pointing in the positive x direction.

A. Find the magnitude of the electric field.

B. Find the direction of the electric field.

C. Find the magnitude of the magnetic field.

D. Find the direction of the magnetic field

Answer 1

given are

F1=8.6*10^-13 N

V=1.6*10^6 m/sec

F2=7.3*10^-13 N

Electric field=Force/Q

Q charge of proton =1.6*10^-19 C

E=8.6*10^-13/1.6*10^-19

E=5.37*10^6 N/C—answer

b)in the x direction

c)F=qvB

B=F/qv

F= 7.3x10^-13-8.6x10^-13

=-1.3*10^-13 N

B=1.3*10^-13/(1.6*10^-19)*(1.6*10^6)

B=0.507T—answer b

d) in z direction