Answer to Question #120148 in Electricity and Magnetism for Jody Dhaliwal

Question

An electron, with a mass of 9.1 x 10–31 kg and charge of 1.6 x 10–19 C, is
accelerated to a velocity of 4.0 x 106
m/s then enters a uniform magnetic field
of 5.0 x 10–3 T at an angle of 90° to the direction of travel.
a. What is the radius of the circular path it follows?
b. Through what potential difference was the electron accelerated?

Answer 1

As the electron moves in the magnetic field, two forces of equal magnitude are acting on it:

$$qvB=\frac{mv^2}{R},\\ \space\\ R=\frac{mv}{qB}=4.55\text{ mm}.$$

Apply the work-energy theorem to find the potential difference:

$$\frac{mv^2}{2}=q\phi,\\ \space\\ \phi=\frac{mv^2}{2q}=45.5\text{ V}.$$