Question

When at rest, a proton experiences a net electromagnetic force of magnitude 9.0×10^?13 Npointing 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

Part C: Find the magnitude of the magnetic field.

Part D: Find the direction of the magnetic field.

Answer 1

here,

charge on proton , q = 1.6 * 10^-19 C

a)

let the magnitude of electric feild be E

E * q = Fi

E * 1.6 * 10^-19 = 9 * 10^-13

E = 5.6 * 10^6 N/C

b)

the direction of force is in positive x- direction

so, the electric feild points in positive x direction

c)

let the magnitude of magnetic feild be B

F = q * ( v X B)

(9 * 10^-13 - 7 * 10^-13) = 1.6 * 10^-19 * 1.4 * 10^6 * B

B = 0.89 T

d)

as the forces decreases, i.e the magnetic force directs in -x direction

velocity is in +y direction

using right hand rule

the magnetic feild is in -z direction