An electron and a proton are each moving at 830 km/s in perpendicular paths as s

Question

An electron and a proton are each moving at 830 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 = 4.05 nm and dy = 5.10 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 direction magnitude T direction magnetic force: magnitude N direction electric force: magnitude N direction ° counterclockwise from the +x-axis

Explanation / Answer

magnetic filed due to a moving charge = B = mu/4pi * q(-v)sin(phi)/r^2

part a ) Bp = mu/4pi * 1.6 x 10^-19 x 830x 10^3 / (4.05 x 10^-9)^2

here phi = 90 sin90 = 1

Bp = 0.81 mT

Be = mu/4pi * 1.6 x 10^-19 x 830x 10^3 / (5.10 x 10^-9)^2

Be = 0.51 mT

B = Bp + Be

B = 0.81 mT + 0.51 mT = 1.32 mT

direcion = into the page

part b ) B = mu/4pi * qvbsin(phi) / r^2

r = sqrt (rp^2 + re^2 )

rp = distance of proton from origin

re = distance of electron from origin

phi = 180 - tan^-1(5.10/4.05) = 128.5 degree

r = sqrt ( 5.10nm^2 + 4.05nm^2 )

B = mu/4pi * 1.6x10^-19 * 830 x 10^3 * sin(128.5) / 42.4125 x 10^-18

B = 2.45 x 10^-4 T

direction into the page

part c )

Fmag = qvBsintheta

theta = 90 sin90 =1

Fmag = qVB = 1.6 x 10^-19 x 830 x 10^3 x 2.54 x 10^-4

Fmag = 3.373 x 10^-17 N

in the + x direction

Felec = kq1q2/r^2

Felc = 9 x 10^9 x (1.6 x 10^-19)^2 / 42.4125 x 10^-18

Felc = 5.43 x 10^-12 N

direction = tan^-1(5.10/4.05 ) = 51.5 degree from clockwise

anticlock wise = 360 - 51.5 = 308.5 degree counter clock wise from the +x axis

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