An electron and a proton are each moving at 860 km/s in perpendicular paths as s

Question

An electron and a proton are each moving at 860 km/s in perpendicular paths as shown in the following figure. Consider the instant when they are at the positions shown in the figure. (Let dx = 3.55 nm and dy = 5.35 nm.)

(a) Find the magnitude and direction of the total magnetic field they produce at the origin.

(b) Find the magnitude and direction of the magnetic field the electron produces at the location of the proton.

(c) Find the magnitude and direction of the total electrical force and the total magnetic force that the electron exerts on the proton.

magnitude _______ mT An electron and a proton are each moving at 860 km/s in perpendicular paths as shown in the following figure. Consider the instant when they are at the positions shown in the figure. (Let dx = 3.55 nm and dy = 5.35 nm.) An electron and a proton are each moving at 8 (a) Find the magnitude and direction of the total magnetic field they produce at the origin. (b) Find the magnitude and direction of the magnetic field the electron produces at the location of the proton. (c) Find the magnitude and direction of the total electrical force and the total magnetic force that the electron exerts on the proton. magnetic force: electric force: magnitude direction counterclockwise from the +x-axis

Explanation / Answer

v = 860 km/s, dx = 3.55 nm, dy = 5.35 nm, q =1.6*10-19 C
a) B= (?0/4?)*qv/dy2 +(?0/4?)*qv/dx2 = 1.57 mT
   Direction: into the page
b) B =(?0/4?)*qvsin?/(dx2+ dy2), where ? = tan-1(dy/dx) = 56.43o
    B = 2.78*10-4 T
    Direction: into the page
c) Fm = qvB = 3.826*10-17 N, along the +x-axis.

     electric field: E = kq2/(dx2 +dy2) = 5.5888*10-12 N/C
Force due to electric field: Fe = qE = 8.94*10-31 N
Direction: 360o - 56.43o = 303.57o

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