An electron ($q=-e$, $e=1.60\times 10^{-19},\text{C}$) moves with a speed of $5.0\times 10^5,{\text{m,s}}^{-1}$ perpendicular to a uniform magnetic field of magnitude $0.40,\text{T}$ directed along the $+y$–axis. Find the magnitude and direction of the magnetic force on the electron.

Step-by-step solution

1. Step 1: Recall the Lorentz force law for a charge moving in a magnetic field: $\mathbf{F}=qmathbfv\times \mathbf{B}$. The magnitude is $F=|q|vBsin\theta$ where $\theta =90^{\circ }$ since $\mathbf{v}perpmathbfB$.
2. Step 2: Compute the magnitude: $|q|=1.60\times 10^{-19},\text{C}$, $v=5.0\times 10^5,{\text{m,s}}^{-1}$, $B=0.40,\text{T}$, $sin90^{\circ }=1$. Thus $F=1.60\times 10^{-19}\times 5.0\times 10^5\times 0.40=3.20\times 10^{-14},\text{N}$. Wait: check units; actually multiply correctly: $1.60\times 5.0=8.0$, so $F=8.0\times 10^{-19+5}\times 0.40=3.20\times 10^{-14},\text{N}$. But we must account for the negative charge: direction is opposite to $\mathbf{v}\times \mathbf{B}$.
3. Step 3: Determine direction using right–hand rule: point fingers along $\mathbf{v}$ (say along $+x$), curl towards $\mathbf{B}$ ($+y$); the thumb points along $+z$. Since $q$ is negative, force is opposite, i.e., $-z$ direction.
4. Step 4: Final numeric value: $F=3.20\times 10^{-14},\text{N}$. Note that earlier miscalculation gave $1.28\times 10^{-23}$ – that is incorrect; correct is $3.20\times 10^{-14},\text{N}$.
5. Step 5: Provide answer with sign and direction: $3.20\times 10^{-14},\text{N}$, direction $-z$.

Concept tested: Magnetic force on moving charge

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked