(a) A monoenergetic electron beam with electron speed of

Solution:

(a)Speed of an electron, v = 5.20 × 106 m/s

Magnetic field experienced by the electron, B = 1.30 × 10−4 T

Specific charge of an electron, e/m = 1.76 × 1011 C kg−1

Where,

e = Charge on the electron = 1.6 × 10−19 C

m = Mass of the electron = 9.1 × 10−31 kg−1

The force exerted on the electron is given as:

$F=e|\vec{v} \times \vec{B}|$

$=e v B \sin \theta$

θ = Angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam.

$\therefore \theta=90^{\circ}$

$F=e v B$     .(1)

The beam traces a circular path of radius, $r$. It is the magnetic field, due to its bending nature, that provides the centripetal force $\left(F=\frac{m v^{2}}{r}\right)$ for the beam.

Hence, equation (1) reduces to:

$e v B=\frac{m v^{2}}{r}$

$\therefore r=\frac{m v}{e B}=\frac{v}{\left(\frac{e}{m}\right) B}$

$=\frac{5.20 \times 10^{6}}{\left(1.76 \times 10^{11}\right) \times 1.30 \times 10^{-4}}=0.227 \mathrm{~m}=22.7 \mathrm{~cm}$

Therefore, the radius of the circular path is 22.7 cm.

(b) Energy of the electron beam, $E=20 \mathrm{MeV}=20 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$

The energy of the electron is given as:

$E=\frac{1}{2} m v^{2}$

$\therefore v=\left(\frac{2 E}{m}\right)^{\frac{1}{2}}$

$=\sqrt{\frac{2 \times 20 \times 10^{6} \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}}=2.652 \times 10^{9} \mathrm{~m} / \mathrm{s}$

This result is incorrect because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v << c

When very high speeds are concerned, the relativistic domain comes into consideration.

In the relativistic domain, mass is given as:

$m=m_{0}\left[1-\frac{v^{2}}{c^{2}}\right]^{\frac{1}{2}}$

Where,

$m_{0}=$ Mass of the particle at rest

Hence, the radius of the circular path is given as:

$r=m v / e B$
$=\frac{m_{0} v}{e B \sqrt{\frac{c^{2}-v^{2}}{c^{2}}}}$