Question

Example 4.3 What is the radius of the path of an electron (mass
9x10 ^-31 kg and charge 1.6 x 10^-19 C) moving at a speed of 3 x10^7 m/s in
a magnetic field of 6 X 10^-4 T perpendicular to it? What is its
frequency? Calculate its energy in keV. (1 eV = 1.6x10^-19 J).​

Answer 1

Answer:

$R=\frac{9}{32}m\\ \\ f=1.7*10^7Hz\\ \\E=2.5KeV$

Explanation:

1) We know,

Centripetal force in moving electron is provided by magnetic force .

Hence,

$F_{mag}=F{r}\\ \\=>q(v \times B)=\frac{mv^2}{R}\\ \\=>R=\frac{mv}{qB}\\ \\=\frac{9*10^{-31}*3*10^7}{1.6*10^{-19}*6*10^{-4}}=\frac{9}{32}m$

2) Now,

Frequency,

$f=\frac{1}{T}=\frac{v}{2\pi r}=\frac{3*10^7}{2*\pi*9/32}=1.7*10^7 Hz$

3) Energy,

$E=\frac{1}{2}mv^2=\frac{1}{2}*9*10^{-31}*(3*10^7)^2\\ \\ =4.05*10^{-16}J \\ \\ =\frac{4.05*10^{-16}}{1.6*10^{-19}} \approx 2.5*10^3eV=2.5KeV$