Question

A magnetic field, which has an intensity of 0.950 Tesla, is directed vertically downward (along the -y axis). A beta particle (electron) is moving horizontally through the field from left to right along the +x axis at 12,000 m/s as shown in the diagram above. What will be the magnitude of the magnetic force acting on this electron?

Answer 1

The magnitude of the magnetic force acting on the electron is 2.304 x 10^-15 N.

Step-by-step explanation:

The magnitude of the magnetic force acting on the electron can be calculated using the formula: F = qvBsinθ, where F is the force, q is the charge of the electron (-1.6 x 10^-19 C), v is the velocity of the electron (12,000 m/s), B is the magnetic field intensity (0.950 Tesla), and θ is the angle between the velocity and the magnetic field.

In this case, since the electron is moving horizontally through the field, the angle between the velocity and the magnetic field is 90 degrees, and sinθ is equal to 1.

Using the formula, we can calculate the force: F = (-1.6 x 10^-19 C)(12,000 m/s)(0.950 Tesla)(1) = -2.304 x 10^-15 N.

Therefore, the magnitude of the magnetic force acting on the electron is 2.304 x 10^-15 N.