An electron enters a uniform magnetic field of magnitude $0.20\text{T}$ perpendicular to its velocity. Its speed is $2.0\times 10^6{\text{m s}}^{-1}$. Find the radius of its circular path (in meters).

Step-by-step solution

1. Use the formula for the radius of the circular motion of a charge in a magnetic field: $r=\frac{mv}{|q|B}$.
2. Insert the electron mass $m=9.11\times 10^{-31}\text{kg}$, speed $v=2.0\times 10^6{\text{m s}}^{-1}$, charge magnitude $|q|=1.60\times 10^{-19}\text{C}$ and $B=0.20\text{T}$.
3. Calculate the numerator $mv=9.11\times 10^{-31}\times 2.0\times 10^6=1.822\times 10^{-24}{\text{kg m s}}^{-1}$.
4. Calculate the denominator $|q|B=1.60\times 10^{-19}\times 0.20=3.20\times 10^{-20}\text{C T}$.
5. Divide: $r=\frac{1.822\times 10^{-24}}{3.20\times 10^{-20}}=5.69\times 10^{-5}\text{m}$.
6. Round to two significant figures: $5.7\times 10^{-5}\text{m}$.
7. This is the required radius.

Concept tested: Radius of charge in magnetic field

Published 5 September 2026 · How this solution was prepared and checked