EMI

EMI

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Electromagnetic Induction (EMI) is the generation of an electromotive force (emf) in a circuit when the magnetic flux linked with it changes with time. The governing relations are Faraday’s law ε = −dΦ/dt and Lenz’s law, which fixes the negative sign: the induced current opposes the change in flux that produced it. For a straight rod of length l moving with velocity v perpendicular to a uniform field B, the motional emf is ε = Blv. A coil with self-inductance L obeys ε = −L dI/dt, and energy stored in an inductor is U = ½LI². JEE Main tests these through rod-rail numericals, solenoid/toroid inductance (L = μ₀N²A/l), and LR transient growth/decay with time constant τ = L/R.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Faraday’s Law and Lenz’s Law

Magnetic flux through a surface is Φ = ∫B·dA; units are weber (Wb). When Φ changes, an emf is induced: ε = −dΦ/dt. The minus sign encodes Lenz’s law — the induced current creates a magnetic field that opposes the original flux change. Dropping the sign loses the direction of induced current, the single most common JEE trap.

Motional EMF

For a conducting rod of length l moving with velocity v through a uniform magnetic field B*,** the free charges experience qv × B, producing ε = Blv when v, B, and l are mutually perpendicular. In a rod-rail setup (rod slides on parallel rails in a uniform B), the motional emf drives a current I = Blv/R, with a magnetic braking force F = B²l²v/R opposing the motion — an energy-conservation consequence of Lenz’s law.

Self and Mutual Inductance

A coil’s self-inductance L is defined by ε = −L dI/dt. For a long solenoid of N turns, area A, length l: L = μ₀N²A/l. Mutual inductance M between two coils: ε₂ = −M(dI₁/dt); for two tightly coupled solenoids sharing a core, M = μ₀N₁N₂A/l. When two inductors are in series, Lₛ = L₁ + L₂ ± 2M; the sign depends on whether their fluxes add (aiding) or cancel (opposing). Parallel combination: 1/Lₚ = 1/L₁ + 1/L₂ ∓ 2M. Treating M = 0 (ignoring coupling) is a frequent error.

LR Circuits and Stored Energy

On connecting a battery V across an LR series circuit, current grows as I(t) = I₀(1 − e^(−t/τ)) with steady value I₀ = V/R and time constant τ = L/R. The energy stored in the inductor’s magnetic field is U = ½LI², transferred entirely from the battery (half is dissipated in R).

Eddy Currents

In bulk conductors moving through non-uniform B, circulating eddy currents dissipate energy as heat. They are suppressed by lamination (thin insulated sheets) and exploited in induction furnaces, brakes, and metal detectors.

Quantity	Formula	Units
Magnetic flux	Φ = BA cosθ	Wb
Faraday’s law	ε = −dΦ/dt	V
Motional emf (rod)	ε = Blv	V
Self-inductance of solenoid	L = μ₀N²A/l	H
LR time constant	τ = L/R	s
Inductor energy	U = ½LI²	J

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Vector Care

The motional emf expression ε = ∫(v × B)·dl is general; ε = Blv is its special case when v ⊥ B ⊥ l. For an angled flux through a coil, use Φ = BA cosθ — replacing it with BA loses the rotation-based emf in AC generators. In rotating-coil generators of area A rotating at ω in field B, peak emf is ε₀ = NBAω, and instantaneous emf is ε = NBAω sin(ωt).

Transient Details

During LR growth, the voltage across the inductor is V_L = V e^(−t/τ), falling exponentially, while V_R = V(1 − e^(−t/τ)) rises. At steady state dI/dt = 0, so V_L = 0 and the inductor behaves like a plain wire — opposite to a capacitor at steady state. This duality shows up in JEE AC problems (XL = ωL vs XC = 1/ωC).

Energy Perspective of Lenz’s Law

The braking force F = B²l²v/R on a rod-rail system can be derived by power balance: mechanical power input Fv must equal I²R = (Blv/R)²·R = B²l²v²/R, giving F = B²l²v/R. This is why Lenz’s law is not optional — without it energy is not conserved.

Common Mistakes

• Writing ε = dΦ/dt (sign error → wrong current direction).
• Adding inductances directly in series without considering mutual coupling and winding sense.
• Confusing flux linkage NΦ with flux Φ when applying ε = −dΦ/dt to multi-turn coils (use NΦ).
• Using ε = Blv when v and B are not perpendicular (use |v × B|·l).
• Forgetting that an open rod still has motional emf across its ends — current flows only if a closed path exists.

Adjacent Topics

EMI links directly to AC circuits (XL = ωL), transformers (V₁/V₂ = N₁/N₂, ideal), electromagnetic waves (Maxwell’s displacement current as the time-varying electric flux that completes Ampère’s law), and the working of electric motors/generators.

Practice Prompts

1. A 50 cm rod moves at 10 m/s perpendicular to B = 0.4 T on rails of resistance 2 Ω (rest of circuit negligible). Find induced emf, current, magnetic force on the rod, and power dissipated.
2. Two solenoids share a core of μᵣ = 1000, l = 20 cm, A = 4 cm², with N₁ = 500, N₂ = 1000. Compute L₁, L₂, and M, then find the equivalent inductance when connected in series aiding and opposing.

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