# (a) A monoenergetic electron beam with electron speed of

Question:

(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s−1 is subject to a magnetic field of 1.30 × 10−4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg−1.

(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]

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:

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