Question

When at rest, a proton experiences a net electromagnetic force of magnitude 8.1×10⁻¹³ 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 7.0×10⁻¹³ N , still pointing in the positive x direction.

Part A: Find the magnitude of the electric field. Express your answer using two significant figures.

Part B: Find the direction of the electric field. (Positive or negative in either x,y,z direction)

Part C: Find the magnitude of the magnetic field. Express your answer using two significant figures.

Part D: Find the direction of the electric field. (Positive or negative in either x,y,z direction)

Answer 1

A.

Charge on proton

Electric field E = Force/charge

= 8.1×10-13/(1.6×10-19) = 5.0625×106 N/C

B.

Direction = +x direction

C.

As the charge in moving in +y direction

And force is reduced by = 8.1×10-13 - 7×10-13= 1.1×10-13 N

So force due to magnetic field is in - x direction because it decreases the net force. So magnetic field must be in +z direction.

Magnitude of magnetic field

F = eVB

1.1×10-13 = 1.6×10-19×1.6×106​​​​​​×B

B = 0.43 T

d.

Direction of magnetic field = +z direction