Physics · Moving Charges And Magnetism · NEET
When v is perpendicular to B, the magnetic force has magnitude F = qvB and always points toward the centre of the path (perpendicular to v). A force that is always perpendicular to velocity cannot change speed, only direction — this is exactly the condition for uniform circular motion. So this magnetic force plays the role of the centripetal force. Setting them equal: qvB = mv^2/r. Cancel one v from both sides: qB = mv/r, which rearranges to r = mv/qB. This is why the derivation is just 'magnetic force = centripetal force'.
Two different quantities behave differently, and NEET loves this trap. The RADIUS does depend on speed: r = mv/qB, so if v doubles, r doubles. But the TIME PERIOD T = 2 pi m/(qB) does NOT depend on speed at all — a faster particle travels a bigger circle but also moves faster, so the time to go around stays the same. Rule to remember: radius scales with v; period and frequency do not. Keep these separate on the next page about cyclotron frequency.
Rewrite the formula. Since momentum p = mv, we have r = p/(qB) — radius is directly proportional to momentum. For kinetic energy K = (1/2)mv^2, momentum p = sqrt(2mK), so r = sqrt(2mK)/(qB). So r is proportional to the square root of kinetic energy. Quick checks: same momentum, different charge means r is inversely proportional to q; same kinetic energy means r depends on sqrt(m)/q. Choosing the right form (p or K) before plugging numbers saves time in NEET.
Because the magnetic force qvB is always perpendicular to the velocity, it never speeds the charge up or slows it down — it only bends the direction by the same amount everywhere. A constant-magnitude force that is always sideways to the motion traces a perfect circle at constant speed. It becomes a spiral (helix) only when v has a component ALONG B; then that parallel part is unaffected and the particle drifts forward while circling. Pure circle needs v exactly perpendicular to B.
Use r = mv/qB. Same v and same B, so r is proportional to m/q. The proton is about 1836 times heavier than the electron, and both have the same charge magnitude e. So the proton's radius is about 1836 times larger. The electron makes a very tight circle, the proton a wide one. Also note they curve in opposite senses because their charges have opposite signs.
Ionized hydrogen atoms and alpha-particles with same momenta enter perpendicular to a constant magnetic field B. The ratio of the radii of their paths r_H : r_alpha will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
r = mv/(qB), where m is mass, v is the speed perpendicular to the field, q is the charge and B is the magnetic field strength. It comes from setting the magnetic force qvB equal to the centripetal force mv^2/r.
Only the perpendicular component of velocity, v_perp, sets the radius: r = m v_perp/(qB). The parallel component makes the path a helix, but it does not affect the circle's radius.
Using p = mv, r = p/(qB). Using kinetic energy K = p^2/2m, momentum p = sqrt(2mK), so r = sqrt(2mK)/(qB). Radius grows with momentum and with the square root of kinetic energy.
It links directly to velocity selectors, mass spectrometers and the cyclotron. Many questions compare radii of two particles (same speed, same momentum or same energy), so knowing which quantity is held equal decides the answer instantly.
Smaller. Since r = mv/qB, r is inversely proportional to B. A stronger field bends the particle more sharply, giving a tighter circle.
