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.

 

[tomcue]

I would like to clarify that the direction of the net force on the charged particle is not in the positive direction as indicated in the attempt at a solution. The direction of the net force can be found using the right hand rule, where the thumb points in the direction of the velocity, the index finger points in the direction of the magnetic field, and the middle finger points in the direction of the net force. In this case, the net force would be in the negative x direction, as the electric field is also pointing in the negative x direction. Therefore, the correct answer for the direction of the net force would be a negative angle of 0.0029 degrees. Additionally, it is important to note that the magnitude of the net force can be found using the Pythagorean theorem, as both the electric and magnetic forces are acting on the particle simultaneously. The magnitude of the net force would be approximately 2.28x10^-8 N.