CUET UG Physics Magnetism Magnetic Force on Moving Charge Lorentz Force Circular Motion in Field

Must-Know

• The Lorentz force on a charged particle moving in electric and magnetic fields is given by F = q(E + v × B), where q is charge, v is velocity, B is magnetic field, and E is electric field. Example: A proton moving in +x direction with 10⁴ m/s in a uniform 0.1 T magnetic field along +z experiences a force in +y direction.
• Magnetic force on a moving charge is F = qvB sinθ, where θ is angle between v and B; force is maximum when θ = 90°. Example: Electron moving perpendicular to 0.2 T field at 10⁶ m/s experiences force = (1.6×10⁻¹⁹)(10⁶)(0.2) = 3.2×10⁻¹⁴ N.
• Magnetic force does zero work on a charged particle because it is always perpendicular to velocity; hence, kinetic energy remains constant.
• Direction of magnetic force on positive charge is given by right-hand rule for cross product v × B; for negative charge (e.g., electron), direction is opposite.
• In a uniform magnetic field, if v is perpendicular to B, the charged particle undergoes uniform circular motion with centripetal force provided by magnetic force.
• Radius of circular path: r = mv/(qB). Example: Proton (m = 1.67×10⁻²⁷ kg, q = 1.6×10⁻¹⁹ C) moving at 2×10⁶ m/s perpendicular to 0.5 T field has r = (1.67×10⁻²⁷×2×10⁶)/(1.6×10⁻¹⁹×0.5) ≈ 0.042 m.
• Time period of revolution in circular path: T = 2πm/(qB) — independent of speed and radius. Example: For electron in 0.1 T field, T ≈ (2×3.14×9.1×10⁻³¹)/(1.6×10⁻¹⁹×0.1) ≈ 3.58×10⁻¹⁰ s.
• Frequency of revolution (cyclotron frequency): f = qB/(2πm). Example: For proton in 1 T field, f ≈ (1.6×10⁻¹⁹×1)/(2×3.14×1.67×10⁻²⁷) ≈ 1.5×10⁷ Hz.
• If velocity has a component parallel to B, the motion is helical; radius depends on perpendicular component, pitch on parallel component.
• Pitch of helical path: p = v_∥ × T = v cosθ × (2πm)/(qB). Example: If α-particle (He²⁺) moves at 30° to B with speed 2×10⁶ m/s in 0.4 T field, pitch = (2×10⁶×cos30°)×(2π×6.64×10⁻²⁷)/(3.2×10⁻¹⁹×0.4) ≈ 0.11 m.
• Lorentz force is not conservative because it depends on velocity and cannot be derived from a scalar potential.
• In crossed electric and magnetic fields (velocity selector), only particles with speed v = E/B pass undeflected. Example: If E = 1000 V/m and B = 0.1 T, selected speed = 10⁴ m/s.
• Cyclotron uses perpendicular magnetic field and oscillating electric field to accelerate charged particles; frequency of electric field matches cyclotron frequency.
• Cyclotron cannot accelerate electrons because their relativistic mass increase at high speeds causes loss of resonance with fixed-frequency electric field.
• Cyclotron cannot accelerate neutrons because they are uncharged and experience no Lorentz force.
• Magnetic force on a moving charge is measured in newtons (N); SI unit of B is tesla (T), where 1 T = 1 N/(A·m).
• 1 tesla = 10⁴ gauss; Earth’s magnetic field ≈ 0.3–0.6 gauss = 3–6×10⁻⁵ T.
• The direction of force on a negative charge moving in magnetic field is opposite to that predicted by right-hand rule for v × B.
• When θ = 0° or 180° (parallel or antiparallel), magnetic force is zero. Example: Proton moving along magnetic field lines experiences no force.
• The Lorentz force equation combines both electric and magnetic forces and is fundamental in electromagnetism as per NCERT Class 12 Physics, Chapter 4.

Difficulty Level

Intermediate — requires understanding of vector cross product, motion dynamics, and application of formulas in different configurations (circular, helical, velocity selector), but direct numerical problems are common and formula-based.

Common CUET Traps

• Trap: Assuming magnetic force can change speed or do work. Avoid: Remember magnetic force is always perpendicular to velocity → changes direction only, not speed.
• Trap: Using total velocity in radius formula when motion is helical. Avoid: Use only the component of velocity perpendicular to B in r = mv⊥/(qB).
• Trap: Applying cyclotron to electrons or neutrons. Avoid: Cyclotron works only for positively charged particles like protons, α-particles; not for electrons (relativistic effects) or neutrons (no charge).

Practice MCQs

1. A proton moves perpendicular to a uniform magnetic field of 0.2 T with a speed of 3×10⁶ m/s. What is the radius of its circular path? (Mass of proton = 1.67×10⁻²⁷ kg, charge = 1.6×10⁻¹⁹ C)
   A. 0.156 m
   B. 0.312 m
   C. 0.078 m
   D. 0.624 m
   Answer: A
   Explanation: r = mv/(qB) = (1.67×10⁻²⁷×3×10⁶)/(1.6×10⁻¹⁹×0.2) = 0.156 m.
   Why others fail: Option C results from halving the correct value, possibly due to miscalculation or wrong substitution.

2. Which of the following is true about the magnetic force on a moving charged particle?
   A. It can increase the kinetic energy of the particle
   B. It is always parallel to the magnetic field
   C. It is independent of the direction of velocity
   D. It is zero if the particle moves parallel to the field
   Answer: D
   Explanation: When v is parallel to B, sinθ = 0 → force = 0.
   Why others fail: Option A is tempting because forces often change energy, but magnetic force does no work.

3. A charged particle enters a region of uniform magnetic field with a velocity at an angle of 60° to the field. The path of the particle will be:
   A. Circular
   B. Parabolic
   C. Helical
   D. Straight line
   Answer: C
   Explanation: Velocity has both parallel and perpendicular components → helical motion.
   Why others fail: Option A is tempting if student assumes perpendicular entry only.

4. In a velocity selector, the electric field is 2000 V/m and the magnetic field is 0.1 T. What should be the speed of the charged particle to pass undeflected?
   A. 2×10³ m/s
   B. 2×10⁴ m/s
   C. 4×10⁴ m/s
   D. 5×10³ m/s
   Answer: B
   Explanation: v = E/B = 2000 / 0.1 = 2×10⁴ m/s.
   Why others fail: Option A results from dividing by 1 T instead of 0.1 T — unit error.

5. A deuteron and an alpha particle have the same kinetic energy and move in a uniform magnetic field perpendicular to their velocities. What is the ratio of their radii of circular paths? (Mass of deuteron = 2m, charge = q; mass of alpha = 4m, charge = 2q)
   A. 1:1
   B. 1:√2
   C. √2:1
   D. 2:1
   Answer: C
   Explanation: r = √(2mK)/(qB); ratio = [√(2×2m×K)/(qB)] / [√(2×4m×K)/(2qB)] = (√4m / q) × (2q / √8m) = (2/1) × (2 / 2√2) = √2:1.
   Why others fail: Option A assumes same momentum or ignores charge and mass differences.

Last‑Minute Revision

• ⚠️ Lorentz force: F = q(E + v × B) — vector sum.
• ⚠️ Magnetic force: F = qvB sinθ; max at 90°, zero at 0°.
• ⚠️ Magnetic force does no work — KE constant.
• ⚠️ Direction for negative charge: opposite to v × B.
• ⚠️ Circular motion if v ⊥ B: centripetal force = qvB.
• ⚠️ Radius: r = mv/(qB) — higher mass → larger radius.
• ⚠️ Time period: T = 2πm/(qB) — independent of v.
• ⚠️ Frequency: f = qB/(2πm) — called cyclotron frequency.
• ⚠️ Helical path if v has ∥ and ⊥ components.
• ⚠️ Pitch: p = v cosθ × (2πm)/(qB).
• ⚠️ Velocity selector: v = E/B — only selected speed passes.
• ⚠️ Cyclotron: uses perpendicular B and oscillating E.
• ⚠️ Cyclotron frequency must match particle revolution frequency.
• ⚠️ Cyclotron cannot accelerate electrons — relativistic mass change.
• ⚠️ Cyclotron cannot accelerate neutrons — no charge → no force.
• ⚠️ 1 T = 10⁴ G; Earth’s field ≈ 5×10⁻⁵ T.
• ⚠️ Right-hand rule gives force direction for positive charge.
• ⚠️ Force is zero when charge moves parallel to B.
• ⚠️ Magnetic field unit: tesla (T) = N/(A·m).
• ⚠️ Mnemonic: "Fleming’s Left Hand Rule" — Thumb = Force, Index = Field, Middle = Current (for positive charges).