Special theory of relativity > Motion of relativistic particles in electric and magnetic fields > Problem 14.4.25

Statement

$14.4.25$

Estimate at what minimum energy electrons located at an altitude of h =
1000 km will be able to reach the Earth’s surface in the equator region, if the
Earth’s magnetic field induction B = 30 µT?

Solution

Physical criterion

At the magnetic equator, the field $\mathbf{B}$ is horizontal.
An electron moving toward Earth experiences a perpendicular magnetic force that curves its trajectory.
For it not to become trapped in a spiral, the Larmor radius must be at least of the order of the height h:

$r_L = \frac{p}{eB} \gtrsim h \quad\Rightarrow\quad p \approx e B h$.

Here p is the relativistic momentum of the electron

Calculation of the momentum

Substituting the values

$p = (1.6\times 10^{-19})(3.0\times 10^{-5})(10^6) = 4.8\times 10^{-18}\ \text{kg·m/s}$

We multiply by c to work in energy units:

$pc = 4.8\times 10^{-18} \times 3.0\times 10^8 = 1.44\times 10^{-9}\ \text{J}$

Convert to MeV $(1\ \text{MeV} = 1.6\times 10^{-13}\ \text{J})$

$pc = \frac{1.44\times 10^{-9}}{1.6\times 10^{-13}}\ \text{MeV} \approx 9.0\ \text{MeV}$

Kinetic energy

The electron is clearly ultra‑relativistic $(pc \gg m_e c^2)$ The exact relation between momentum and total energy is:

$E_{\text{total}} = \sqrt{(pc)^2 + (m_e c^2)^2}$

and the kinetic energy is obtained by subtracting the rest mass:

$K = E_{\text{total}} - m_e c^2$

Substituting the values:

$K = \sqrt{9.0^2 + 0.511^2} - 0.511
\approx \sqrt{81 + 0.261} - 0.511
\approx 9.014 - 0.511
\approx 8.5\ \text{MeV}$

$\boxed{K_{\min} \approx 8.5\ \text{MeV}}$

Electrons need a kinetic energy of the order of several MeV to cross the equatorial magnetic field and reach the ground. Less energetic electrons become trapped in the radiation belts.

Answer

$\boxed{K_{\min} \approx 8.5\ \text{MeV}}$