Question

A velocity selector uses a 66 mT magnetic field perpendicular to a 25 kN/C electric field.

At what speed will charged particles pass through the selector undeflected? (m/s)

Answer 1

magnetic field = B = 66 x10^-3 T

electric field = E = 25 X 10^3 N/C

for undeflected   Eq = q v B

speed = v = E / B = 25 X10^3 / 66 X 10^-3 = 3.79 X 10^5 m/ sec

                                                                  = 3.8 x 10^5 m/sec