A particle with charge $+2.5\times 10^{-6},\text{C}$ moves with a speed of $2.0\times 10^5,\text{m/s}$ at an angle of $30^{\circ }$ to a uniform magnetic field of magnitude $0.15,\text{T}$. What is the magnitude of the magnetic force on the particle?

Step-by-step solution

1. Compute the component of velocity perpendicular to the field: $v_{\perp }=vsin\theta =2.0\times 10^5,\text{m/s}\times \sin 30^{\circ }=1.0\times 10^5,\text{m/s}$.
2. Use the magnetic force formula $F=q{v}_{perp}B$.
3. Substitute: $F=(2.5\times 10^{-6},\text{C})(1.0\times 10^5,\text{m/s})(0.15,\text{T})$.
4. Multiply: $2.5\times 10^{-6}\times 1.0\times 10^5=0.25$. Then $0.25\times 0.15=0.0375$.
5. Result: $F=3.75\times 10^{-2},\text{N}$.
6. The force acts perpendicular to both velocity and field (to the right if using right‑hand rule).

Concept tested: Magnetic force on moving charge

Published 7 October 2026 · How this solution was prepared and checked