Question

An electron with kinetic energy 2.9 keV moving along the positive direction of an x axis enters a region in which a uniform electric field of magnitude 7.5 kV/m is in the negative direction of the y axis. A uniform magnetic field is to be set up to keep the electron moving along the xaxis, and the direction of the field is to be chosen to minimize the required magnitude of the field. What is the magnitude of the magnetic field in mT?

Answer 1

The electric force on the electron is opposite in direction to the electric field E. E points in the -y direction, so the electric force will point in the +y direction. The magnitude of the electric force is given by:

F = Eq

F = electric force, E = electric field strength, q = electron charge

We need to set up a magnetic field such that the magnetic force on the electron balances out the electric force. Since the electric force points in the +y direction, we need the magnetic force to point in the -y direction. Using the  reversed right hand rule, the magnetic field must point in the -z direction for this to happen. Since the direction is perpendicular to the +x direction of the electron's velocity, the magnetic force is given by:

F = qvB

F = magnetic force, q = charge, v = velocity, B = magnetic field strength

The electric force must equal the magnetic force.

Eq = qvB

Do some algebra to isolate B:

E = vB

B = E/v

Let's solve for the electron's velocity. Its kinetic energy is given by:

KE = 0.5mv²

KE = kinetic energy, m = mass, v = velocity

Given values:

KE = 2.9keV = 4.6×10⁻¹⁶J

m = 9.1×10⁻³¹kg

Plug in and solve for v:

4.6×10⁻¹⁶ = 0.5(9.1×10⁻³¹)v²

v = 3.2×10⁷m/s

B = E/v

Given values:

E = 7500V/m

v = 3.2×10⁷m/s

Plug in and solve for B:

B = 7500/3.2×10⁷

B = 0.00023T

B = 0.23mT