What is the magnetic force acting on an electron if its speed is 3.0 × 106 meters/second and the direction is perpendicular to a magnetic field of 0.020 teslas? The value of q = -1.6 × 10-19 coulombs.
F = 0 newtons
F = -6.0 × 10-15 newtons
F = -9.6 × 10-15 newtons
F = -3.0 × 10-16 newtons
F = -3.2 × 10-21 newtons

Question

What is the magnetic force acting on an electron if its speed is 3.0 × 106 meters/second and the direction is perpendicular to a magnetic field of 0.020 teslas? The value of q = -1.6 × 10-19 coulombs.
F = 0 newtons
F = -6.0 × 10-15 newtons
F = -9.6 × 10-15 newtons
F = -3.0 × 10-16 newtons
F = -3.2 × 10-21 newtons

amonoconnor · Answer

Fb= qvBsin(theta)

So, 
q = 1.602 X 10^(-19)C
v = 3 X 10^6m/s
B = .02T
theta= 90 (do you understand where the theta came from?)

Fb = qvBsin(theta)
     = (1.602X10^(-19))(3X10^6)(.02)(sin(90))
     = (4.806X10^(-13))(.02)(1)
     = 4.806X10^(-13)(.02)
     = 9.612 X 10^(-15)

Do you follow all of that ? :)

amonoconnor · Answer

And I left the negative out of my calculations, but yes, the force is negative because the charge is.

anonymous · Answer

yeah i got it now, thank you!!

amonoconnor · Answer

You bet! :)