Velocity Selector And Crossed Fields

NCERT Class 12, Chapter 4 (Moving Charges and Magnetism) describes a velocity selector in which mutually perpendicular electric and magnetic fields are arranged so that their forces on a moving charge oppose each other. A particle passes undeflected only when the electric force balances the magnetic force: $qE=qvB$, which is independent of charge and gives the selected speed $v=\frac{E}{B}$. Substituting $E=3\times 10^4$ V/m and $B=0.06$ T: $v=\frac{3\times 10^4}{0.06}=5\times 10^5$ m/s. The value $1.8\times 10^3$ m/s comes from wrongly multiplying $E$ by $B$. The value $2\times 10^6$ m/s and $5\times 10^6$ m/s result from decimal-place errors in dividing by $0.06$. Plausibility check: units give $\frac{\text{V/m}}{\text{T}}=\frac{\text{N/C}}{\text{Ncdotps/(Ccdotpm)}}=\text{m/s}$, correctly a speed, and $5\times 10^5$ m/s is a realistic selected speed for charged particles in such an apparatus. Note that because both the electric and magnetic forces scale with the same charge $q$, the balance condition is independent of the sign and magnitude of the charge, so the selector transmits the chosen speed regardless of the particle species; this charge-independence is the very feature that makes it a clean velocity filter feeding a mass spectrometer.