Magnetic Field and Moving Charge

[runfast220]

Homework Statement

The drawing shows a charged particle (q=2.80x10^-6C) moving along the +y axis with a speed of 4.80X10^6 m/s. A magnetic field of magnitude 3.35x10^-5 T is directed along the +z axis, and an electric field magnitude 123 N/C points along the -x axis. Determine the (a) magnitude and (b) direction of the net force that acts on the particle.

q=2.80x10^-6C V=4.80X10^6 m/s B=3.35x10^-5 T E= 123 N/C

Homework Equations

E=F/q

B=F/(qsin@)

The Attempt at a Solution

2.80x10^-6C / 123 N/C = 2.28X10^-8 N

F=2.28X10^-8 N

Vsin@ = F/Bq 4.80X10^6sin@ = (2.28X10^-8)/ (3.35x10^-5)( 2.80x10^-6)

@= .0029deg in the positive direction

I think I am doing the problem right, but I think my algebra might be bad?

 

[rl.bhat]

Force due to magnetic field F = qVBsinθ.
Find the direction of the magnetic filed by Flemming's left hand rule.
Find the direction of the electric field. Then find the resultant.