Answer
See answers.
Work Step by Step
a. Assume that the magnetic field is perpendicular to the velocity of the electrons. For the electrons to be undeflected, the magnitude of the magnetic force equals the magnitude of the electric force.
$$qvB=qE$$
$$B=\frac{E}{v}=\frac{12000V/m}{4.8\times10^6 m/s}=2.5\times10^{-3}T$$
b. The electric field is pointing south, and since the electron has negative charge, the force on the electrons is to the north. To balance, the magnetic force must be to the south.
Use the right-hand rule, remembering the negative charge on the electrons. The magnetic field points vertically upward, toward the sky.
c. After the electric field is turned off, the electrons moving in a uniform magnetic field will follow a circular path. The frequency is the cyclotron frequency for a charged particle in a field, equation 20–5.
$$f=\frac{qB}{2\pi m}=\frac{qE}{2\pi mv}=\frac{(1.60\times10^{-19}C)(12000V/m)}{2\pi (9.11\times10^{-31}kg)(4.8\times10^6 m/s)}$$
$$f=7.0\times10^7 Hz$$
