Answer to Question #168407 in Astronomy | Astrophysics for Bethv

Question

1) The magnitude of the Earth’s magnetic field is about 0.5 Gauss near Earth's surface. What’s the maximum possible force on an electron near the Earth’s surface with kinetic energy of 1 keV? How does this compare to the gravitational force on the electron near the Earth’s surface?

Answer 1

The maximum modulus of the Lorentz force is $F_L = q V B$. We may obtain velocity from the kinetic energy.

$E = \dfrac{m_eV^2}{2}, \;\; V = \sqrt{\dfrac{2E}{m_e}} = \sqrt{\dfrac{2\cdot 10^3\cdot1.6\cdot10^{-19}\,\mathrm{J}}{9.1\cdot10^{-31}\,\mathrm{kg}}} = 1.9\cdot10^7\,\mathrm{m/s}$

The modulus of the Lorentz force is $F_L = 1.6\cdot10^{-19}\,\mathrm{C}\cdot 1.9\cdot10^{7}\,\mathrm{m/s}\cdot 0.5\cdot10^{-4}\,\mathrm{T}= 1.52\cdot 10^{-16}\,\mathrm{N}.$

The gravitational force is $F_G = \dfrac{GM_{\oplus}m_e}{R_{\oplus}^2} = m_e\cdot g = 9.1\cdot10^{-31}\,\mathrm{kg}\cdot 9.81\,\mathrm{N/kg} = 8.9\cdot10^{-30}\,\mathrm{N}.$

We can see that the Lorentz force is significantly greater than the gravitational force.