# Homework Help: Magnetic Field and Moving Charge

1. Jul 29, 2009

### runfast220

1. The problem statement, all variables and given/known data
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

2. Relevant equations

E=F/q

B=F/(qsin@)

3. 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?

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2. Jul 29, 2009

### 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.