A charge of $+1.2\times 10^{-6},\text{C}$ moves with a speed of $1.5\times 10^5,\text{m/s}$ perpendicular to a uniform magnetic field of magnitude $0.25,\text{T}$. What is the magnitude of the magnetic force acting on the charge?

Step-by-step solution

1. Identify that the angle between the velocity and the magnetic field is $90^{\circ }$, hence $sin\theta =1$.
2. Compute the product $vB=(1.5\times 10^5,\text{m/s})(0.25,\text{T})=3.75\times 10^4,\text{(m/s)⋅T}$.
3. Apply the magnetic force formula $F=qvBsin\theta$ to get $F=(1.2\times 10^{-6},\text{C})(3.75\times 10^4)=4.5\times 10^{-2},\text{N}$.
4. State the final result $F=4.5\times 10^{-2},\text{N}$.

Concept tested: Magnetic force on a moving charge

Published 21 September 2026 · How this solution was prepared and checked