Lorentz Force & Charged Particle Motion Lorentz Force & Charged Particle Motion A magnetic field changes the direction of a moving charge, not its speed. The rule is simple: a charge moving with velocity v in a magnetic field B feels a force F that is perpendicular to both v and B . This sideways push bends the path. If v is exactly across B , the charge takes a circle. If it makes some angle with B , the motion becomes a helix around the field lines. Because the force is always at right angles to the motion, it cannot do work, so the kinetic energy and the speed remain constant. This is why magnets can steer beams of electrons without heating them up. It is also the core idea behind devices like cyclotrons and mass spectrometers that select, bend, and accelerate charged particles predictably. Direction matters. For a positive charge, F follows the right-hand rule for the cross product v B . For a negative charge, the force reverses. The magnitude depends on the angle between v and B as F = q v B . The largest bending happens when = 90 (pure circle), and there is no magnetic force when = 0 or 180 (straight line). These clear cases let you quickly classify motion in typical NEET questions. Beyond direction and magnitude, the motion connects to uniform circular motion (UCM). In a uniform B , the magnetic force provides the necessary centripetal force for the perpendicular component of motion. This gives a neat set of results: radius r = m v q B , angular speed = q B m , period T = 2 m q B , and frequency f = q B 2 m , independent of speed. The parallel component v just continues unchanged, building the helix with pitch p = v T . Applications flow directly. Velocity selectors use crossed electric and magnetic fields to pass only charges with v = E/B . Cyclotrons exploit the constant cyclotron frequency f = qB/(2 m) to accelerate particles to high energies: the magnetic field bends the path into semicircles inside two D-shaped electrodes while an alternating potential speeds them up at each gap. Mass spectrometers use the radius r p qB (magnetic rigidity) to separate isotopes. Throughout, watch the boundary conditions: uniform B , non-relativistic speeds (so mass m is constant), and absence of electric fields unless stated. Changing these assumptions changes the motion (e.g., with E 0 , you can get drift; at relativistic speeds, cyclotron frequency decreases due to relativistic mass increase). For NEET-level problems, stick to the clean ideal cases unless the question adds extras explicitly. Big picture: Magnetic force steers, not speeds. It is always perpendicular to motion, so it bends the path without changing the speed. Purely across B → circle; partly along B → helix; along B → straight. remember Magnetic part: F = q ,( v B ) . Perpendicular to both v and B ; magnitude F = q v B . Lorentz Force Right-hand rule Point fingers along v , curl toward B ; thumb gives F for positive charge. Reverse direction for negative charge. Gyroradius (Larmor radius) Radius of circular projection of a charge in a magnetic field: r = m v q B . Angular frequency = qB m and frequency f = qB 2 m of circular motion in uniform B . Cyclotron frequency Pitch of helix Axial distance advanced in one period: p = v T = 2 m v qB . Crossed E and B fields arranged so qE = qvB ; only particles with v = E/B pass undeviated. Velocity selector p/(qB) , proportional to bending radius for a given B . Larger rigidity → less curvature. Magnetic rigidity Motion at constant speed in a circle; needs centripetal force F c = m v 2 / r toward the center. Uniform circular motion (UCM) Direction by right-hand rule; for negative charges, reverse the direction. Lorentz force (magnetic part) is the angle between v and B . Magnitude of magnetic force This force determines the total resulting force on a charge, combining the electric push and the magnetic deflection based on velocity angle. When = 90 , the force is maximum and acts as a perfect centripetal force for the perpendicular component of motion. When = 0 or 180 , the force vanishes and the motion is a straight line along the field. For intermediate , decompose v into v and v relative to B . The v part undergoes circular motion; v remains constant and gives the axial advance that creates a helix. Uniform magnetic field B Non-relativistic speeds (mass m constant) No electric field Motion decomposed into v and v components Radius, frequency, and period in a uniform magnetic field r = m v q B , = qB m , T = 2 m qB , f = qB 2 m , p = 2 m v qB Magnetic force equals centripetal force for circular motion of the perpendicular component. Gyroradius depends on v only. Angular speed independent of speed and radius. Period and frequency are constants for given q, m, B . Pitch of the helix. Magnetic force does zero work; speed and kinetic energy remain constant. No work by magnetic force This formula applies to any charged particle moving with a constant velocity in a uniform magnetic field. The result P=0 is the key reason why speed is constant in a pure magnetic field. In combined fields, an electric field E can change speed because F E = q E does work. Practical devices exploit this separation: B steers and confines, E accelerates. Velocity selector condition Only particles with speed v = E/B pass undeflected. This condition applies to a velocity selector where a charged particle moves undeflected due to the perfect cancellation of electric and mag Sign convention: Treat q with its sign inside q ,( v B ) . For magnitudes, use |q| and fix direction at the end. Always decompose v into v and v relative to B before plugging into r = m v q B . tip Force magnitude F arbitrary Sine dependence: F ∝ sin θ. Zero at 0° and 180°, maximum at 90°. Magnetic force varies as sin θ. control dependent Angle θ between v and B (degrees) deg custom 0° No force Maximum force 90° 180° No force (opposite direction) Helical path: circular projection due to v⊥ and uniform advance along B due to v∥. Charge moving in a helix around magnetic field lines; components v perp and v parallel shown. Case analysis helps in fast identification: (i) = 0 or 180 → straight line along B . (ii) = 90 → circle with r = m v qB . (iii) 0 < < 90 → helix with radius using v = v and pitch using v = v . Remember that period and frequency depend only on q, m, B , not on speed. Charged Particle in B-Field Entry Angle ( θ ) Path Shape Velocity Magnitude Pitch Formula Force Calculation Zero force is straight, ninety is round, anything else spirals like a screw in the ground. = 90 Uniform Circular Constant v No Pitch ( P = 0 ) Maximum Force: F = qvB = 0 or 180 Straight Line Constant v Infinite/Not Applicable Minimum Force: F = 0 0 < < 90 Helical Constant v Pitch P = 2 mv qB Intermediate Force: F = qvB = 30 Helical Constant v Pitch P = 3 ,mv qB Half Max Force: F = 0.5qvB = 45 Helical Constant v Pitch P = 2 ,mv qB Force F = qvB 2 = 60 Helical Constant v Pitch P = mv qB Force F = 3 2 qvB charged particle in b field neet-alert Classic trap: In r = m v q B , use only the perpendicular component v = v . Using total v for oblique entry gives the wrong radius. In a cyclotron, the RF source flips polarity each half period to accelerate the particle at the gap. Cyclotron frequency and energy gain per gap Applies to charged particles undergoing circular motion in a uniform magnetic field (cyclotron motion) and accelerating through a potential Time for one full circle in uniform B. Time for a semicircle inside one dee. RF must flip every half-turn: frequency equals twice the cyclotron orbital frequency. This keeps the particle in phase with the accelerating field at the gap. Uniform magnetic field between dees Non-relativistic particle ( m constant) RF voltage flips every half revolution Particle crosses the gap each half-turn Cyclotron resonance condition f RF = qB m (flip rate) , f orbit = qB 2 m Magnetic bending relates momentum and charge. Since r = p qB for v c with p = mv , doubling B halves the radius at fixed p/q . This is why spectrometers can separate isotopes: same q but different m give different radii in the same B . Gyroradius grows linearly with perpendicular speed. Radius r Linear dependence r ∝ v⊥ for fixed B and q/m. v⊥ control dependent custom Speed v⊥ m/s Rest → no circle Electron: m = 9.11 10 -31 kg , q = 1.60 10 -19 C (magnitude) B = 0.50 T v = 3.0 10 6 m/s = 90 v = v easy Pattern: NEET 2017-type magnetic bending Plug values with SI units. Compute to two significant figures: 3.4 10 -5 m . Period independent of speed. Two significant figures: 7.2 10 -11 s . Radius r and period T An electron enters a uniform magnetic field B = 0.50 T with speed v = 3.0 10 6 m/s perpendicular to B . Find the radius and the time period of its circular motion. m, s Use r = m v q B and T = 2 m q B . Note that the electron’s direction of rotation will be opposite to that of a positive charge because q is negative. But radius and period calculations use |q| in magnitudes; handle direction at the end with the right-hand rule. Use r = m v qB and p = v T with T = 2 m qB . Helix radius r and pitch p A proton enters a uniform magnetic field B of magnitude 0.20 T making 60 with the field direction. Its speed is 2.0 10 6 m/s . Find the helix radius and pitch. medium Pattern: NEET 2015/2019 helical motion Compute perpendicular component. Plug into radius formula. Two significant figures: 9.0 10 -2 m . Period independent of speed and angle. Two significant figures: 3.3 10 -7 s . Parallel component. Two significant figures: 0.33 m . Proton: m = 1.67 10 -27 kg , q = 1.60 10 -19 C B = 0.20 T v = 2.0 10 6 m/s = 60 v = v 60 , v = v 60 neet-alert Do not use degrees directly inside a calculator expecting radians for trigonometric functions. Convert if needed, or ensure your calculator mode matches the angle unit in the problem. Direction of motion around the field lines: For q>0 and B pointing toward you, v to the right gives force upward (right-hand rule). For q<0 , flip the direction: the particle curves the other way. Visualizing v B before magnitudes prevents sign mistakes. Right-hand cross product cue v → curl to B → thumb gives F (for +q). For −q, flip the thumb direction. Two forces act on the charge. For straight, undeflected motion, net force is zero. Only charges with this speed pass undeflected; others deflect. v = E B Uniform, mutually perpendicular E and B Fields adjusted so electric and magnetic forces oppose each other Non-relativistic speeds Velocity selector condition v = E/B m/s, m Use v = E/B ; then r = m v qB for v B . Selector speed v and radius r in pure B A velocity selector uses B = 0.30 T and E = 9.0 10 4 V/m . What speed passes undeflected? If a proton of this speed then enters a region with only B = 0.30 T , what radius will it follow? medium Pattern: Velocity selector + bending Speed through selector. Radius in uniform magnetic field. Two significant figures: 1.0 10 -2 m . B = 0.30 T E = 9.0 10 4 V/m Proton: m = 1.67 10 -27 kg , q = 1.60 10 -19 C Cyclotron energy after N gap crossings: Each time the particle crosses the dee gap, it gains energy qV gap . If it crosses N times, K = N q V gap . The final radius r = 2 m K qB follows from p = 2mK and r = p/(qB) , valid for non-relativistic speeds. K = 1.5 keV = 1.5 10 3 e V 1 e V = 1.60 10 -19 J Crossings N = 40 B = 0.80 T m = 3.3 10 -26 kg Charge q = 1.60 10 -19 C Total energy in joules. Compute K. Momentum at exit. Square root evaluated. Final radius. Two significant figures. Pattern: Cyclotron radius from energy hard Final radius r In a cyclotron, a singly charged ion gains K = qV = 1.5 keV energy each gap crossing. The magnetic field is 0.80 T . After 40 crossings, estimate the final radius of its path. Take ion mass m = 3.3 10 -26 kg . Total kinetic energy K = N q V . Use p = 2mK and r = p/(qB) . Boundary conditions: Results here assume uniform B , v c (no relativistic effects), and no electric field unless stated. If B 0 , radii go to infinity (straight line). If q 0 , no magnetic force. If r 0 at finite v , assumptions are violated. tip Magnetic fields speed up charges because a force acts. The magnetic force is always perpendicular to velocity, so it does no work. Speed and kinetic energy remain constant in a pure B . Use total speed v in r = m v/(qB) for any entry angle. Use only the perpendicular component: r = m v /(qB) with v = v . Biot–Savart law gives the magnetic field produced by a current: d B = 0 4 , I , d l r r 2 . Here we treat B as given; this law explains how coils and wires create the uniform fields used to bend charges. Field increases with current and length of the element. Inverse-square fall-off with distance. Only the component of d l perpendicular to r contributes. Combine proportionalities with a constant k. Define the constant to match SI units and experiments. Scalar form; vector form uses the cross product. Direction from d l r by right-hand rule. d B = 0 4 , I , d l r r 2 Steady current (magnetostatics) Field in free space (permeability 0 ) Contribution from a small straight current element d l at distance r Biot–Savart law for the field of a current element Connection: In this chapter, we usually start with a given B (produced by coils/solenoids computed via Biot–Savart or Ampere’s law) and then analyze the particle motion using Lorentz force. Keep the source–effect chain clear: current → B (Biot–Savart) → motion (Lorentz). Plan view showing circular motion of a charge in a perpendicular magnetic field with force arrows. Uniform B-region with a circular orbit projection; centripetal force toward the center equals magnetic force. Quick procedure for charged particle motion in B Draw B and decompose v into v and v . Use right-hand rule on v B to get F direction (flip for −q). If v 0 , set q v B = m v 2 / r to get r . Compute T = 2 m/(qB) and pitch p = v T if helical. Check energy: speed constant unless E present; report direction separately. neet-alert Sign-of-charge trap: The formula for r uses |q| for magnitude, but the sense of rotation (clockwise/anticlockwise) flips with the sign of q. Many errors come from mixing these. Radius r v⊥, B, m, |q| v∥ r = m v⊥/(|q| B) Period T m, |q|, B Speed, angle T = 2π m/(|q| B) Pitch p v∥, m, |q|, B v⊥ p = 2π m v∥/(|q| B) Frequency f B, |q|, m Speed, radius f = |q| B/(2π m) Quantity Depends on Independent of Expression (non-relativistic) Mass spectrometry link: For particles with the same |q| and same speed after a common accelerator, the separation on a detector in a uniform B depends on m . Measuring r yields m (or m/q ) via r = m v |q| B . This is the operational basis for isotope separation and e/m measurement setups. From centripetal balance. Express speed via kinetic energy. Connect to momentum p= 2mK . Uniform magnetic field Non-relativistic particle Perpendicular entry ( v = v ) Relation between radius and kinetic energy r = 2mK |q| B = p |q| B Edge physics (beyond syllabus, but good intuition): At relativistic speeds, effective mass increases as m , so the cyclotron angular frequency becomes = |q|B m , decreasing with speed. Synchrotrons adjust RF frequency or magnetic field to maintain resonance, unlike basic cyclotrons that assume constant m . With =0 , F B=0 and the charge moves in a straight line at constant speed along B . No spiral occurs without some v or other forces. If =0 , the charge will spiral along B slowly. Worked directions: If B is into the page and a positive charge moves to the right, v B points upward (so the path curves upward). For a negative charge with the same v and B , the force is downward. After a quarter-turn, velocity becomes upward for +q (still perpendicular to B ), consistent with uniform circular motion. Compute frequency. Two significant figures: 3.1 MHz . Compute radius. Two significant figures. Pattern: Frequency independent of v medium q = 2 1.60 10 -19 C m = 6.64 10 -27 kg B = 0.40 T v = 1.0 10 6 m/s Use f = |q|B/(2 m) and r = m v/(|q| B) . Hz, m An alpha particle ( q = +2e , m = 6.64 10 -27 kg ) moves perpendicularly to B = 0.40 T with v = 1.0 10 6 m/s . Find its frequency and radius. Frequency f and radius r When formulas apply and when they fail Apply r = m v /(|q|B) only for uniform B and non-relativistic speeds. Use pitch p = 2 m v /(|q|B) only if B is uniform and no E is present along the axis. Cyclotron frequency f = |q|B/(2 m) holds only if mass is constant (non-relativistic). With strong E parallel to B , speed changes and the helix stretches; re-evaluate using energy change from q E v . Speed stays constant in pure B, but momentum direction changes continuously. Think of B as a perfect ‘steering wheel’ for charges. remember Computational tip: In mixed-angle problems, compute v and v numerically first, then treat circular motion and axial advance separately. This reduces algebra mistakes and makes units tracking easier. Free-body diagram logic: The only magnetic force on a single moving charge is F B = q ,( v B ) . If gravity is relevant (slow, heavy ions, weak B ), add m g , but in most NEET cases qvB mg and gravity is neglected. For current-carrying wires, the distributed Lorentz force yields F = I , L B , but here we focus on single charges. Macroscopic analog of Lorentz force per charge ( I= n q A v ). Useful link to later topics. Force on current-carrying conductor (context) Angle evolution in helical motion: Although the speed components v and v remain constant in magnitude, the direction of v keeps rotating with angular speed = |q|B/m . Thus v traces out a helix with constant pitch and radius. The projection on a plane perpendicular to B is a uniform circle. x = r ( t + 0), y = r ( t + 0), z = v t Parametric equations for helical motion (axis along z) Uniform B = B z Initial velocity has components v in x–y plane and v along z Circular motion in x–y plane with r = m v /(|q|B) . Uniform motion along the field. Cyclotron angular frequency. Energy–radius scaling quick checks: If kinetic energy quadruples at fixed B and q , the radius doubles because r K . If mass doubles at fixed K , r m increases by 2 . Such proportional reasoning speeds up MCQ solving. Cyclotron limit: As particle speed approaches a significant fraction of c, mass increase lowers cyclotron frequency. If you still use f = |q|B/(2 m) with constant m, you overestimate the resonant frequency and acceleration fails. neet-alert Units sanity: B in tesla ( T ), q in coulomb ( C ), v in m/s , r in m . Check that r = m v/(|q|B) has units kg ,m/s / ( C ,T ) = N ,s /( C ,T ) = (C ,V ,s)/ (C ,(V ,s/m 2)) = m . Sign of curvature: Facing along B , a positive charge rotates anticlockwise if its initial v is to your right; a negative charge rotates clockwise. State the sense of rotation explicitly in answers when asked. Orbital frequency f = |q|B/(2 m) (speed independent). RF flip frequency at the gap is f RF = 2f (semicircle timing). Energy gain per crossing K = |q| V gap . Exit radius set by final momentum: r = p/(|q|B) . Cyclotron essentials at a glance Velocity selector Steers (deflects) off-speed particles Balances magnetic force at v = E/B Selects a single speed Cyclotron Bends path semicircularly Accelerates at the gap Frequency independent of speed (non-relativistic) Mass spectrometer (magnetic sector) Separates by radius r ∝ p/(qB) Pre-accelerates beam to known p Measures m/q via r Device Role of B field Role of E field Speed dependence Problem-solving rhythm: (1) Draw vectors; (2) Decompose velocities; (3) Use qv B = m v 2/r ; (4) Use time/frequency formulas; (5) Translate to energy or momentum if convenient; (6) State direction and sense of rotation clearly. Use r = 2mK |q| B . Radius r A singly charged ion ( q = +e ) with K = 4.0 keV enters a 0.25 T magnetic field perpendicularly. If its mass is 6.6 10 -26 kg , find the radius. Pattern: e/m via radius Convert eV to J. Compute momentum. Square root. Divide by |q|B. Two significant figures. hard K = 4.0 keV = 4.0 10 3 1.60 10 -19 J B = 0.25 T q = 1.60 10 -19 C m = 6.6 10 -26 kg Direction wrap-up example: If B is along +z and a positive ion initially moves along +x, the force is along +y, so it curves toward +y. After a quarter period, velocity is along +y; after half a period, along −x; after three quarters, along −y; and back to +x in one full period. Multiple-choice hack: Before computing, predict trends. Increasing B should decrease r and T but increase f. If your number goes the other way, recheck algebra or units. Experimental scales: Laboratory magnets of 0.1–1 T are common; MRI machines go up to a few tesla. Earth’s field is about 5 10 -5 T , far too weak to bend fast charges into small circles, which is why cosmic rays trace very large arcs in the geomagnetic field. Sign-sensitive reporting: Always report both magnitude and sense (clockwise/anticlockwise) of rotation, and whether the motion is circle, helix, or line. This typically earns full marks even when the question is primarily numerical. Lorentz force F = q ,( v B ) magnetic force Larmor radius r = m v⊥/(|q|B) Gyroradius f = |q|B/(2πm) Cyclotron frequency Pitch p = 2π m v∥/(|q|B) Velocity selector Passes only v = E/B p/(|q|B) Magnetic rigidity p/(|q|B) Key terms recap