a)
Given that
A monoenergetic electron beam speed v = 5.20 * 10^{6} m/s
Magnetic field B = 1.30 * 10^{-4} T
Specific charge of electron \frac{e}{m} = 1.76*10^{11} C/kg
Magnetic field is perpendicular to the beam velocity
So \theta = 90 \degree
Step 1: Set up an equation for the radius of the circular path of charged particle
Formulas required
Force acting on electron in magnetic field F = evB \sin \theta
Where, e - charge, B - magnetic field, v - velocity, \theta - angle between the velocity and the magnetic field.
F = evB .........................(1) \because \theta = 90 \degree
A charged particle moving in a circular path, so it acquires centripetal force
Centripetal force F = \frac{mv^2}{r} ...............(2)
Where, r - radius of the circular path
From equation(1) and (2) we can write as follows
evB = \frac{mv^2}{r}
r = \frac{mv^2}{evB}
r = \frac{mv}{eB}
Hence, radius of the circular path of charged particle r = \frac{mv}{qB} ................(3)
Step 2: Plug in the given values in the equation (3) for radius of circular path
r = \frac{v}{\frac{e}{m} B}
r = \frac{5.20 * 10^{6}}{1.76*10^{11} * 1.30 * 10^{-4}}
r = 0.227 m
r = 22.7 cm
Therefore, radius of the circular path of charged particle r = 22.7 cm
b)
Given that
Energy of the electron beam E = 20 MeV = 20* 10^{6} eV
E = 20 * 10^{6} * 1.6 * 10^{-19} J
Step 1: Set up an equation for the velocity of the electron
Kinematic energy K.E = \frac{1}{2} mv^2.
v^2 = \frac{2 K.E}{m}
v = \sqrt{\frac{2 K.E}{m}}
Step 2: Substitute the given values in the velocity equation
Mass of an electron m = 9.1 * 10^{-31} kg
v = \frac{2 * 20 * 10^{6} * 1.6 * 10^{-19}}{9.1 * 10^{-31}}
v = 2.65 * 10^{9} m/s
Step 3: Observation
Speed of light c = 38 10^{8} m/s
Speed of electron is greater than the speed of the light which is practically not possible.
K.E = \frac{1}{2} mv^2. is valid only for v << c
So modify the formula of radius by replacing the relativistic mass
Modified mass m=m_0*(1-\frac{v^2}{c^2})^{-\frac{1}{2}}
m_o - mass of particle at rest position
Thus, valid formula for calculating the radius of circular path is as follows
Radius r=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}\cdot\frac{v}{eB}