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runfast220
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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?