Question

In an experiment to determine the specific charge (e/m) of an electron, electrons where accelerated by an electric field strength of 10N/C and the magnetic field strength of 0.1H is used to deflect them. Find the speed of the accelerated electrons. What is the radius of the path described by the electron?

Answer 1

a) Using the Newton’s Second Law of Motion we can write:

$$evB=\dfrac{mv^2}{r},$$ $$r=\dfrac{mv}{eB}.$$

Let's assume that both fields are adjusted so that remains undeflected and arrives at the point at the center of the screen. Then, we can write:

$$F_e=F_m,$$ $$Bev=eE,$$ $$v=\dfrac{E}{B}=\dfrac{E}{\mu_0H}=\dfrac{10\ \dfrac{N}{C}}{4\pi\cdot10^{-7}\ \dfrac{H}{m}\cdot0.1\ H}=7.96\cdot10^7\ \dfrac{m}{s}.$$

Then, we can find the radius of the path described by the electron:

$$r=\dfrac{9.1\cdot10^{-31}\ kg\cdot7.96\cdot10^7\ \dfrac{m}{s}}{1.6\cdot10^{-19}\ C\cdot4\pi\cdot10^{-7}\ \dfrac{H}{m}\cdot0.1\ H}=3602\ m.$$