Question

Crossed electric and magnetic fields are established over a certain region. The magnetic field is 0.635 T vertically downward. The electric field is $2.68 \times 10^{6} \mathrm{V} / \mathrm{m}$ horizontally east. An electron, traveling horizontally northward, experiences zero net force from these fields and so continues moving in a straight line. What is the electron's speed?

Short answer

Answer: The speed of the electron is 4.22 x 10^6 m/s.

Step-by-step solution

Write down the given values

The given values are: Magnetic field strength, \(B = 0.635\, T\) (vertical, downward) Electric field strength, \(E = 2.68 \times 10^6\, \mathrm{V/m}\) (horizontal, east) Electron charge, \(q = -1.6 \times 10^{-19}\) C Desired force on the electron, \(F_{net} = 0\)

Equate the electric and magnetic forces

The net force on the electron is zero. Therefore, the electric force and the magnetic force should be equal and opposite: \(F_E = F_B\) Since \(F_E=qE\) and \(F_B=qvB\sin\theta\), we can write: \(qE = qvB\sin\theta\)

Determine the angle between the velocity and the magnetic field

In this problem, the electron is traveling horizontally northward, and the magnetic field is vertically downward. The angle between these two directions is 90 degrees, i.e., \(\theta = 90^{\circ}\) or \(\sin\theta = 1\).

Solve for the electron's speed

Now we can solve for the electron's speed, v: \(E = vB\sin\theta\) \(2.68 \times 10^6\, \mathrm{V/m} = v (0.635\, T)(1)\) \(v = \frac{2.68 \times 10^6\, \mathrm{V/m}}{0.635\, T}\) \(v \approx 4.22 \times 10^6\, \mathrm{m/s}\) Therefore, the speed of the electron is \(\boxed{4.22 \times 10^6\, \mathrm{m/s}}\).