Question

"Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were accelerated by a voltage of 3.0 kV; the beam was then steered to different points on the screen by coils of wire that produced a magnetic field of up to 0.55"

Answer 1

Answer:

Speed of electrons, v = $3.24\times 10^7\ m/s$

Explanation:

Given that,

The electrons in the beam were accelerated by a voltage of 3.0 kV, V = 3000 volts

Magnetic field, B = 0.55 T

Let us assume to find the speed of the electrons in the beam. In this case, final kinetic energy is equal to the electric potential energy lost moving in potential V

$\dfrac{1}{2}mv^2=qV$

v is the speed of electron

$v=\sqrt{\dfrac{2qV}{m}}$

m and e are mass and charge on electron

$v=\sqrt{\dfrac{2qV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 3000}{9.1\times 10^{-31}}} \\\\v=3.24\times 10^7\ m/s$

So, the speed of the electron is $3.24\times 10^7\ m/s$. Hence, this is the required solution.