Question

An electron of kinetic energy 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.

Answer 1

Given:
Kinetic energy of an electron = 100 eV
Radius of the circle = 10 cm
$\frac{1}{2}m{v}^2$ = 100 eV = 1.6 × 10⁻¹⁷ J (1 eV = 1.6 × 10⁻¹⁹ J)
Here,
m is the mass of an electron and v is the speed of an electron. Thus,
$\frac{1}{2}$ × 9.1 × 10⁻³¹ × v² = 1.6 × 10⁻¹⁷ J
⇒ v² = 0.35 × 10¹⁴
v = 0.591 × 10⁷ m/s
Now,
$$\begin{gathered} r=\frac{mv}{eB} \\ \Rightarrow B=\frac{mv}{er} \end{gathered}$$
$=\frac{9.1\times {10}^{-31}\times 0.591\times {10}^7}{1.6\times {10}^{-19}\times 0.1}$
B = 3.3613 × 10⁻⁴ T
Therefore, the applied magnetic field = 3.4 × 10⁻⁴ T
Number of revolutions per second of the electron,
$f=\frac{1}{T}$
$T=\frac{2\pi r}{v}=\frac{2\pi m}{eB}$
$T=\frac{2\mathrm{\pi }m}{Be}$
$$\begin{gathered} f=\frac{Be}{2\mathrm{\pi }m} \\ =\frac{3.4\times {10}^{-4}\times 1.6\times {10}^{-19}}{2\times 3.14\times 9.1\times {10}^{-31}} \end{gathered}$$

= 0.094 × 10⁸
= 9.4 × 10⁶
f = 9.4 × 10⁶