Question 8.3: A charged particle moves with a uniform velocity 4ax m/s in ...

If the particle moves with a constant velocity, it is implied that its acceleration is zero. In other words, the particle experiences no net force. Hence

0=F=ma=Q(E+u\times B)

0=Q(20a_{y}+4a_{x}\times B_{o}a_{z})

or

-20a_{y}=-4B_{o}a_{y}

Thus B_{o}=5.

This example illustrates an important principle employed in a velocity filter shown in Figure 8.3. In this application, E, B, and u are mutually perpendicular so that Qu\times B is directed opposite to QE, regardless of the sign of the charge. When the magnitudes of the two vectors are equal

QuB=QE

or

u=\frac{E}{B}

This is the required (critical) speed to balance out the two parts of the Lorentz force. Particles with this speed are undeflected by the fields; they are “filtered” through the aperture. Particles with other speeds are deflected down or up, depending on whether their speeds are greater or less than this critical speed.