Total force on a charge q in fields E and B: F = q(E + v × B). Magnetic force is always perpendicular to v; does no work on charge.
-- NCERT Class 12 Physics, Ch. 4, p. 109Lorentz Force on Charge
8 MCQs4 revision cards9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 27 Sep 2026
Lorentz Force on Charge, explained for NEET
A common confusion on this topic: treating the magnetic force like any other force and asking how much work it does. It does none. In F = q(E + v × B), the magnetic term is a cross product with v, so it is perpendicular to the velocity at every instant. A force perpendicular to displacement transfers no energy. Speed is constant; only direction changes. The electric term qE has no such constraint — it is along E regardless of motion, and it does change speed.
NCERT Class 12 Physics, Chapter 4, page 108 states the Lorentz force as the sum of these two parts. Read the vector product carefully: the magnetic force vanishes when v is parallel or antiparallel to B, and is maximum when they are perpendicular.
That perpendicular case is the one NEET asks about. With v ⊥ B and no electric field, the force qvB stays perpendicular to v and has constant magnitude — the definition of uniform circular motion. Equating qvB to mv²/r gives
r = mv / (qB), T = 2πm / (qB)
NCERT page 110 notes the consequence worth memorising: the period does not contain v. A faster particle traces a bigger circle in the same time. This is why a cyclotron can use a fixed-frequency oscillator.
When v has a component along B, that component is untouched by the magnetic force and the path becomes a helix: circular in the plane perpendicular to B, uniform along it.
Watch out for three things in the exam hall. First, r = mv/(qB) uses momentum mv — if a question gives kinetic energy, convert via mv = √(2mK) before substituting. Second, the radius depends on the charge magnitude; sign only sets the sense of rotation. Third, when both fields act, check whether qE and qvB oppose — a velocity selector passes only v = E/B, and that condition is independent of charge and mass.
Can you answer these Lorentz Force on Charge MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the Lorentz force expression F = q(E + v × B), the work done by the magnetic part of the force on a moving charge over any interval is
Show answer and why every option is right or wrong
Answer: A. A is correct. The magnetic term q(v × B) is perpendicular to v by the definition of the cross product, so it is perpendicular to the displacement at every instant and does zero work — stated in NCERT Class 12 Physics, Chapter 4, page 108.
Why B is wrong: B is wrong because the magnetic force changes no kinetic energy at all; any change in kinetic energy must come from the electric term qE or from some non-magnetic agent.
Why C is wrong: C is wrong because it confuses the sign of the charge with the sign of the work. Reversing q reverses the direction of the force, but the force stays perpendicular to v, so the work stays zero either way.
Why D is wrong: D is wrong because it multiplies a force magnitude by a path length as though the two were parallel. Work requires the component of force along the displacement, which here is zero.
MCQ 2Easy RecallPractice
A charged particle moves through a region containing a uniform magnetic field B. The magnetic force on it is zero. Which condition on the velocity guarantees this, for a non-zero speed?
Show answer and why every option is right or wrong
Answer: C. C is correct. The magnitude of the magnetic force is qvB sin θ, which vanishes when θ is 0° or 180° — that is, when v lies along B. NCERT Class 12 Physics, Chapter 4, page 108 gives the cross-product form from which this follows.
Why A is wrong: A is wrong because perpendicular velocity gives sin θ = 1, the maximum magnetic force qvB, not zero. This option inverts the condition.
Why B is wrong: B is wrong because at 45° the force is qvB/√2, which is non-zero for a non-zero speed.
Why D is wrong: D is wrong because it imports the velocity-selector condition, which concerns the balance of electric and magnetic forces. With no electric field present, the magnetic force does not vanish at any particular speed.
MCQ 3Easy RecallPractice
For a charged particle in uniform circular motion in a uniform magnetic field perpendicular to its velocity, the period T = 2πm/(qB) is independent of
Show answer and why every option is right or wrong
Answer: B. B is correct. The speed v does not appear in T = 2πm/(qB); a faster particle moves on a proportionally larger circle and completes it in the same time. NCERT Class 12 Physics, Chapter 4, page 110 records this independence.
Why A is wrong: A is wrong because m appears in the numerator of T — a heavier particle of the same charge takes longer per revolution.
Why C is wrong: C is wrong because B appears in the denominator; increasing the field shortens the period.
Why D is wrong: D is wrong because q appears in the denominator; a larger charge magnitude shortens the period.
MCQ 4Direct ApplicationPractice
A particle of mass 9.0 × 10⁻³¹ kg carrying a charge of 1.6 × 10⁻¹⁹ C moves at 4.0 × 10⁶ m/s perpendicular to a uniform magnetic field of 0.20 T. The radius of its circular path is closest to
Show answer and why every option is right or wrong
Answer: C. C is correct. Applying r = mv/(qB): r = (9.0 × 10⁻³¹ × 4.0 × 10⁶) / (1.6 × 10⁻¹⁹ × 0.20) = 3.6 × 10⁻²⁴ / 3.2 × 10⁻²⁰ = 1.1 × 10⁻⁴ m. The relation is developed in NCERT Class 12 Physics, Chapter 4, page 110.
Why A is wrong: A is wrong because it multiplies by the field instead of dividing: mvB/q = 4.5 × 10⁻⁶ m. A stronger field bends the path more tightly, so B belongs in the denominator.
Why B is wrong: B is wrong because it doubles the correct value, as though r = 2mv/(qB). No factor of 2 belongs in the radius; the 2π belongs to the period.
Why D is wrong: D is wrong because it drops the speed: m/(qB) = 2.8 × 10⁻¹¹ m. Check that every one of m, v, q and B appears exactly once.
MCQ 5Direct ApplicationPractice
A proton enters a region of crossed uniform fields in which E and B are mutually perpendicular and both perpendicular to the proton's velocity. The electric field magnitude is 3.0 × 10⁴ V/m and the magnetic field magnitude is 0.15 T. For the proton to pass through undeflected, its speed must be
Show answer and why every option is right or wrong
Answer: C. C is correct. Undeflected passage requires the electric and magnetic forces to cancel: qE = qvB, so v = E/B = 3.0 × 10⁴ / 0.15 = 2.0 × 10⁵ m/s. This balance follows directly from the Lorentz force given in NCERT Class 12 Physics, Chapter 4, page 108.
Why A is wrong: A is wrong because it multiplies E by B instead of dividing. A quick unit check rules it out: (V/m) × T does not give m/s.
Why B is wrong: B is wrong because it inverts the ratio, computing B/E. The resulting speed is absurdly small for a charged particle in a 3.0 × 10⁴ V/m field.
Why D is wrong: D is wrong because it uses E/B but slips a power of ten, reading 3.0 × 10⁴ as 6.75 × 10⁴ or mis-dividing by 0.15. Dividing by 0.15 multiplies by about 6.7, so 3.0 × 10⁴ gives 2.0 × 10⁵, not 4.5 × 10⁵.
MCQ 6Direct ApplicationPractice
A charged particle enters a uniform magnetic field with its velocity at 30° to the field direction. Which description of the subsequent path is correct?
Show answer and why every option is right or wrong
Answer: C. C is correct. Resolve v into components parallel and perpendicular to B. The parallel component experiences no magnetic force and stays constant; the perpendicular component drives uniform circular motion. Superposing the two gives a helix. NCERT Class 12 Physics, Chapter 4, page 110 discusses this decomposition.
Why A is wrong: A is wrong on two counts: the circular part of the motion lies in the plane perpendicular to B, not the plane containing it, and the parallel velocity component prevents the path from closing into a circle at all.
Why B is wrong: B is wrong because a non-zero perpendicular component produces a non-zero magnetic force, which curves the path. A straight line requires the velocity to be entirely along B.
Why D is wrong: D is wrong because it assumes the magnetic force drains energy. It does no work, so the speed and hence the helix radius stay constant.
MCQ 7CalculationPractice
Two particles carry the same charge magnitude and are accelerated from rest through the same potential difference before entering the same uniform magnetic field perpendicular to their velocities. Particle X has mass m and particle Y has mass 4m. The ratio of their circular-path radii r_Y : r_X is
Show answer and why every option is right or wrong
Answer: D. D is correct. Acceleration through the same potential difference gives both particles the same kinetic energy K = qV, so momentum mv = √(2mK) scales as √m. Since r = mv/(qB) ∝ √m, quadrupling the mass doubles the radius: r_Y : r_X = 2 : 1. The radius relation is from NCERT Class 12 Physics, Chapter 4, page 110.
Why A is wrong: A is wrong because it inverts the dependence, treating r as proportional to 1/√m. The mass sits in the numerator of mv, and although a heavier particle moves more slowly at equal energy, the net effect is r increasing with √m.
Why B is wrong: B is wrong because it uses r ∝ m directly, forgetting that equal accelerating potential means equal kinetic energy rather than equal speed — the heavier particle is slower, which partly offsets its greater mass.
Why C is wrong: C is wrong because it combines both earlier errors, applying r ∝ 1/m. It is the reciprocal of the r ∝ m answer and doubly wrong.
MCQ 8CalculationPractice
An alpha particle (charge +2e, mass 4u) and a proton (charge +e, mass 1u) move with the same speed perpendicular to the same uniform magnetic field. Taking r for radius and T for period, which pair of ratios is correct?
Show answer and why every option is right or wrong
Answer: A. A is correct. Both r = mv/(qB) and T = 2πm/(qB) depend on the ratio m/q. For the alpha particle m/q = 4u/2e = 2u/e; for the proton m/q = 1u/e. Both ratios are therefore 2 : 1. NCERT Class 12 Physics, Chapter 4, page 110 gives both relations.
Why B is wrong: B is wrong because it uses the mass ratio 4 : 1 while ignoring that the alpha particle also carries twice the charge. The charge in the denominator halves the result to 2 : 1.
Why C is wrong: C is wrong because it inverts the m/q ratio, dividing charge by mass instead. A heavier particle at the same speed traces a larger circle, not a smaller one.
Why D is wrong: D is wrong because it gets the radius ratio right but then asserts the periods are equal. Period is independent of speed, not of m/q — since the two particles have different m/q, their periods differ.
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Lorentz Force on Charge: quick recall before you leave
How do you solve a Lorentz Force on Charge question? A worked example
Pattern: P.PHY.U13.MOVING_CHARGE_RADIUS (topic-specific: PHY.U13.LORENTZ_FORCE_ON_CHARGE)
- 1
Given
An electron is accelerated from rest through a potential difference of 2.00 × 10² V and then enters a uniform magnetic field of magnitude 5.00 × 10⁻³ T, moving perpendicular to the field.
Electron mass m = 9.11 × 10⁻³¹ kg, charge magnitude e = 1.60 × 10⁻¹⁹ C. - 2
Required
The radius of the electron's circular path.
- 3
Concept
The accelerating stage is electric: the work done, eV, appears entirely as kinetic energy. The field stage is magnetic: with v ⊥ B, the force qvB is perpendicular to v and constant in magnitude, producing uniform circular motion. Note which field does what — the electric field sets the speed, the magnetic field then bends the path without changing that speed.
- 4
Formula
Speed from the accelerating stage: eV = ½mv², so v = √(2eV/m).
Radius in the field: r = mv/(qB).
Combining, r = √(2mV/e) / B. - 5
Substitution
Using the combined form:
r = √(2 × 9.11 × 10⁻³¹ × 2.00 × 10² / 1.60 × 10⁻¹⁹) / (5.00 × 10⁻³) - 6
Calculation
Numerator inside the root: 2 × 9.11 × 10⁻³¹ × 2.00 × 10² = 3.644 × 10⁻²⁸.
Divide by e: 3.644 × 10⁻²⁸ / 1.60 × 10⁻¹⁹ = 2.278 × 10⁻⁹.
Square root: 4.772 × 10⁻⁵.
Divide by B: 4.772 × 10⁻⁵ / 5.00 × 10⁻³ = 9.54 × 10⁻³.
The 2 in ½mv² and the 2π in the period formula are exact counting/mathematical constants and do not contribute to the significant-figure count; the three-significant-figure result is set by the given data (m, V, e, B all to three figures). - 7
Final answer
r = 9.54 × 10⁻³ m (about 9.54 mm).
Cross-check on the speed: v = √(2eV/m) = √(2 × 1.60 × 10⁻¹⁹ × 2.00 × 10² / 9.11 × 10⁻³¹) ≈ 8.38 × 10⁶ m/s. Then r = mv/(eB) = (9.11 × 10⁻³¹ × 8.38 × 10⁶)/(1.60 × 10⁻¹⁹ × 5.00 × 10⁻³) = 9.54 × 10⁻³ m. Consistent. - 8
Common trap
The pattern's documented distractor is dropping m or q from r = mv/(qB) — tempting because the combined form √(2mV/e)/B hides them inside a root and it is easy to substitute the accelerating-stage numbers without re-checking which symbol is which. A second trap specific to this two-stage setup: using the potential difference V as if it were a speed, or writing r = mV/(qB). V is in volts, not m/s. If a unit check on your substitution does not deliver metres, you have mixed the two stages.
- 9
Similar NEET-style question
A proton is accelerated from rest through 1.00 × 10³ V and enters a uniform magnetic field of 0.100 T perpendicular to its velocity. Taking m_p = 1.67 × 10⁻²⁷ kg and e = 1.60 × 10⁻¹⁹ C, find the radius of its path, then state by what factor the radius changes if the accelerating potential is raised to 4.00 × 10³ V with the field unchanged. *(The factor is 2 — radius scales as √V, since r ∝ √(2mV/e).)*
What to remember before solving Lorentz Force on Charge questions
Charge q moving with v ⊥ B undergoes circular motion. r = mv/(qB) (radius); T = 2πm/(qB) (period, independent of v); ω = qB/m (cyclotron frequency).
-- NCERT Class 12 Physics, Ch. 4, p. 112Which Lorentz Force on Charge formulas do you need for NEET?
2 formulas — click to collapse
Cyclotron radius / period
Charge q moves in circle of radius r in uniform B perpendicular to v. Period independent of v.
| Symbol | Quantity | SI Unit |
|---|---|---|
| r | radius | m |
| T | period | s |
| m | mass | kg |
| v | speed | m/s |
| q | charge | C |
| B | field | T |
Valid when
- v ⊥ B
- Uniform B
- Non-relativistic
Lorentz force
Total force on charge q in electric and magnetic fields. Magnetic part always perpendicular to v.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | force | N |
| q | charge | C |
| E | electric field | V/m |
| v | velocity | m/s |
| B | magnetic field | T |
Valid when
- Point particle
- Non-relativistic
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 9 formulas · 4 question patterns from its other lessons.
Lorentz Force on Charge questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
1E⃗ is parallel to B⃗ and its magnitude is 27 × 10⁴ V m⁻¹
2E⃗ is perpendicular to B⃗ and its magnitude is 27 × 10⁴ V m⁻¹
3E⃗ is perpendicular to B⃗ and its magnitude is 27 × 10² V m⁻¹
4E⃗ is parallel to B⃗ and its magnitude is 27 × 10² V m⁻¹
NTA Answer: Option 3(final)
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
How does NEET ask about Lorentz Force on Charge?
1 recurring pattern from past papers — click to collapse
Charge in uniform B (perpendicular to v): r = mv/(qB). Compute radius or speed.
Multi StepMedium
Common distractors
forgets charge or mass in formula
Drops m or q from r formula
Sources
NCERT refs: Class 12 Physics Chapter 4, p.108
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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