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.Select one of the options below as your answer:
A. F = 0 newtons
B. F = -6.0 × 10-15 newtons
C. F = -9.6 × 10-15 newtons
D. F = -3.0 × 10-16 newtons
E. F = -3.2 × 10-21 newtons

Answer 1

Force on electron is...
\[F=qvBsin \theta \]
here \[\theta=90^{0}\]

Answer 2

would i just plug it in as is?

Answer 3

yeah...B-magnetic field
q-charge (only magnitude )
v-velocity

Answer 4

I couldnt figure it out...

Answer 5

F=(1.6*10^-19)*(3.0*10^6)*(0.020)
now you can calculate it..

Answer 6

looked nothing like mine lol

Answer 7

c ty