An electron and a proton are each moving at 870 km/s in perpendicular paths as s

Question

An electron and a proton are each moving at 870 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 = 3.90 nm and dy = 5.20 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.

please make sure post right answer

Explanation / Answer

Fmag = qvBsinphi

B = uo/4pi * qvsinphi/r^2

part a )

both fields are into the page, so their magnitudes add

B = Be + Bp

phi = 90 degree = sin90 =1

B = uo/4pi (ev/re^2 + ev/rp^2)

B = uo/4pi * (1.6 x 10^-19)(870 *10^3 ) [1/(5.2*10^-9)^2 + 1/(3.9 *10^-9)^2

B = 1.43 T

into the page

part B )

B = uo/4pi * qvsinphi/r^2

r = 6.5 nm

phi = 180 - tan^-1(5.2/3.9) = 126.87 degree

B = 2.6357 x 10^-4 T

into the page

part c )

Fmag = qvBsin90 = 1.6 *10^-19 * 870 * 10^3 x 2.6357*10^-4 = 3.67 x 10^-17 N

in +x direction

Felec = kq^2/r^2

r^2 = 4.225 x 10^-17 m^2

Felec = 5.45 x 10^-12 N

theta = 306.87 degree counterclockwise from the +x axis

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