An electron with speed $4.0\times 10^5;{\text{m s}}^{-1}$ moves perpendicular to a uniform magnetic field of magnitude $0.30;\text{T}$ directed towards the observer. Find the magnitude and direction of the magnetic force on the electron.

Step-by-step solution

1. Use the magnitude formula for magnetic force: $F=|q|vB$.
2. Take the charge of electron $|q|=1.602\times 10^{-19};\text{C}$, $v=4.0\times 10^5;{\text{m s}}^{-1}$, $B=0.30;\text{T}$.
3. Compute $F=(1.602\times 10^{-19})(4.0\times 10^5)(0.30)=1.9224\times 10^{-14};\text{N}$, round to $1.92\times 10^{-14};\text{N}$.
4. Determine direction using the right–hand rule for a negative charge: the force is opposite to the direction given by the rule.
5. For a field directed towards the observer and a velocity perpendicular to it, the rule gives a force towards the observer; therefore the electron experiences a force away from the observer.
6. Thus the force is $1.92\times 10^{-14};\text{N}$ directed away from the observer.

Concept tested: Force on moving charge in magnetic field

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked