Question

At some instant, an alpha particle(q=+2e) has the coordinates(3.0,0.0,0.0)(nm) and is moving along the positive x-axis with 500km/s while an electron at (0.0,4.0,0.0)(nm) is moving along the negative y-axis with 1000km/s.
A)SHOW ALL WORK
Find the magnitude and direction of the total magnetic field at the location(3.0,4.0,0.0)(nm)
B)SHOW ALL WORK
Find the magnitude and direction of the magnetic forces felt by the electron and alpha particle.

Answer 1

magnetic field by a charge particle is given as ::

for alpha particle ::

q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

V = 500 km/s = 500 x 10³ m/s

r = 4 nm = 4 x 10^-9 m

so B_alpha = (u_o/4) (qV/r²)

B_alpha = (10^-7) ((3.2 x 10^-19 )(500 x 10³ )/(4 x 10^-9 )²)

B_alpha = 10^-3 T           outward of the page

for electron ::

q = e = 1.6 x 10^-19 C = 1.6 x 10^-19 C

V = 1000 km/s = 10⁶ m/s

r = 3 nm = 3 x 10^-9 m

so B_elec = (u_o/4) (qV/r²)

B_elec = (10^-7) ((1.6 x 10^-19 )(10⁶ )/(4 x 10^-9 )²)

B_elec = 1.78 x 10^-3 T          inward of the page

net magnetic field = B_elec - B_alpha

B_net = 1.78 x 10^-3 - 10^-3

B_net = 0.78 x 10^-3 T

B)

magnetic force by the alpha particle at the location of electron ::

B_alpha =10^-3 T  

q = e = 1.6 x 10^-19 C = 1.6 x 10^-19 C

V = 1000 km/s =10⁶ m/s

magnetic force felt by the electron ::

F = q V B Sin90

F = (1.6 x 10^-19 ) (10⁶) (10^-3)

F = 1.6 x 10^-16 N toward positive X-direction

magnetic force by the electron particle at the location of alpha particle ::

B_elec = 1.78 x 10^-3 T  

q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

V = 500 km/s = 500 x 10³ m/s

magnetic force felt by the alpha particle ::

F = q V B Sin90

F = (3.2 x 10^-19 ) (500 x 10³ ) (1.78 x 10^-3)

F = 2.85 x 10^-16 N

F = 2.85 x 10^-16 N    towards positive y-direction