Magnetism, EMI and AC JEE Main PYQ — The LCR Pythagorean Setups and Sign Traps That Decide This Block (2015-2026)
Magnetism, EMI and AC JEE Main PYQ analysis (2015-2026). Series LCR is the top NVQ source, the 3-4-5 impedance setup, the deleted syllabus, Biot-Savart geometries, motional EMF, galvanometer conversions, and 14 exam-level PYQs with traps.
Magnetism, EMI and AC JEE Main PYQ Analysis (2015–2026): The LCR Pythagorean Setups and Sign Traps That Decide This Block
One block, three chapters, and an answer key built on 3-4-5 triangles.
This block carries three linked chapters: Moving Charges and Magnetism, Magnetism and Matter, and the Electromagnetic Induction with Alternating Current pair. Together they give about 3 questions a shift, near an eighth of JEE Main Physics. The marks are steady and the patterns are rigid, which makes the block very trainable.
The format is what you must respect. Series LCR is the top NVQ source in the whole physics paper. The reason is arithmetic: examiners pick R, X_L and X_C so the impedance lands on a clean integer through a 3-4-5 or 5-12-13 triangle. Spot the triangle and the sum is fast. Miss it and you grind decimals under the clock.
The danger is the compulsory NVQ section. Since 2025 you cannot skip the calculation-heavy questions, and there are no options to check against. One wrong sign or one reactance added like a resistance gives a wrong integer and −1. This block rewards clean procedure over clever insight.
We analysed every Magnetism, EMI and AC question across JEE Main shifts from 2015 to 2026. This is part of Logic Bloom's JEE Main analysis series, after Mechanics, Optics, and SHM and Waves.
| 🎯 We mapped every JEE Main Magnetism, EMI and AC question of the decade. The app has them all, ready to play and practice. | |
|---|---|
| This block is won by drilling multi-step setups until the LCR triangle and the sign rules are automatic. Logic Bloom's Playground turns it into practice: add reactances as phasors and watch the impedance triangle form, slide a rod along rails and read the motional EMF, convert a galvanometer to a voltmeter and see the series resistor fall out. Then drill every PYQ, including the compulsory NVQ type. When a reactance-as-resistance slip or a Lenz sign error catches you, TarQ teaches the fix, and your Mistake Book logs it. | Get the app → Free to start. |
The decade at a glance: where the marks sit, by format
Across 2015 to 2026 the block held near 3 questions per shift. What changed is the split between MCQ and NVQ. Some sub-topics give clean integers and migrated to Section B. Others carry π or vectors and stayed in Section A. Knowing which is which tells you how to solve each one.
| Sub-topic | Decade frequency | Format lean | Why that format |
|---|---|---|---|
| Series LCR and resonance | ~150 | Heavily NVQ | Impedance lands on clean integers through 3-4-5 triangles. |
| Biot-Savart and Ampere geometries | ~135 | Heavily MCQ | Answers carry π and μ₀, which do not give neat integers. |
| Motional EMF and flux rate | ~115 | Split | Graphs go MCQ; pull-force and heat-power go NVQ. |
| Moving coil galvanometer | ~95 | Heavily NVQ | Shunt and series formulas give whole numbers. |
| Lorentz force on a charge | ~90 | Heavily MCQ | Radius and pitch carry roots and trig, so options fit. |
| Torque and dipole moment | ~65 | MCQ | Compares shapes of equal wire length. |
| Self and mutual inductance | ~50 | NVQ | Energy ½LI² is pure substitution. |
The plan follows the table. Spend the most time on series LCR, because it is the biggest NVQ source and the most mechanical once you see the triangle. Lock the galvanometer conversions, since they are easy integer marks. Keep Biot-Savart sharp for the MCQ section, where field directions and π placement decide the answer.
Chapter 1: Moving Charges and Magnetism
This chapter is the foundation. Nearly everything rests on two laws. Biot-Savart gives the field from a current. The Lorentz force gives what a field does to a moving charge. Master these and the geometry questions become routine.
What JEE Main asks here
The field questions rarely ask for a plain formula. They give a shape built from arcs and straight wires and ask for the field at a point. The force questions test circular and helical paths, where the radius is r = mv/qB. The instrument question is the moving-coil galvanometer, converted to an ammeter with a shunt or a voltmeter with a series resistor.
The traps JEE Main repeats here
Trap 1: loop field vs straight-wire field. The field at the centre of a loop is μ₀I/2R. The field of a long straight wire is μ₀I/2πR. The straight wire carries a π, the loop does not. In a mixed-geometry problem, dropping or adding a π gives a distractor that looks right. Keep the two forms separate.
Trap 2: the straight part of a Biot-Savart figure often gives zero. When the point lies on the line of a straight segment, the current element and the position vector are parallel. Their cross product is zero, so that segment adds no field. Students waste time computing it, or worse, add a spurious term. Check the geometry first.
Trap 3: the charge sign in the Lorentz force. The force is F = q(v × B). For an electron q is negative, which flips the direction. Students find the right-hand-rule direction and forget to reverse it. The exact opposite direction is always a waiting option.
| 🧠 The field shortcuts, kept apart | |
|---|---|
| Long straight wire | B = μ₀I / 2πR. Has the π. |
| Centre of a loop | B = μ₀I / 2R. No π. |
| Inside a solenoid | B = μ₀nI, where n is turns per metre. |
| Circular-path radius | r = mv/qB = √(2mK)/qB. |
A worked JEE Main question
Q. A wire carrying current I is bent into a semicircle of radius R, with two long straight segments running along the diameter line on each side. What is the magnetic field at the centre of the semicircle?
(A) μ₀I/4R, into the page (B) μ₀I/2R, into the page (C) μ₀I/4πR, into the page (D) zero
Solution. The centre lies on the line of both straight segments, so each has the current element parallel to the position vector. Their field contribution is zero. A full loop gives μ₀I/2R at the centre, so a semicircle gives half of that, which is μ₀I/4R. The right-hand rule sets the direction into the page.
Answer: (A). Trap: (B) uses a full loop; (C) wrongly injects the straight-wire π.
Chapter 2: Magnetism and Matter
This chapter was cut hard by the rationalisation. What remains is small and mostly theory. The bar magnet acts as a dipole, with a magnetic moment and a torque in a field. The material types, diamagnetic, paramagnetic, and ferromagnetic, are tested as property-matching MCQs. That is nearly the whole chapter now.
What is gone, so do not study it: Earth's magnetism, meaning declination, dip, and the horizontal and vertical components. Also gone are Curie's law, the hysteresis loop, and the bar-magnet-as-solenoid derivation. These once carried numericals. They are dead content now. Confirm against your current NCERT edition.
The one live numerical idea: torque on a dipole, τ = MB sinθ, with M = NIA. A common MCQ gives a fixed length of wire, bends it into different shapes, and asks which has the largest magnetic moment. A single circular loop wins, because a circle encloses the most area for a given perimeter.
| 🧠 Dia vs para vs ferro, the one-line tells | |
|---|---|
| Diamagnetic | Weakly repelled. Susceptibility small and negative. No unpaired electrons. |
| Paramagnetic | Weakly attracted. Susceptibility small and positive. Has unpaired electrons. |
| Ferromagnetic | Strongly attracted. Susceptibility large and positive. Keeps magnetisation. |
Chapter 3: Electromagnetic Induction
EMI is where the block turns numerical again. The core idea is simple. A changing magnetic flux makes an EMF. Faraday gives the size, Lenz gives the direction. From there the questions build into force, power, and inductance.
What JEE Main asks here
Motional EMF leads this chapter. A rod slides on rails, or a rod rotates about one end. The sliding case gives e = BvL. The rotating case gives e = ½BωL². From the EMF, examiners ask for the pull force F = B²L²v/R or the heat power, which makes clean NVQs. Inductance questions test the stored energy ½LI² by direct substitution.
The traps JEE Main repeats here
Trap 1: the rotating rod uses the half. A rod rotating about one end has EMF e = ½BωL², not BωL². Students reuse the sliding formula and substitute ω for v. The factor of half and the L-squared both get lost. This single error ruins a whole NVQ.
Trap 2: a 180-degree flip doubles the flux change, not cancels it. When a coil turns through 180 degrees, the flux goes from +BA to −BA. The change is 2BA, not zero. Students think the coil "looks the same" and write ΔΦ = 0. The average EMF is 2NBA over the time taken.
Trap 3: maximum energy needs peak current. In an AC circuit the energy in an inductor swings. Maximum energy uses the peak current, U = ½LI₀². Students plug in the rms current and land one option too low. Read whether the question asks for maximum or average.
A worked JEE Main question
Q. A rod of length 1.0 m rotates at 400 rad/s about an axis through one end, perpendicular to the rod. A field of 0.5 T runs parallel to the axis. The rod has resistance 10 Ω. Find the power dissipated as heat, in watts.
Solution. Use the rotating-rod EMF: e = ½BωL² = ½ × 0.5 × 400 × 1² = 100 V. Then heat power P = e²/R = 100²/10 = 1000 W.
Answer: 1000 W. Trap: using e = BωL gives 200 V and a wrong 4000 W.
Chapter 4: Alternating Current
This is the densest NVQ ground in the whole physics paper. The reason is pure arithmetic. The series-LCR impedance uses a right-triangle sum, and examiners rig the numbers so the answer is a clean integer. If you see the triangle, the question is fast.
What JEE Main asks here
The staple is the series-LCR circuit. You find the two reactances, combine them into the impedance Z, then get current, power, or power factor. Resonance is the special case where X_L equals X_C, so Z drops to R and current peaks. The transformer gives a simple ratio question. The LC-oscillation differential equation is deleted, so skip it.
The traps JEE Main repeats here
Trap 1: reactance is not resistance, so do not just add. With R = 30 Ω and X_L = 40 Ω, the impedance is not 70 Ω. You add them as a right triangle: Z = √(R² + X_L²) = 50 Ω. Simple addition is the single most common AC error, and it ruins every downstream value.
Trap 2: a stated AC voltage is the rms value. "A 220 V AC source" means V_rms = 220 V, not the peak. Treating it as the peak introduces a √2 factor that is always a waiting distractor. Only an explicit V = V₀ sin(ωt) gives you the peak directly.
Trap 3: power needs the power factor. Average power is P = V_rms I_rms cosφ, with cosφ = R/Z. Multiplying just V_rms by I_rms gives the apparent power, in volt-amperes, which NTA plants as the trap. Power is only dissipated in R, so P = I_rms² R is the clean check.
| ⚡ The AC checklist, in order | |
|---|---|
| Reactances | X_L = ωL, X_C = 1/ωC. Compute both first. |
| Impedance | Z = √(R² + (X_L − X_C)²). Look for the 3-4-5 triangle. |
| Resonance | X_L = X_C, so Z = R, current is maximum. f₀ = 1/(2π√LC). |
| Power | P = V_rms I_rms cosφ = I_rms² R. Never the apparent power. |
| rms vs peak | A bare "220 V" is rms. V_rms = V₀/√2. |
A worked JEE Main question
Q. An AC source of ω = 1000 rad/s and V_rms = 200 V drives a series circuit with R = 40 Ω, L = 50 mH, and C = 50 μF. Find the average power dissipated, in watts.
Solution. X_L = ωL = 1000 × 0.05 = 50 Ω. X_C = 1/(ωC) = 1/(1000 × 50×10⁻⁶) = 20 Ω. So X_L − X_C = 30 Ω. Then Z = √(40² + 30²) = √2500 = 50 Ω, a 3-4-5 triangle. Current I_rms = 200/50 = 4 A. Power P = I_rms² R = 16 × 40 = 640 W.
Answer: 640 W. Trap: adding 40 + 30 = 70 for Z gives a wrong 2.86 A and a wrong power.
The formula and constant sheet you need cold
JEE Main gives about two minutes a question. Deriving any of these mid-paper loses the question. Have them at reflex speed. μ₀ = 4π × 10⁻⁷ T·m/A.
| Quantity | Formula | Note |
|---|---|---|
| Field, straight wire | B = μ₀I / 2πR | Has the π. |
| Field, loop centre | B = μ₀I / 2R | No π. |
| Field, solenoid | B = μ₀nI | n is turns per metre. |
| Lorentz force | F = q(v × B) | Flip direction for an electron. |
| Circular-path radius | r = mv/qB = √(2mK)/qB | From kinetic energy K. |
| Torque on a loop | τ = MB sinθ, M = NIA | Circle gives the largest M for fixed wire. |
| Ammeter shunt | S = I_g R_g / (I − I_g) | Low resistance, in parallel. |
| Voltmeter series R | R = V/I_g − R_g | High resistance, in series. |
| Motional EMF, sliding | e = BvL | Straight rod on rails. |
| Motional EMF, rotating | e = ½BωL² | Note the half and the L². |
| Inductor energy | U = ½LI² | Use peak current for maximum. |
| Reactances | X_L = ωL, X_C = 1/ωC | ω = 2πf. |
| Impedance, series LCR | Z = √(R² + (X_L − X_C)²) | Right-triangle sum, not plain. |
| Resonant frequency | f₀ = 1/(2π√LC) | Here X_L = X_C and Z = R. |
| Average power | P = V_rms I_rms cosφ = I_rms² R | cosφ = R/Z. |
| Transformer ratio | V_s/V_p = N_s/N_p = I_p/I_s | Ideal, no losses. |
What to expect in JEE Main 2027
The decade pattern plus the compulsory-NVQ shift points to four clear moves. Prepare for these, not for deleted content.
1. Series LCR stays the NVQ engine. Integer answers force examiners toward impedance and resonance. Train yourself to spot 3-4-5 and 5-12-13 triangles so you skip the decimal grind. Expect resonant-frequency questions that test clean exponent handling.
2. Motional EMF chained to mechanics. A rod falling under gravity through a field reaches a terminal velocity v = mgR/B²L². This joins EMI, resistance, and forces in one chain, which fits the Main level without crossing into Advanced.
3. The galvanometer is the only instrument left. With Earth's magnetism gone, the moving-coil galvanometer carries all the instrumentation marks. Shunt and series conversions are reliable NVQ integers. Know current and voltage sensitivity cold.
4. Lorentz force may shift to an NVQ magnitude. It was MCQ because directions use unit vectors. A likely new form asks for the magnitude of the force, which squares the components into a testable integer. Practice the full cross-product then the magnitude.
Do not revise: Earth's magnetism, hysteresis, Curie's law, the cyclotron derivation, the toroid, eddy currents, the solenoid-equivalent derivation, LC-oscillation maths, and the Q-factor sharpness of resonance. All deleted. Confirm against your current NCERT edition.
| 🎯 Drill the whole decade of this block, sorted by format and trap. | |
|---|---|
| Reading a trap once is not the same as beating it with no options and a −1 penalty. Logic Bloom's Playground has every JEE Main Magnetism, EMI and AC PYQ from 2015 to 2026, tagged by chapter, format, and the exact trap it hides. Miss the reactance-as-resistance slip or the rotating-rod half and TarQ walks you through the fix, then your Mistake Book brings it back until it sticks. Deleted topics are stripped out, so you never waste a rep. | Get the app → Free to start. |
14 must-attempt JEE Main PYQs, with the traps named
These fourteen cover the real spread, MCQ and NVQ, across all four chapters. Each is written at JEE Main level, two to four chained steps. Try each before reading the answer.
1. (MCQ) An electron moves with velocity v = (2î + 3ĵ) × 10⁶ m/s in a field B = 0.5k̂ T. Find the magnitude of the magnetic force. (e = 1.6 × 10⁻¹⁹ C)
(A) 1.6 × 10⁻¹³ N (B) 2.88 × 10⁻¹³ N (C) 3.2 × 10⁻¹³ N (D) 2.4 × 10⁻¹³ N
Work. v × B = (2î + 3ĵ) × 0.5k̂ × 10⁶. Using î×k̂ = −ĵ and ĵ×k̂ = î, this is (1.5î − ĵ) × 10⁶. Multiply by |q|: F = 1.6×10⁻¹⁹ × 10⁶ × √(1.5² + 1²) = 1.6×10⁻¹³ × √3.25 ≈ 2.88 × 10⁻¹³ N.
Answer: (B). Trap: (D) keeps only one component of the cross product.
2. (NVQ) A galvanometer of resistance 50 Ω gives full-scale deflection at 2 mA. It is converted to a voltmeter reading up to 10 V by a series resistance R. Find R in ohms.
Work. V = I_g(R_g + R), so R = V/I_g − R_g = 10/0.002 − 50 = 5000 − 50 = 4950 Ω.
Answer: 4950. Trap: writing 5000 by forgetting to subtract R_g gives zero marks.
3. (NVQ) A 50 mH inductor is connected to V = 100 sin(200t) V. Find the maximum energy stored in the inductor, in milli-joules.
Work. X_L = ωL = 200 × 0.05 = 10 Ω. Peak current I₀ = V₀/X_L = 100/10 = 10 A. Maximum energy U = ½LI₀² = ½ × 0.05 × 100 = 2.5 J = 2500 mJ.
Answer: 2500. Trap: using the rms current (I₀/√2) halves the energy to a wrong value.
4. (MCQ) A circular coil of radius 10 cm and 50 turns sits in a field of 0.2 T, perpendicular to its plane. It is flipped 180° about a diameter in 0.1 s. Find the average induced EMF.
(A) 6.28 V (B) 3.14 V (C) 1.57 V (D) zero
Work. Initial flux per turn = BA = 0.2 × π × 0.01 = 0.002π Wb. A 180° flip reverses it, so ΔΦ per turn = 2 × 0.002π. For 50 turns, total change = 50 × 0.004π = 0.2π. EMF = 0.2π/0.1 = 2π ≈ 6.28 V.
Answer: (A). Trap: (D) assumes the flip gives no change.
5. (NVQ) A rod of length 2 m slides at 5 m/s on rails in a field of 0.4 T, perpendicular to the plane. The circuit resistance is 8 Ω. Find the force needed to keep the rod moving at constant speed, in newtons.
Work. EMF e = BvL = 0.4 × 5 × 2 = 4 V. Current I = e/R = 4/8 = 0.5 A. Force F = BIL = 0.4 × 0.5 × 2 = 0.4 N. (Equal to B²L²v/R = 0.16 × 4 × 5/8 = 0.4 N.)
Answer: 0.4. Trap: dropping one factor of B or L gives a wrong value.
6. (NVQ) In a series circuit R = 30 Ω and X_L = 40 Ω with an AC source of V_rms = 100 V. Find the rms current, in amperes.
Work. Z = √(R² + X_L²) = √(900 + 1600) = √2500 = 50 Ω. Current I = V/Z = 100/50 = 2 A.
Answer: 2. Trap: adding R + X_L = 70 gives a wrong 1.43 A.
7. (MCQ) A given length of wire carries current I. It is bent once into a single circular loop, then into a double loop of half the radius. What is the ratio of the magnetic moment of the single loop to the double loop?
(A) 2 : 1 (B) 1 : 1 (C) 1 : 2 (D) 4 : 1
Work. Single loop: radius r, M₁ = I × πr². Double loop of the same wire: 2(2πr') = 2πr, so r' = r/2, and M₂ = I × N × πr'² = I × 2 × π(r/2)² = I × πr²/2. Ratio M₁ : M₂ = πr² : πr²/2 = 2 : 1.
Answer: (A). Trap: (C) forgets the turn count N = 2 in M₂.
8. (NVQ) A series LCR circuit has L = 2 H, C = 8 μF, and R = 10 Ω. Find the resonant angular frequency, in rad/s.
Work. At resonance ω₀ = 1/√(LC) = 1/√(2 × 8×10⁻⁶) = 1/√(16×10⁻⁶) = 1/(4×10⁻³) = 250 rad/s.
Answer: 250. Trap: using f₀ = 1/(2π√LC) when the question asks for angular frequency adds a stray 2π.
9. (MCQ) A proton and an alpha particle enter the same magnetic field with the same kinetic energy, moving perpendicular to the field. What is the ratio of the radius of the proton's path to the alpha's?
(A) 1 : 1 (B) 1 : √2 (C) √2 : 1 (D) 1 : 2
Work. r = √(2mK)/qB. Same K and B, so r ∝ √m/q. Alpha has m = 4m_p and q = 2q_p. Ratio r_p : r_α = (√m_p/q_p) : (√(4m_p)/2q_p) = (√m_p/q_p) : (2√m_p/2q_p) = 1 : 1.
Answer: (A). Trap: (D) uses mass alone and ignores the charge.
10. (NVQ) A series LCR circuit has R = 50 Ω, X_L = 120 Ω, X_C = 0 Ω, driven by V_rms = 130 V. Find the power dissipated, in watts.
Work. Z = √(50² + 120²) = √(2500 + 14400) = √16900 = 130 Ω, a 5-12-13 triangle. Current I = 130/130 = 1 A. Power P = I²R = 1 × 50 = 50 W.
Answer: 50. Trap: using P = V_rms × I_rms = 130 W reports apparent power, not real power.
11. (MCQ) Two long parallel wires 0.1 m apart carry currents 3 A and 4 A in the same direction. Find the force per unit length between them. (μ₀ = 4π × 10⁻⁷)
(A) 2.4 × 10⁻⁵ N/m, attractive (B) 2.4 × 10⁻⁵ N/m, repulsive (C) 1.2 × 10⁻⁵ N/m, attractive (D) 4.8 × 10⁻⁵ N/m, attractive
Work. F/L = μ₀I₁I₂/2πd = (4π×10⁻⁷ × 3 × 4)/(2π × 0.1) = (2×10⁻⁷ × 12)/0.1 = 2.4×10⁻⁵ N/m. Same-direction currents attract.
Answer: (A). Trap: (B) flips the force direction; parallel currents attract, antiparallel repel.
12. (NVQ) A transformer steps 2200 V down to 220 V. The output delivers 10 A to the load. Assuming it is ideal, find the input current, in amperes.
Work. Ideal transformer: V_p I_p = V_s I_s. So I_p = V_s I_s / V_p = 220 × 10 / 2200 = 1 A.
Answer: 1. Trap: using the voltage ratio the wrong way gives 100 A.
13. (MCQ) Two statements are given. Statement I: In a purely inductive AC circuit, the average power over a cycle is zero. Statement II: In a pure inductor, the current lags the voltage by 90°, so the power factor is zero. Choose the correct option.
(A) Both true, II explains I (B) Both true, II does not explain I (C) I true, II false (D) I false, II true
Work. In a pure inductor the phase difference is 90°, so cosφ = cos90° = 0. Average power P = V_rms I_rms cosφ = 0. Statement I is true, Statement II is true, and the zero power factor is exactly why the power is zero.
Answer: (A). Trap: (B) accepts both but misses that II is the reason for I.
14. (NVQ) A square loop of side 0.2 m and resistance 2 Ω lies in a field that increases uniformly from 0 to 0.5 T in 0.1 s, perpendicular to the loop. Find the induced current, in amperes.
Work. Area A = 0.2² = 0.04 m². Flux change ΔΦ = A ΔB = 0.04 × 0.5 = 0.02 Wb. EMF e = ΔΦ/Δt = 0.02/0.1 = 0.2 V. Current I = e/R = 0.2/2 = 0.1 A.
Answer: 0.1. Trap: using the side length instead of the area drops a factor and ruins the integer.
How to actually clear this block
The plan is mechanical because the block is mechanical. Lock the series-LCR chain first: reactances, then the triangle for Z, then power through the power factor. Train your eye to spot 3-4-5 and 5-12-13 so the arithmetic is instant. Then secure the galvanometer conversions and the two motional-EMF forms. That covers most of the NVQ marks.
Reading traps in an article is step one. Beating them with no options and a −1 penalty is step two, and that only comes from reps. That is what Logic Bloom's Playground is built for: every PYQ mapped by chapter, format, and trap, TarQ to teach the fix the moment you slip, and a Mistake Book that brings your weak spots back until they hold.
| 🚀 Turn this analysis into a real score. | |
|---|---|
| You now know where the marks sit and which traps to fear. Close the gap between knowing and scoring. Logic Bloom's Playground drills every JEE Main Magnetism, EMI and AC PYQ of the decade, and Battleground lets you test the block in 1v1 duels against other aspirants. Understand through games. Score through practice. | Get the app → Free to start. |