A scientist observes aurora bands at an altitude of $100$ km. The Earth's magnetic field at that latitude is $5\times 10^{-5}$ T. If the charged particles causing the aurora have kinetic energy $1$ keV, calculate their speed. Assume the particles are electrons.

Step-by-step solution

1. Step 1: Convert kinetic energy to joules: $1\text{ keV}=1000\text{ eV}=1000\times 1.602\times 10^{-19}\text{ J}=1.602\times 10^{-16}\text{ J}$.
2. "Step 2: Use kinetic energy formula $K=\frac{1}{2}m{v}^2$ for an electron ($m=9.11\times 10^{-31}$ kg).
3. "Step 3: Solve for $v$: $v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\times 1.602\times 10^{-16}}{9.11\times 10^{-31}}}$.
4. "Step 4: Compute inside the square root: $\frac{3.204\times 10^{-16}}{9.11\times 10^{-31}}=3.515\times 10^{14}$.
5. "Step 5: Take the square root: $vapprox1.88\times 10^7$ m/s, rounded to $1.9\times 10^7$ m/s.

Concept tested: Aurora particle speed

Published 10 October 2026 · How this solution was prepared and checked