Question

A 6.60 −μC particle moves through a region of space where an electric field of magnitude 1350 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction. If the net force acting on the particle is 6.21×10−3 N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Answer 1

Charge on the particle (q) = 6.6*10^-6 C
Magnitude of electric field (E) = 1350 N/C along + x axis  
magnitude of magnetic field (B) = 1.25 T along +z axis
now it is given that the net force is in +x direction
Since charge is positive and electric field is having its direction along + x axis therefore force on the charge due to electric field will be + X axis
Force due to electric field (F_E) =qE = 6.6*10^-6*1350 = 8.91*10^-3 N
Now force due to the magnetic field must be in the + x axis direction only to make the net force in + x axis.
hence the
F_net = F_B + F_E
6.21*10^-3 = F_B + 8.91*10^-3
F_B = -2.7*10^-3 N
-ve sign shows that the assumed direction is opposite then actual.
hence the F_B will be along - x axis direction.
now using right hand rule the direction of the velocity will be toward - y direction.
We know that
F_B = qVB
-2.7*10^-3 = 6.6*10^-6*V*1.25
V = -327.27 m/s
- sign shows that the direction is -y
hence the component of velocity will be
V_X = 0
V_Y = -327.27 m/s
V_Z = 0