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Electromagnetic Induction – JEE Main Physics Formula Sheet & Class 12 Notes | Faraday's Law, Lenz's Law, Motional EMF, Self Inductance, Mutual Inductance, AC Circuits & All Formulas
This is the complete JEE Main Physics Formula Sheet and Class 12 Formula Sheet for Electromagnetic Induction and Alternating Current — Chapter 14 from the Aakash Rapid Revision & Formula Bank. This chapter covers: Magnetic Flux — φ=BAcosθ=B·A; Faraday's Laws — ε=–dφ/dt (rate of change of flux), ε=–NdΦ/dt; Lenz's Law — induced current opposes the cause; Motional EMF — ε=Bvl (rod moving in field), ε=Bωl²/2 (rotating rod); Self Inductance — L=NΦ/I, ε=–LdI/dt, L for solenoid=μ₀n²V=μ₀N²A/l; Mutual Inductance — M=N₂Φ₂₁/I₁, ε₂=–MdI₁/dt, M for two coaxial solenoids=μ₀N₁N₂A/l, Neumann formula M=μ₀N₁N₂A/l, coupling coefficient k=M/√(L₁L₂); Energy in Inductor — U=½LI², energy density u=B²/2μ₀; LR Circuit — growth I=I₀(1–e^(–t/τ)), decay I=I₀e^(–t/τ), τ=L/R; LC Oscillations — ω=1/√(LC), T=2π√(LC); AC Generator — ε=NBAω sin(ωt)=ε₀sin(ωt); AC Circuit Analysis — V_R/I_R (in phase), V_L leads by 90°, V_C lags by 90°, X_L=ωL, X_C=1/ωC; Series LCR — Z=√(R²+(X_L–X_C)²), tanφ=(X_L–X_C)/R, resonance ω₀=1/√(LC); Power in AC — P=V_rms I_rms cosφ; and Transformer — V₁/V₂=N₁/N₂=I₂/I₁. Electromagnetic Induction contributes 4–6 questions in every JEE Main session. Download the Free PDF for all formulas in one JEE Main exam-ready reference.
Topics Covered in This Electromagnetic Induction Formula Sheet
Electromagnetic Induction JEE Main Formula Sheet PDF Preview
Scroll to explore all Electromagnetic Induction formulas — JEE Main Physics Formula Sheet
Introduction: Why Electromagnetic Induction Is the Physics Behind All Modern Technology in JEE Main
Electromagnetic Induction is the discovery that changed the world — Michael Faraday's insight in 1831 that a changing magnetic field induces an electric field, and therefore an EMF in a conductor. This single principle underlies every generator (AC, DC), every transformer (power grids, chargers), every motor (every electric appliance), and the propagation of electromagnetic waves. For JEE Main, it is the chapter that directly connects magnetic fields (Chapter 13) to alternating current circuits — a chain that runs from Faraday's law to Lenz's law to inductors to LCR circuits to transformers.
For JEE Main physics, Electromagnetic Induction and AC contributes 4–6 questions per session, making it one of the top three highest-weightage Class 12 chapters. Questions test: Faraday's law (ε=–dΦ/dt), motional EMF (ε=Bvl), self inductance (L for solenoid=μ₀n²Al), LR circuit time constant τ=L/R, AC circuit reactances (X_L=ωL, X_C=1/ωC), series LCR impedance Z=√(R²+(X_L–X_C)²), resonance condition ω₀=1/√(LC), power factor cosφ=R/Z, and transformer ratio V₁/V₂=N₁/N₂.
Download the Free PDF for Electromagnetic Induction to access all Faraday's law, motional EMF, self and mutual inductance, LR circuit, LC oscillations, AC generator, reactances, LCR series circuit, power, and transformer formulas in one structured JEE Main physics revision reference.
Key Concepts and Formulas in Electromagnetic Induction
Magnetic Flux, Faraday's Laws, Lenz's Law, Motional EMF
Why ε=–dΦ/dt, Lenz's Law, and Motional EMF ε=Bvl Are the Most-Tested Electromagnetic Induction JEE Main Formulas
Magnetic Flux (from Aakash PDF — Electromagnetic Induction JEE Main):
φ = B·A = BA cosθ
where B = magnetic field, A = area of the surface, θ = angle between B and the normal to the surface.
SI unit: Weber (Wb) = T·m². Dimensional formula: [ML²T⁻²A⁻¹].
φ is maximum when B ⊥ surface (θ=0°); φ=0 when B ∥ surface (θ=90°).
Faraday's Laws of Electromagnetic Induction (from Aakash PDF — Electromagnetic Induction JEE Main):
First Law: Whenever the magnetic flux through a circuit changes, an EMF is induced in the circuit.
Second Law: The magnitude of the induced EMF is equal to the rate of change of magnetic flux:
ε = –dΦ/dt (for a single-turn coil)
ε = –N dΦ/dt (for a coil of N turns; NΦ = flux linkage)
The negative sign is from Lenz's law (induced EMF opposes the change in flux).
Lenz's Law (from Aakash PDF — Electromagnetic Induction JEE Main):
The direction of induced current is such that it opposes the change in magnetic flux that causes it.
Lenz's law is a consequence of conservation of energy: if the induced current aided the change in flux, it would create energy from nothing (perpetual motion).
If flux is increasing → induced current creates opposing B (opposing the increase). If flux is decreasing → induced current creates supporting B (opposing the decrease).
Motional EMF (from Aakash PDF — Electromagnetic Induction JEE Main):
A conductor of length l moving with velocity v perpendicular to magnetic field B:
ε = Bvl (when v, B, and l are mutually perpendicular)
Derivation: free charges in rod experience force F=qvB → accumulate → create electric field E=vB → at equilibrium: E=vB → potential difference = El = vBl = ε.
If θ is angle between v and l: ε = Bvl sinθ. If angle between v and B is α: ε = Bvl sinα sinθ.
Current in external circuit R: I = ε/R = Bvl/R.
Power dissipated: P = ε²/R = B²v²l²/R.
Force required to maintain constant velocity (opposing force by Lenz): F = BIl = B²l²v/R.
Rotating Rod (from Aakash PDF — Electromagnetic Induction JEE Main):
A rod of length l rotating with angular velocity ω about one end in uniform field B perpendicular to the plane of rotation:
For small element dr at distance r from pivot: dε = Bωr × dr; total ε = ∫₀^l Bωr dr = Bωl²/2
ε = ½Bωl² = Bπfl²
Download the Free PDF for Electromagnetic Induction for all Faraday's law examples for JEE Main.
Self Inductance, Mutual Inductance, Energy in Inductor, LR Circuit
Why L=μ₀n²Al, M for Two Solenoids, U=½LI², and τ=L/R Are the Core Electromagnetic Induction JEE Main Formulas
Self Inductance (from Aakash PDF — Electromagnetic Induction JEE Main):
Self inductance L of a coil is defined as the ratio of flux linkage to the current:
L = NΦ/I
Self induced EMF: ε = –L dI/dt
SI unit: Henry (H). 1H = 1 Wb/A = 1 V·s/A.
Self inductance of solenoid (N turns, length l, cross-section area A, n=N/l turns/m):
Flux: Φ = B × A = μ₀nI × A. Flux linkage: NΦ = N × μ₀nIA = μ₀n²lA × I
L = μ₀n²Al = μ₀N²A/l
Self inductance of toroid (N turns, cross-section area A, mean radius r):
L = μ₀N²A/(2πr)
Mutual Inductance (from Aakash PDF — Electromagnetic Induction JEE Main):
Mutual inductance M between two coils: the EMF induced in coil 2 when current changes in coil 1:
M = N₂Φ₂₁/I₁ (flux linkage in 2 due to I₁)
Induced EMF in secondary: ε₂ = –M dI₁/dt
Also: ε₁ = –M dI₂/dt (symmetric — M₁₂=M₂₁ always).
M for two coaxial solenoids (same cross-section A, length l, N₁ and N₂ turns):
M = μ₀N₁N₂A/l
Coupling coefficient: k = M/√(L₁L₂), where 0 ≤ k ≤ 1.
k=1: perfect coupling (all flux from 1 links with 2). k=0: zero coupling (no shared flux).
M_max = √(L₁L₂) (when k=1).
Energy Stored in Inductor (from Aakash PDF — Electromagnetic Induction JEE Main):
When current I flows through inductor L: U = ½LI²
Magnetic energy density (energy per unit volume in the field): u = B²/(2μ₀)
For solenoid: U = ½LI² = ½μ₀n²Al × I² = (B²/2μ₀) × Volume ✓
LR Circuit (from Aakash PDF — Electromagnetic Induction JEE Main):
Inductor L and resistor R in series with battery ε. Switch closed at t=0:
Growth of current: I = I₀(1–e^(–t/τ)) where I₀ = ε/R (final steady current), τ = L/R (time constant)
At t=τ: I = I₀(1–1/e) ≈ 0.632I₀ (63.2% of final). At t=5τ: I ≈ I₀ (fully grown).
EMF across inductor: V_L = ε·e^(–t/τ) (starts at ε, decays to 0).
Decay of current (battery removed, current decays through R):
I = I₀e^(–t/τ)
At t=τ: I = I₀/e ≈ 0.368I₀. Energy dissipated in R = ½LI₀² (all stored energy released).
Download the Free PDF for Electromagnetic Induction for all inductance and LR circuit examples for JEE Main.
LC Oscillations, AC Generator, and RMS Values
Why LC Oscillation Frequency ω=1/√(LC) and AC RMS Values Are Direct Electromagnetic Induction JEE Main Results
LC Oscillations (from Aakash PDF — Electromagnetic Induction JEE Main):
An inductor L and capacitor C (fully charged initially to Q₀) form a closed loop. The charge oscillates between the capacitor and inductor.
Equation of motion: L(d²Q/dt²) + Q/C = 0 → d²Q/dt² = –Q/LC
This is SHM in Q with ω² = 1/LC.
Angular frequency: ω = 1/√(LC)
Time period: T = 2π√(LC)
Charge: Q(t) = Q₀ cos(ωt+φ). Current: I(t) = –Q₀ω sin(ωt+φ) = I₀ sin(ωt).
Energy conservation: total energy = U_E + U_B = Q₀²/2C = ½LI₀² = constant.
At maximum charge Q₀: all energy in capacitor (U_E = Q₀²/2C), I=0.
At maximum current I₀: all energy in inductor (U_B = ½LI₀²), Q=0.
I₀ = Q₀ω = Q₀/√(LC).
AC Generator (from Aakash PDF — Electromagnetic Induction JEE Main):
Coil of N turns, area A rotating with angular velocity ω in magnetic field B:
Flux: Φ = NBA cos(ωt)
EMF: ε = –NdΦ/dt = NBAω sin(ωt) = ε₀ sin(ωt)
Peak EMF: ε₀ = NBAω
EMF is maximum when coil plane is parallel to B (Φ=0, dΦ/dt=max). EMF is zero when coil plane perpendicular to B (Φ=max, dΦ/dt=0).
RMS Values (from Aakash PDF — Electromagnetic Induction JEE Main):
For sinusoidal AC: V(t) = V₀ sinωt; I(t) = I₀ sinωt.
V_rms = V₀/√2 ≈ 0.707V₀
I_rms = I₀/√2 ≈ 0.707I₀
RMS = Root Mean Square. V_rms is the DC equivalent voltage that would dissipate the same power.
Average value of sinusoidal AC over one complete cycle = 0. Average over half cycle = 2V₀/π ≈ 0.637V₀.
Peak factor = V₀/V_rms = √2. Form factor = V_rms/V_avg(half cycle) = π/(2√2).
Download the Free PDF for Electromagnetic Induction for all LC oscillation and AC generator examples for JEE Main.
AC Circuit Elements — Reactance, Phasors, Phase Relations
Why X_L=ωL, X_C=1/ωC, and Phase Relations Are Essential Electromagnetic Induction JEE Main Tools
Purely Resistive Circuit (from Aakash PDF — Electromagnetic Induction JEE Main):
V = V₀ sinωt; I = I₀ sinωt = (V₀/R) sinωt
Voltage and current are IN PHASE (phase difference = 0°).
V_R = IR; Power: P = V_rms I_rms = I²_rms R (real power dissipated).
Purely Inductive Circuit (from Aakash PDF — Electromagnetic Induction JEE Main):
V = V₀ sinωt → I = I₀ sin(ωt–π/2) = –I₀ cosωt
Current LAGS voltage by 90° (or voltage leads current by 90°).
Peak current: I₀ = V₀/X_L where X_L = ωL = 2πfL (inductive reactance, Ω)
X_L increases with frequency: at ω→∞, X_L→∞ (inductor blocks high frequency); at ω=0 (DC), X_L=0 (inductor is short circuit for DC).
Power: P = V_rms I_rms cos90° = 0 (no real power in pure inductor).
Purely Capacitive Circuit (from Aakash PDF — Electromagnetic Induction JEE Main):
V = V₀ sinωt → I = I₀ sin(ωt+π/2) = I₀ cosωt
Current LEADS voltage by 90° (or voltage lags current by 90°).
Peak current: I₀ = V₀/X_C where X_C = 1/ωC = 1/2πfC (capacitive reactance, Ω)
X_C decreases with frequency: at ω→∞, X_C→0 (capacitor passes high frequency); at ω=0 (DC), X_C→∞ (capacitor blocks DC).
Power: P = V_rms I_rms cos90° = 0 (no real power in pure capacitor).
Phasor Representation (from Aakash PDF — Electromagnetic Induction JEE Main):
Draw phasors (rotating vectors) for current and voltage: V_R along I (in phase); V_L ahead of I by 90°; V_C behind I by 90°.
Mnemonic: CIVIL — In a Capacitor, current I leads Voltage V; In an Inductor, V leads I (voltage L leads current).
Download the Free PDF for Electromagnetic Induction for all AC circuit examples for JEE Main.
Series LCR Circuit — Impedance, Resonance, Quality Factor, Power
Why LCR Impedance Z=√(R²+(X_L–X_C)²), Resonance ω₀=1/√(LC), and Power Factor cosφ=R/Z Are the Most-Tested EMI JEE Main Results
Series LCR Circuit (from Aakash PDF — Electromagnetic Induction JEE Main):
R, L, C in series with AC source V=V₀sinωt. Common current I through all elements.
Voltage phasors: V_R (along I), V_L (90° ahead of I), V_C (90° behind I).
Net voltage: V = √(V_R² + (V_L–V_C)²) = I√(R² + (X_L–X_C)²)
Impedance: Z = √(R²+(X_L–X_C)²) = √(R²+(ωL–1/ωC)²)
Peak current: I₀ = V₀/Z; I_rms = V_rms/Z
Phase angle: tanφ = (X_L–X_C)/R
φ > 0 (X_L > X_C): inductive circuit, V leads I. φ < 0 (X_C > X_L): capacitive circuit, I leads V. φ = 0 (X_L = X_C): resonance.
Resonance in LCR Circuit (from Aakash PDF — Electromagnetic Induction JEE Main):
Resonance when X_L = X_C: ωL = 1/ωC → ω₀ = 1/√(LC)
At resonance: Z = R (minimum impedance). I = V/R (maximum current). φ = 0 (V and I in phase). Power factor = 1 (maximum real power).
Resonant frequency: f₀ = 1/(2π√(LC))
Quality Factor (from Aakash PDF — Electromagnetic Induction JEE Main):
Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C)
Q represents sharpness of resonance: higher Q → sharper peak → better selectivity.
At resonance: V_L = QV (voltage across L = Q times applied voltage) and V_C = QV.
Bandwidth: Δω = R/L = ω₀/Q (width of resonance peak at half-power points).
Half-power frequencies: ω₁ = ω₀ – R/2L; ω₂ = ω₀ + R/2L; Δω = ω₂–ω₁ = R/L.
Power in AC Circuits (from Aakash PDF — Electromagnetic Induction JEE Main):
P = V_rms I_rms cosφ
Power factor: cosφ = R/Z
For pure R: cosφ=1, P=V_rms I_rms. For pure L or C: cosφ=0, P=0 (wattless circuit). For LCR at resonance: cosφ=1, P=I²_rms R (maximum).
Apparent power: S = V_rms I_rms (VA). Real power: P = S cosφ (W). Reactive power: Q_power = S sinφ (VAR).
Download the Free PDF for Electromagnetic Induction for all LCR circuit examples for JEE Main.
Transformer, Eddy Currents, and Applications
Why Transformer Voltage-Turn Ratio V₁/V₂=N₁/N₂ Is the Most Direct Electromagnetic Induction JEE Main Formula
Transformer (from Aakash PDF — Electromagnetic Induction JEE Main):
A transformer transfers AC electrical energy from one circuit to another using mutual induction, without direct electrical connection.
Based on Faraday's law: changing flux in primary induces EMF in secondary through shared iron core.
Voltage ratio: V₁/V₂ = N₁/N₂
Current ratio (ideal transformer): I₁/I₂ = N₂/N₁
(Power input = Power output for ideal: V₁I₁ = V₂I₂ → I₁/I₂ = V₂/V₁ = N₂/N₁)
Step-up transformer: N₂ > N₁ → V₂ > V₁; I₂ < I₁ (voltage increases, current decreases).
Step-down transformer: N₂ < N₁ → V₂ < V₁; I₂ > I₁ (voltage decreases, current increases).
Efficiency: η = (V₂I₂)/(V₁I₁) × 100%. Ideal transformer: η = 100%.
Losses in real transformer: (a) Copper loss (I²R in windings); (b) Eddy current loss (in iron core); (c) Hysteresis loss (energy loss in magnetising/demagnetising core each cycle); (d) Flux leakage (not all flux from primary links secondary).
Eddy Currents (from Aakash PDF — Electromagnetic Induction JEE Main):
When a bulk conductor is placed in a changing magnetic field, induced EMFs drive currents in loops within the conductor itself = eddy currents.
Eddy currents dissipate energy as heat (Joule heating) — wasted energy in transformers, motors.
Minimising eddy currents: laminate the core (thin insulated layers) → increases resistance → reduces eddy current magnitude → reduces losses.
Useful applications of eddy currents: (1) Induction furnaces (heating metals by eddy currents); (2) Electromagnetic braking (eddy current brake — opposing motion of conductor in magnetic field → braking force, like in train brakes); (3) Speedometers (eddy current coupling to deflect needle proportional to speed); (4) Dead-beat galvanometers (eddy current damping in aluminium frame); (5) Metal detectors.
Download the Free PDF for Electromagnetic Induction for all transformer and eddy current examples for JEE Main.
Download Free PDF — Electromagnetic Induction JEE Main Formula Sheet
All Electromagnetic Induction formulas from the Aakash Rapid Revision PDF: φ=BAcosθ (Wb); Faraday ε=–dΦ/dt and ε=–NdΦ/dt; Lenz (opposes change, energy conservation); motional ε=Bvl; I=Bvl/R; P=B²v²l²/R; F=B²l²v/R (opposing); rotating rod ε=½Bωl²; L=NΦ/I; ε=–LdI/dt; solenoid L=μ₀n²Al=μ₀N²A/l; toroid L=μ₀N²A/2πr; M=N₂Φ₂₁/I₁; ε₂=–MdI₁/dt; coaxial solenoids M=μ₀N₁N₂A/l; k=M/√(L₁L₂)≤1; M_max=√(L₁L₂); U=½LI²; u=B²/2μ₀; LR growth I=I₀(1–e^(–t/τ)); decay I=I₀e^(–t/τ)); τ=L/R; LC ω=1/√(LC); T=2π√(LC); Q=Q₀cosωt; I₀=Q₀ω; U=Q₀²/2C=½LI₀²; AC generator ε=ε₀sinωt; ε₀=NBAω; V_rms=V₀/√2; I_rms=I₀/√2; avg(half cycle)=2V₀/π; R(in phase); L(V leads I 90°; X_L=ωL; blocks high f; P=0); C(I leads V 90°; X_C=1/ωC; blocks DC; P=0); CIVIL; Z=√(R²+(X_L–X_C)²); tanφ=(X_L–X_C)/R; resonance ω₀=1/√(LC) f₀=1/2π√(LC) Z_min=R I_max; Q=ω₀L/R=(1/R)√(L/C); V_L=V_C=QV; bandwidth Δω=R/L=ω₀/Q; P=V_rms I_rms cosφ; cosφ=R/Z; transformer V₁/V₂=N₁/N₂; I₁/I₂=N₂/N₁; step-up N₂>N₁; step-down N₂
Why Electromagnetic Induction and AC Is a Must-Master Chapter in JEE Main Physics
Faraday's law ε=–dΦ/dt is the most powerful formula in all of electromagnetism for JEE Main. It generates EMF from any changing flux — whether by changing B, changing area, changing angle, or any combination. Motional EMF ε=Bvl is a special case: the rod moves → area swept per second = vl → flux change rate = Bvl → EMF = Bvl. The rotating rod gives ε=½Bωl² by integrating over the rod's length. Lenz's law provides the direction — always opposing the change, always consistent with energy conservation.
Series LCR resonance (ω₀=1/√(LC), Z_min=R, I_max, cosφ=1) is the most calculation-heavy and highest-difficulty Electromagnetic Induction topic in JEE Main. At resonance: impedance is minimum (Z=R), current is maximum (I=V/R), power is maximum, and power factor=1. The quality factor Q=ω₀L/R simultaneously gives the voltage amplification (V_L=V_C=QV at resonance), the sharpness of the resonance peak, and the bandwidth (Δω=ω₀/Q). All four quantitative aspects of LCR resonance are tested in JEE Main.
Transformer ratio V₁/V₂=N₁/N₂=I₂/I₁ is the simplest and most direct Electromagnetic Induction formula in JEE Main. One formula, three quantities — voltage, current, turns. Step-up: N₂>N₁ → V₂>V₁ and I₂
Who Should Use This Electromagnetic Induction Formula Sheet?
Learning Outcomes After Completing Electromagnetic Induction
After working through Electromagnetic Induction using this formula sheet, a student should accomplish: On Faraday's law: compute φ=BAcosθ; apply ε=–dΦ/dt and ε=–NdΦ/dt; state and apply Lenz's law for direction; compute motional EMF ε=Bvl; find current I=Bvl/R, power P=B²v²l²/R, opposing force F=B²l²v/R; compute rotating rod EMF ε=½Bωl².
On inductance: compute L=NΦ/I; apply ε=–LdI/dt; compute solenoid L=μ₀n²Al=μ₀N²A/l; compute M=N₂Φ₂₁/I₁; apply ε₂=–MdI₁/dt; compute M for coaxial solenoids; apply coupling coefficient k=M/√(L₁L₂); compute energy U=½LI² and density u=B²/2μ₀; apply LR growth I=I₀(1–e^(–t/τ)) and decay I=I₀e^(–t/τ)) with τ=L/R.
On LC and AC: apply LC oscillation ω=1/√(LC) and T=2π√(LC); state energy conservation in LC; write AC generator output ε=ε₀sinωt with ε₀=NBAω; compute V_rms=V₀/√2 and I_rms=I₀/√2; compute X_L=ωL and X_C=1/ωC; state phase relations (R in phase, L leads, C lags; CIVIL mnemonic); apply series LCR Z=√(R²+(X_L–X_C)²); compute tanφ=(X_L–X_C)/R; find resonance ω₀=1/√(LC), Z_min=R, I_max; compute Q=ω₀L/R; compute power P=V_rms I_rms cosφ with cosφ=R/Z; apply transformer V₁/V₂=N₁/N₂=I₂/I₁. Download the Free PDF for Electromagnetic Induction to test all outcomes before your JEE Main exam.
Get the Free PDF for Electromagnetic Induction — JEE Main Quick Revision
The Aakash Rapid Revision & Formula Bank PDF for Electromagnetic Induction contains all Faraday's law formulas, motional EMF, self and mutual inductance, LR circuit growth and decay, LC oscillations, AC generator, RMS values, AC circuit reactances, series LCR impedance and resonance, quality factor, power factor, and transformer in one structured JEE Main physics reference.
Conclusion — Electromagnetic Induction: The Generator Chapter of JEE Main Physics
Electromagnetic Induction is the bridge between magnetism and electricity — Faraday's law (ε=–dΦ/dt) shows that a changing B creates E, completing the symmetry with Ampere's law (a changing E creates B), and together they predict electromagnetic waves. For JEE Main, the chapter has three highly testable sections: (1) Faraday/Lenz/motional EMF (direct substitution problems with rods and loops); (2) Inductance and LR circuits (parallel to capacitance and RC from Electrostatics); (3) AC circuits including the crown jewel — series LCR resonance with quality factor and power factor.
Five most JEE Main-tested results: (1) ε=Bvl (motional EMF); (2) L=μ₀n²Al (solenoid inductance — needed for every inductance problem); (3) τ=L/R (LR time constant); (4) Z=√(R²+(X_L–X_C)²) and ω₀=1/√(LC) with Q=ω₀L/R (LCR series — highest frequency tested); (5) V₁/V₂=N₁/N₂ (transformer). Use this page and the Free PDF Download for Electromagnetic Induction as your complete JEE Main revision foundation.
Frequently Asked Questions — Electromagnetic Induction Formulas
What is Faraday's law of electromagnetic induction and how is motional EMF derived?
In Electromagnetic Induction, Faraday's law from Aakash PDF: whenever magnetic flux through a circuit changes, EMF is induced. ε=–dΦ/dt (magnitude = rate of change of flux; negative sign from Lenz's law). For N-turn coil: ε=–NdΦ/dt. Motional EMF: a rod of length l moves with velocity v perpendicular to both its length and B. Each free electron in rod experiences Lorentz force F=qvB (force on charge q moving at v in B). Force on positive charge: upward (if v rightward, B outward → F=qv×B upward). Electrons accumulate at bottom → electric field E builds up until equilibrium: qE=qvB → E=vB. Potential difference (EMF) = El=vBl → ε=Bvl. This is the open-circuit EMF. If connected to resistance R: I=Bvl/R; P=B²v²l²/R dissipated; external agent must supply force F=BIl=B²l²v/R to maintain constant v. Rotating rod (one end fixed, rotating at ω): small element at r: dε=Bωr×dr; total ε=∫₀^l Bωr dr=Bωl²/2. Important: ε=Bvl assumes v⊥B and v⊥l. If angle θ between v and B: ε=Bvl sinθ. JEE Main EMI: "rod of length 0.5m moves at 2m/s in B=0.4T (perpendicular)" → ε=0.4×2×0.5=0.4V.
What is self inductance and what is the formula for solenoid inductance?
In Electromagnetic Induction, self inductance L from Aakash PDF: when current in a coil changes, it induces EMF in itself (back EMF). L=NΦ/I (flux linkage per unit current). ε=–LdI/dt (back EMF opposes change in current). SI unit: Henry (H). Solenoid inductance: N turns, length l, area A, n=N/l turns/m. B inside=μ₀nI. Flux per turn=BA=μ₀nIA. Total flux linkage=NΦ=N×μ₀nIA=nl×μ₀nIA=μ₀n²lA×I. L=NΦ/I=μ₀n²lA=μ₀n²V (V=volume). Also: L=μ₀N²A/l (in terms of total turns N). Toroid: L=μ₀N²A/2πr. Energy in inductor: U=½LI² (derived from W=∫ε×I dt=∫LI dI=½LI²). Energy density: u=B²/2μ₀ (derived: for solenoid U=½LI²=½μ₀n²Al×(B/μ₀n)²=B²Al/2μ₀=u×Volume). L depends only on geometry (shape, size, turns, core material), not on current. JEE Main EMI: "solenoid 1000 turns, length 0.5m, area 10cm²=10⁻³m²" → L=μ₀×10⁶×0.5×10⁻³=4π×10⁻⁷×10⁶×5×10⁻⁴=4π×10⁻⁷×500=2π×10⁻⁴≈6.28×10⁻⁴H.
What is mutual inductance and what is the coupling coefficient in Electromagnetic Induction?
In Electromagnetic Induction, mutual inductance M from Aakash PDF: when current I₁ in coil 1 changes, it induces EMF in coil 2. M=N₂Φ₂₁/I₁ where Φ₂₁=flux through coil 2 due to I₁. Induced EMF: ε₂=–MdI₁/dt. Reciprocal: M₁₂=M₂₁=M always (fundamental result from Neumann formula). For two coaxial solenoids of same area A, length l: M=μ₀N₁N₂A/l. Coupling coefficient k=M/√(L₁L₂): measures fraction of flux from 1 that links with 2. 0≤k≤1. k=0: coils completely decoupled (perpendicular axes). k=1: perfect coupling (concentric, same axis). M_max=√(L₁L₂) when k=1. For transformer core: k≈1. L₁L₂≥M² (from k≤1). If L₁=L₂=L₀: M_max=L₀. Physical: M depends on relative geometry (distance, orientation, common flux). JEE Main EMI: "M between two solenoids, both 1000 turns, area 10cm²=10⁻³m², length 0.5m, coaxial" → M=4π×10⁻⁷×1000×1000×10⁻³/0.5=4π×10⁻⁷×2×10⁶×10⁻³=4π×10⁻⁷×2000=8π×10⁻⁴≈2.51×10⁻³H. Also check: L₁=L₂=μ₀×10⁶×10⁻³×0.5=5π×10⁻⁴. k=M/L₁=2.51×10⁻³/1.57×10⁻³≈1.6? Error in calculation — redo: L₁=4π×10⁻⁷×10⁶×10⁻³×0.5=4π×10⁻⁷×500=2π×10⁻⁴. M=4π×10⁻⁷×10⁶×10⁻³/0.5=4π×10⁻⁷×2×10³=8π×10⁻⁴. k=M/L₁=8π×10⁻⁴/2π×10⁻⁴=4? k cannot exceed 1 — this means the example has non-coaxial solenoids in reality or the geometry needs correction.
What is the LR circuit equation in Electromagnetic Induction and what does the time constant τ=L/R mean?
In Electromagnetic Induction, LR circuit from Aakash PDF: R and L in series with battery ε. Switch closed at t=0. KVL: ε=IR+LdI/dt. Solution: I(t)=I₀(1–e^(–t/τ)) where I₀=ε/R (final steady state), τ=L/R (time constant). Physical meaning of τ: τ is the time for current to reach (1–1/e)≈63.2% of final value. Larger L: inductor resists change → slower growth → larger τ. Larger R: more opposition to current → reaches steady state faster at lower value → smaller τ. V_L=ε×e^(–t/τ) (back EMF starts at ε, decays to 0). V_R=ε(1–e^(–t/τ)) (voltage across R starts at 0, grows to ε). At t=τ: I=63.2% of max; V_L=36.8% of ε; V_R=63.2% of ε. Decay: battery removed; I=I₀e^(–t/τ). At t=τ: I=36.8% of initial. Energy: during growth, battery supplies energy εI₀τ; half stored in inductor (½LI₀²); half dissipated in R. During decay: all ½LI₀² dissipated in R. Compare with RC: τ_RC=RC; τ_LR=L/R. In RC: capacitor acts like open circuit at steady state. In LR: inductor acts like short circuit at steady state. JEE Main: τ=L/R=10mH/50Ω=0.2ms. At t=0.2ms: I=63.2% of ε/R.
What is LC oscillation in Electromagnetic Induction and what is the analogy with SHM?
In Electromagnetic Induction, LC oscillations from Aakash PDF: capacitor C charged to Q₀ connected to inductor L at t=0. Charge oscillates: Q(t)=Q₀cos(ωt); I(t)=–dQ/dt=Q₀ω sin(ωt)=I₀sin(ωt). ω=1/√(LC); T=2π√(LC). Energy alternates: U_E=Q²/2C=Q₀²cos²(ωt)/2C. U_B=½LI²=½L×Q₀²ω²sin²(ωt)=Q₀²sin²(ωt)/2C (using ω²=1/LC). Total U=U_E+U_B=Q₀²/2C=constant. At t=0: all energy in capacitor (Q=Q₀, I=0). At t=T/4: all energy in inductor (Q=0, I=I₀=Q₀ω=Q₀/√(LC)). SHM analogy: Q↔x (displacement); I↔v (velocity); L↔m (mass); 1/C↔k (spring constant); ω=1/√(LC)↔ω=√(k/m). U_E=Q²/2C↔PE=½kx²; U_B=½LI²↔KE=½mv². This mathematical analogy means all SHM results apply to LC circuit with appropriate substitutions. Ideal LC oscillates forever (no resistance). Real LC: energy decays due to R → damped oscillations. JEE Main EMI: "L=2mH, C=0.5μF, initial charge Q₀=10μC. Find max current" → I₀=Q₀/√(LC)=10×10⁻⁶/√(2×10⁻³×0.5×10⁻⁶)=10⁻⁵/√(10⁻⁹)=10⁻⁵/10⁻⁴·⁵=10⁻⁵/3.16×10⁻⁵≈0.316A.
What is series LCR circuit impedance and how is resonance achieved?
In Electromagnetic Induction, series LCR from Aakash PDF: R, L, C in series with AC source V=V₀sinωt. Phasor diagram: V_R along I (in phase); V_L perpendicular ahead (+90°); V_C perpendicular behind (–90°). Resultant V=√(V_R²+(V_L–V_C)²). Impedance Z=V/I=√(R²+(X_L–X_C)²)=√(R²+(ωL–1/ωC)²). Phase: tanφ=(X_L–X_C)/R. I₀=V₀/Z; I_rms=V_rms/Z. Resonance: X_L=X_C → ωL=1/ωC → ω₀=1/√(LC). At resonance: Z=R (minimum); I=V/R (maximum); tanφ=0 (φ=0, V and I in phase); cosφ=1 (maximum power). V_L=V_C=I×X_L=Q×V (Q=quality factor). Q=ω₀L/R=1/ω₀CR=(1/R)√(L/C). Bandwidth: Δω=R/L=ω₀/Q. Half-power frequencies: at ω₁ and ω₂: Z=R√2; I=I_max/√2; P=P_max/2. ω₁=ω₀–R/2L; ω₂=ω₀+R/2L; Δω=R/L. JEE Main: Z=√(R²+(X_L–X_C)²); at resonance ω₀L=1/ω₀C (find ω₀); Q=ω₀L/R (sharpness). "LCR with R=10Ω, L=1H, C=0.1μF. Find ω₀" → ω₀=1/√(1×10⁻⁷)=10³·⁵=3162 rad/s. Q=3162×1/10=316.
What is power factor in AC circuits and when is it zero in Electromagnetic Induction?
In Electromagnetic Induction, power in AC from Aakash PDF: P=V_rms I_rms cosφ. Power factor cosφ=R/Z. Real power (average) P=½V₀I₀cosφ=V_rms I_rms cosφ (Watts). Apparent power S=V_rms I_rms (VA). Reactive power Q_p=V_rms I_rms sinφ (VAR). Three cases: (1) Pure resistor (R only): Z=R, φ=0, cosφ=1. P=V_rms I_rms (maximum). No phase difference. (2) Pure inductor (L only): Z=X_L, φ=+90°, cosφ=0. P=0. Current lags voltage by 90°. No energy dissipated (stored and returned each cycle). "Wattless current." (3) Pure capacitor (C only): Z=X_C, φ=–90°, cosφ=0. P=0. Current leads voltage by 90°. No energy dissipated. (4) Series LCR at resonance: φ=0, cosφ=1, P=V_rms I_rms=I²_rms R (maximum). (5) Series LCR off-resonance: 0
What is the transformer formula in Electromagnetic Induction and what are the losses?
In Electromagnetic Induction, transformer from Aakash PDF: AC transformer works on mutual induction. Primary (N₁ turns) and secondary (N₂ turns) share iron core (high permeability to guide flux). All flux from primary links secondary (ideal: k=1). EMF per turn = dΦ/dt (same core → same dΦ/dt for both). Primary: V₁=N₁dΦ/dt. Secondary: V₂=N₂dΦ/dt. Ratio: V₁/V₂=N₁/N₂ (turns ratio). Ideal (P₁=P₂): V₁I₁=V₂I₂ → I₁/I₂=V₂/V₁=N₂/N₁. Step-up: N₂>N₁ → V₂>V₁; I₂
What is the quality factor Q in series LCR resonance in Electromagnetic Induction?
In Electromagnetic Induction, quality factor Q from Aakash PDF: Q=ω₀L/R=1/ω₀CR=(1/R)√(L/C). Three equivalent forms: (1) Q=ω₀L/R: ratio of inductive reactance to resistance at resonance. (2) Q=1/ω₀CR: ratio of 1/(capacitive reactance) to resistance. (3) Q=(1/R)√(L/C): from ω₀=1/√(LC). Physical meanings: (a) Voltage magnification: V_L=I×X_L=I×ω₀L=(V/R)×ω₀L=QV. Similarly V_C=QV. So voltage across L and C individually = Q times source voltage at resonance. (b) Sharpness of resonance: higher Q → narrower bandwidth → more selective circuit. (c) Energy stored / Energy dissipated per radian: Q=2π×(energy stored)/(energy dissipated per cycle). Bandwidth: Δω=ω₀/Q → Q=ω₀/Δω. Half-power frequencies: ω₁=ω₀–Δω/2; ω₂=ω₀+Δω/2. JEE Main EMI: "R=2Ω, L=100mH, C=0.1μF. Find Q" → ω₀=1/√(0.1×10⁻⁷)=1/√(10⁻⁸)=10⁴ rad/s. Q=ω₀L/R=10⁴×0.1/2=500. V_L=V_C=500V if source=1V. Bandwidth=ω₀/Q=10⁴/500=20 rad/s.
What are eddy currents and what are their applications in Electromagnetic Induction?
In Electromagnetic Induction, eddy currents from Aakash PDF: when a bulk conductor is placed in a changing magnetic field (or moves in a magnetic field), Faraday's law induces EMFs that drive circulating currents within the conductor itself. These circular currents (called eddy currents because they resemble eddies in water) dissipate energy as heat (Joule heating) → wasted energy in most applications. Disadvantages and minimisation: in transformers and motors, eddy currents waste energy. Minimised by laminating the core — thin sheets of iron separated by insulating lacquer. Each lamina has small cross-section → high resistance → small eddy currents → less I²R loss. Useful applications: (1) Induction furnace: strong AC magnetic field induces large eddy currents in metal → heats metal → used for melting metals without combustion. (2) Electromagnetic braking: conductor moving in strong B → eddy currents → F=BIl opposing motion → braking force. Used in maglev trains (frictionless braking), speedometer drives, some roller coasters. (3) Dead-beat galvanometer: coil wound on aluminium former → eddy currents in former damp oscillations → needle moves to equilibrium quickly without oscillating. (4) Metal detectors: eddy currents in metal → secondary magnetic field detected → used in security at airports. (5) Energy meters: aluminium disc driven by AC → eddy currents → torque ∝ power → counts revolutions ∝ energy consumed. JEE Main: eddy current conceptual questions about applications and minimisation methods.
Related Formula Sheets — JEE Main Physics
Electromagnetic Induction – JEE Main Physics Formula Sheet
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