Question

1.)what is the magnitude of the force experienced by a proton moving at 3 km/s inside magnetic field of strength 0.05T

2.)an electron is moving at 2x10^5 m/s through a uniform magnetic field of 1.4x10^-3 T.what is the magnitude of magnetic force if the velocity of the electron and the field make an angle of 45 degree.

Answer 1

my physics skills are rusty

Answer 2

F = q (V x B)
we need to know the angle between V and B.
I suspect it is 90 degrees

q = 1.6 * 10^-19
V = 3,000
B = .05

then F = 1.6*10^-19 * 3 * 10^3 * 5 * 10^-2 N

second one the same way but use
V x B = |V| |B| sin theta