Answer to Question #103939 in Electric Circuits for Maryam Aminu

Question

Suppose that electron in a uniform magnetic field of flux density 0.03T move in a circle of radius 0.2m what are the velocity and kinetic energy of electrons

Answer 1

Solution. Since an electron in a magnetic field moves in a circle, the magnetic field is perpendicular to the velocityof the electron. The force of Lorentz is equal to

$$F_L=qvB$$

where q=1.6×10^-19C is electron charge; v is velocityof the electron: B=0.03mT is a magnetic field.

A centripetal force acts on an electron moving in a circle. The magnitude of the force

$$F=\frac{mv^2}{r}$$

where m=9.1×10^-31kg is mass electron; r=0.2m is circle radius. Since the electron moves in a circle with a constant speed of force equal.

$$qvB=\frac{mv^2}{r}$$

Therefore

$$v=\frac{qrB}{m}=\frac{1.6×10^{-19}\times 0.2\times 0.00003}{9.1\times 10^{-31}}\approx1\times 10^6\frac{m}{s}$$

The kinetic energy of an electron is equal to

$$E=\frac{mv^2}{2}=\frac{9.1\times 10^{-31}(1\times 10^6)^2}{2}=4.55\times10^{-19}J$$

Answer.

$$v=1\times 10^6\frac{m}{s}$$ $$E=4.55\times10^{-19}J$$