An electron in an electron beam experiences a downward force of 1.9×10^−14 N while traveling in a magnetic field of 6.4 × 10^−2 T west. The charge on a proton is 1.60×10^−19.

a) What is the magnitude of the velocity?

b) What is its direction?

To find the magnitude of the velocity of the electron, we can use the formula for the magnetic force on a moving charged particle:

F = q(vB)

where F is the force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field. We can rearrange this formula to solve for velocity:

v = F / (qB)

a) Plug in the given values:

F = 1.9×10^−14 N
q = charge of an electron = -1.60×10^−19 C (note that the charge is negative for an electron)
B = 6.4 × 10^−2 T

v = (1.9×10^−14 N) / (-1.60×10^−19 C) / (6.4 × 10^−2 T)

Calculating this expression will give us the magnitude of the velocity.

b) The direction of the velocity can be determined by the right-hand rule. If the force on a moving charged particle is downward, and the magnetic field is directed to the west, then by the right-hand rule, the velocity of the electron will be to the south.