nitude $F = BQv\\sin 90^\\circ = BQv$. The force acts perpendicular to both the velocity and the field."},{"text":"$$F = \\frac{BQ}{v}$","correct":false,"explanation":"Dividing by speed incorrectly suggests the force decreases as the particle moves faster. For perpendicular motion, magnetic force is directly proportional to speed, so $F = BQv$."},{"text":"$$F = \\frac{Bv}{Q}$","correct":false,"explanation":"Dividing by charge incorrectly makes particles with smaller charge experience greater force. Magnetic force is proportional to the charge magnitude, giving $F = BQv$ for perpendicular motion."},{"text":"$$F = BQv^2$","correct":false,"explanation":"Squaring the speed confuses magnetic force with expressions involving kinetic energy or centripetal acceleration. Magnetic force is directly proportional to speed, so $F = BQv$."}]},{"id":"590911ac-99ae-4908-aa0e-3f5d1ea435cb","cardType":"basic","front":"Under what specific condition is the equation $F = BQv$ valid?","back":"The particle velocity must be perpendicular (at right angles) to the magnetic field.","points":1,"mcqOptions":[{"text":"The particle velocity must be perpendicular (at right angles) to the magnetic field.","correct":true,"explanation":"The general magnitude is $F = BQv\\sin\\theta$. When velocity is perpendicular to the field, $\\theta = 90^\\circ$ and $\\sin\\theta = 1$, giving $F = BQv$."},{"text":"The particle velocity must be parallel (in the same direction) to the magnetic field.","correct":false,"explanation":"For parallel motion, $\\theta = 0^\\circ$, so $F = BQv\\sin 0^\\circ = 0$. The equation $F = BQv$ applies when velocity is perpendicular to the field."},{"text":"The particle velocity must be antiparallel (in the opposite direction) to the magnetic field.","correct":false,"explanation":"For antiparallel motion, $\\theta = 180^\\circ$, so the magnetic force is zero. The full value $F = BQv$ occurs when velocity is perpendicular to the field."},{"text":"The particle velocity must be inclined (at 45 degrees) to the magnetic field.","correct":false,"explanation":"At 45 degrees, the force is $BQv\\sin45^\\circ = BQv/\\sqrt{2}$, not $BQv$. The stated equation requires velocity perpendicular to the magnetic field."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Motion of charged particles","title":"Motion of charged particles Flashcards","topicSidebar":"$L40"}]}],"topicId":"524648a9-d7da-5582-a354-20115defef48","walled":"$L41"}],"$L42"]