What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric and magnetic fields of magnitude 8100 v/m and 6.0*10?3 t, respectively?

F = qE + qV * B

where force F, electric field E, velocity V, and magnetic field B are vectors and the * operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = - V * B

which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get

|V| = 8.8e3 / 3.7e-3

= 2.4e6 m/sec

Acceleration a = V²/r, where r is the radius of curvature.

a = F/m, where m is the mass of an electron,

so qVB/m = V²/r.

Solving for r yields

r = mV/qB

= 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)

= 3.65e-3 m