The electric field points in the +x direction and the particle's charge is positive, therefore the electric force points in the +x direction.
The electric force acting on the particle is:
F = Eq
F is the electric force, E is the electric field strength, and q is the charge.
Given:
E = 1290N/C
q = 6.58×10⁻⁶C
F = 1290(6.58×10⁻⁶) = 8.49×10⁻³N
The net force points in the +x direction, yet its magnitude is smaller than the magnitude of the electric force, so the magnetic force must point in the -x direction with a magnitude equal to the difference between the electric force and net force:
8.49×10⁻³ - 6.17×10⁻³
= 2.32×10⁻³N
The particle's velocity lies in the x-y plane and the magnetic field points in the +z direction. The velocity is perpendicular to the magnetic field, therefore the magnitude of the magnetic force can be calculated by:
F = qvB
F is the magnetic force, q is the charge, v is the velocity, and B is the magnetic field strength.
Given values:
F = 2.32×10⁻³N (solved for previously)
q = 6.58×10⁻⁶C
B = 1.27T
Plug in the values and solve for v:
2.32×10⁻³ = 6.58×10⁻⁶v(1.27)
v = 278m/s
