Question

When at rest, a proton experiences a net electromagnetic force of magnitude 8.5×10−13 N pointing in the positive x direction. When the proton moves with a speed of 1.4×106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.5×10−13 N , still pointing in the positive x direction.

PART A

Find the magnitude of the electric field.   

PART B

Find the direction of the electric field.

a. positive x direction

b. negative x direction

c. positive y direction

d. negative y direction

e. positive z direction

f. negative z direction

PART C

   Find the magnitude of the magnetic field.

PART D

Find the direction of the magnetic field.

Find the direction of the magnetic field.

A. positive x direction

B. negative x direction

C. positive y direction

D. negative y direction

E. positive z direction

F. negative z direction

PLEASE HELP!

Answer 1

total force on a moving charge in electric and magnetic field

F= qE+qVXB

Part A

at rest v= 0 so force  

F = qE

E = F/q = 8.5×10^-13 / ( 1.6*10^-19)

E = 5312500 N/C

direction of the electric field. = positive x direction

PART C

F= qE+qVXB

7.5*10^-13 = 8.5*10^-13 + 1.6*10^-19*1.4*10^6* B

-1*10^-13 = 1.6*10^-19*1.4*10^6* B

B = -1*10^-13 / (1.6*10^-19*1.4*10^6 )

B = -0.446428571 T

so magnitude of magnetic field = 0.446428571 T answer

direction of the magnetic field.​ = > ​negative z direction

let me know in a comment if something goes wrong