High above the surface of the Earth, charged particles (such as electrons and protons) can become trapped in the Earth's magnetic field in regions known as Van Allen belts. A typical electron in a Van Allen belt has an energy of 40 keV and travels in a roughly circular orbit with an average radius of 210 m. What is the magnitude of the Earth's magnetic field where such an electron orbits?

Question

anonymous · Answer

mv^2 =2* e*40k J/C
v=  sqrt(2* e*40k/m)

mv^2/r = F centripetal = evB
2* e*40k/(210*sqrt(2* e*40k/m)*e) =B

anonymous · Answer

$$ B= \frac{ \sqrt{e*80,000*m} }{ 210 }$$