Physics Electromagnetic Induction and AC

Electromagnetic Induction & AC Formula Sheet

All key Electromagnetic Induction and AC formulas - Faraday's law, motional EMF, inductance, LCR resonance, transformers. JEE Main & Advanced quick revision.

5 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every formula you need for Electromagnetic Induction and AC circuits, grouped by sub-topic for last-minute revision. Headline results are boxed.

Faraday’s Law & Magnetic Flux

The induced EMF equals the negative rate of change of flux linkage.

$$\boxed{\mathcal{E} = -N\frac{d\Phi_B}{dt}}$$

Magnetic flux through a surface:

$$\Phi_B = \int \vec{B}\cdot d\vec{A} = BA\cos\theta$$
QuantityFormulaNotes
Magnetic flux$\Phi_B = BA\cos\theta$$\theta$ = angle between $\vec{B}$ and area vector $\vec{A}$
Induced EMF$\mathcal{E} = -N\dfrac{d\Phi_B}{dt}$$N$ = number of turns
Flux linkage$N\Phi_B$total flux linked with $N$ turns
Rotating coil flux$\Phi_B = NBA\cos(\omega t)$coil spinning at angular velocity $\omega$
Rotating coil EMF$\mathcal{E} = NBA\omega\sin(\omega t)$AC generator principle
Max EMF (AC generator)$\mathcal{E}_{max} = NBA\omega$peak value
Three ways to change flux
Flux $\Phi_B = BA\cos\theta$ changes by varying (1) field $B$, (2) area $A$, or (3) angle $\theta$. Static fields induce no EMF - only a changing flux does. Units: flux in Weber (Wb) = T·m² = V·s.

Lenz’s Law

The induced current always opposes the change in flux that produced it - this is the source of the negative sign in Faraday’s law.

StatementMeaning
Flux increasing ($\frac{d\Phi_B}{dt} > 0$)$\mathcal{E} < 0$; induced field opposes external field
Flux decreasing ($\frac{d\Phi_B}{dt} < 0$)$\mathcal{E} > 0$; induced field supports external field
Magnet approaching coilnear face becomes same pole → repulsion
Magnet receding from coilnear face becomes opposite pole → attraction
Lenz = Energy Conservation

The opposing force means you must do mechanical work to change the flux, and that work becomes the electrical energy: Mechanical work input = Electrical energy output. A magnet dropped through a coil falls slower than g because of this braking.

Motional EMF

For a straight conductor of length $\ell$ moving with velocity $v$ perpendicular to field $B$:

$$\boxed{\mathcal{E} = B\ell v}$$
QuantityFormulaNotes
Straight rod (perpendicular)$\mathcal{E} = B\ell v$“BLV” - $B$, $v$, $\ell$ mutually perpendicular
General angle$\mathcal{E} = B\ell v\sin\theta$$\theta$ = angle between $\vec{v}$ and $\vec{B}$
Vector form$\mathcal{E} = \displaystyle\int (\vec{v}\times\vec{B})\cdot d\vec{\ell}$for non-uniform cases
Rotating rod (about one end)$\mathcal{E} = \frac{1}{2}B\omega L^2$integrate, since $v = \omega r$ varies
Rotating disc (centre to rim)$\mathcal{E} = \frac{1}{2}B\omega R^2$disc generator
Force on current-carrying rod$F = BI\ell = \dfrac{B^2\ell^2 v}{R}$retarding force (Lenz)
Induced current (rod on rails)$I = \dfrac{B\ell v}{R}$closed loop, resistance $R$
Power dissipated$P = \dfrac{(B\ell v)^2}{R} = F_{app}\,v$equals applied mechanical power

Rod released from rest (velocity decay) and terminal velocity on an incline:

$$v(t) = v_0\,e^{-\frac{B^2\ell^2}{mR}t}, \qquad \tau = \frac{mR}{B^2\ell^2}, \qquad v_t = \frac{mgR\sin\theta}{B^2\ell^2}$$

Self Inductance

$$\boxed{N\Phi_B = LI \qquad \mathcal{E} = -L\frac{dI}{dt}}$$
QuantityFormulaNotes
Definition$L = \dfrac{N\Phi_B}{I}$unit: Henry (H) = Wb/A = V·s/A
Back-EMF$\mathcal{E} = -L\dfrac{dI}{dt}$proportional to rate of change of $I$
Solenoid$L = \dfrac{\mu_0 N^2 A}{\ell}$$L\propto N^2$; air core $\mu_0 = 4\pi\times10^{-7}$ H/m
Solenoid (material core)$L = \dfrac{\mu_r\mu_0 N^2 A}{\ell}$$\mu_r$ = relative permeability
Toroid$L = \dfrac{\mu_0 N^2 A}{2\pi r}$$r$ = mean radius
Energy stored$U = \frac{1}{2}LI^2$energy in magnetic field
Energy density$u = \dfrac{B^2}{2\mu_0}$per unit volume

LR circuit (time constant $\tau = L/R$):

$$\text{Growth: } I(t) = I_0\left(1 - e^{-t/\tau}\right), \quad I_0 = \frac{\mathcal{E}_0}{R} \qquad\qquad \text{Decay: } I(t) = I_0\,e^{-t/\tau}$$
Time constant milestones
At $t = \tau$: growth reaches $0.63\,I_0$ (63%), decay falls to $0.37\,I_0$ (37%). Current through an inductor cannot change instantly (would need infinite EMF). In DC steady state an inductor is a short circuit; for AC it blocks high frequencies.

Mutual Inductance

$$\boxed{\mathcal{E}_2 = -M\frac{dI_1}{dt} \qquad M = k\sqrt{L_1 L_2}}$$
QuantityFormulaNotes
Definition$M = \dfrac{N_2\Phi_{21}}{I_1} = \dfrac{N_1\Phi_{12}}{I_2}$reciprocity: $M_{12} = M_{21} = M$
Induced EMF$\mathcal{E}_2 = -M\dfrac{dI_1}{dt}$unit: Henry (H)
Coupling coefficient$M = k\sqrt{L_1 L_2}$$0 \le k \le 1$; $k=1$ is perfect coupling
Solenoid pair$M = \dfrac{\mu_0 N_1 N_2 A}{\ell}$$M\propto N_1 N_2$
Energy of coupled coils$U = \frac{1}{2}L_1 I_1^2 + \frac{1}{2}L_2 I_2^2 \pm M I_1 I_2$$+$ aiding, $-$ opposing

Series combinations:

$$L_{eq} = L_1 + L_2 + 2M \ \text{(aiding)} \qquad L_{eq} = L_1 + L_2 - 2M \ \text{(opposing)} \qquad M = \frac{L_{aiding} - L_{opposing}}{4}$$

Parallel combinations (general, and identical coils $L_1=L_2=L$):

$$\frac{1}{L_{eq}} = \frac{1}{L_1 \pm M} + \frac{1}{L_2 \pm M} \qquad\Rightarrow\qquad L_{eq} = \frac{L \pm M}{2}$$

AC Circuits: R, L, C

AC voltage and current, with RMS (effective) values:

$$V = V_0\sin(\omega t), \qquad V_{rms} = \frac{V_0}{\sqrt{2}}, \qquad I_{rms} = \frac{I_0}{\sqrt{2}}, \qquad \omega = 2\pi f = \frac{2\pi}{T}$$
ElementPhase relationReactance / opposition$I_0$
Pure R$V$ and $I$ in phase$R$ (constant)$I_0 = \dfrac{V_0}{R}$
Pure L$V$ leads $I$ by 90°$X_L = \omega L = 2\pi f L$ ($\propto f$)$I_0 = \dfrac{V_0}{X_L}$
Pure C$I$ leads $V$ by 90°$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$ ($\propto \frac{1}{f}$)$I_0 = \dfrac{V_0}{X_C}$
ELI the ICE man

In an inductor (L): voltage E leads current IELI. In a capacitor (C): current I leads voltage EICE. Average power over a cycle for a pure inductor or capacitor is zero - they only store and return energy. Indian mains: $V_{rms} = 230$ V → $V_0 = 230\sqrt{2} \approx 325$ V at 50 Hz.

Series LCR Circuit & Resonance

$$\boxed{Z = \sqrt{R^2 + (X_L - X_C)^2} \qquad \tan\phi = \frac{X_L - X_C}{R}}$$
QuantityFormulaNotes
Impedance$Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$AC equivalent of resistance
Current$I = \dfrac{V}{Z}$same current through all elements
Phase angle$\tan\phi = \dfrac{X_L - X_C}{R}$$X_L>X_C$ inductive; $X_L
Average power$P = V_{rms}I_{rms}\cos\phi = I_{rms}^2 R$only $R$ dissipates power
Power factor$\cos\phi = \dfrac{R}{Z}$unity at resonance

Resonance ($X_L = X_C$):

$$\omega_r = \frac{1}{\sqrt{LC}}, \qquad f_r = \frac{1}{2\pi\sqrt{LC}}$$

At resonance: $Z = R$ (minimum), $I = \dfrac{V}{R}$ (maximum), $\phi = 0$, $\cos\phi = 1$, power maximum $P = \dfrac{V_{rms}^2}{R}$.

Quality factor and bandwidth:

$$Q = \frac{\omega_r L}{R} = \frac{1}{\omega_r C R} = \frac{1}{R}\sqrt{\frac{L}{C}}, \qquad \Delta\omega = \frac{\omega_r}{Q}, \qquad V_L = V_C = QV \ \text{(at resonance)}$$
PropertySeries LCRParallel LCR
Resonance condition$X_L = X_C$$X_L = X_C$
Resonant frequency$f_r = \frac{1}{2\pi\sqrt{LC}}$$f_r = \frac{1}{2\pi\sqrt{LC}}$
At resonance$Z$ minimum, $I$ maximum$Z$ maximum, $I$ minimum
Useacceptor (passes $f_r$)rejector (blocks $f_r$)
Power-formula trap
Always use $P = I_{rms}^2 R$, never $I_{rms}^2 Z$ - only the resistor consumes power. At resonance $V_L$ and $V_C$ are equal in magnitude but 180° out of phase, so they cancel and can each be $Q$ times the source voltage (voltage magnification).

Transformers

$$\boxed{\frac{V_2}{V_1} = \frac{N_2}{N_1} = \frac{I_1}{I_2} = K}$$
QuantityFormulaNotes
Voltage ratio$\dfrac{V_2}{V_1} = \dfrac{N_2}{N_1} = K$$K>1$ step-up, $K<1$ step-down
Current ratio$\dfrac{I_2}{I_1} = \dfrac{N_1}{N_2} = \dfrac{1}{K}$voltage up → current down
Power conservation (ideal)$V_1 I_1 = V_2 I_2$ideal transformer
Impedance transformation$\dfrac{Z_1}{Z_2} = \dfrac{N_1^2}{N_2^2} = \dfrac{1}{K^2}$transforms as square of turns ratio
Efficiency$\eta = \dfrac{P_{out}}{P_{in}} = \dfrac{V_2 I_2}{V_1 I_1}$typically 95-99% for power transformers
Copper loss$P_{Cu} = I_1^2 R_1 + I_2^2 R_2$load-dependent ($\propto I^2$)
Iron loss$P_{Fe} = P_{hysteresis} + P_{eddy}$hysteresis $\propto f$; eddy $\propto f^2 B^2$
Max-efficiency condition$P_{Cu} = P_{Fe}$variable losses = constant losses
Transmission loss$P_{loss} = I^2 R$step up $V$ to cut $I$ and slash loss
Transformers need AC

A transformer works only with AC. DC gives constant flux, so $\frac{d\Phi_B}{dt} = 0$ and no EMF is induced in the secondary. Laminated soft-iron cores reduce eddy-current losses.

Power-system voltage chain

graph LR
    A["Power plant
11 kV"] -->|Step-up| B["Transmission
400 kV"] B -->|Step-down| C["Substation
11 kV"] C -->|Step-down| D["Home
230 V"]

High transmission voltage means low current, so $P_{loss} = I^2 R$ drops drastically - the reason power is sent at 132-400 kV.

Master Formula Recap

TopicHeadline formula
Faraday’s law$\mathcal{E} = -N\frac{d\Phi_B}{dt}$
Magnetic flux$\Phi_B = BA\cos\theta$
Motional EMF (rod)$\mathcal{E} = B\ell v$
Rotating rod$\mathcal{E} = \frac{1}{2}B\omega L^2$
Self inductance (solenoid)$L = \frac{\mu_0 N^2 A}{\ell}$
Inductor energy$U = \frac{1}{2}LI^2$
LR time constant$\tau = \frac{L}{R}$
Mutual inductance$M = k\sqrt{L_1 L_2}$
RMS value$V_{rms} = \frac{V_0}{\sqrt{2}}$
Reactances$X_L = \omega L,\ X_C = \frac{1}{\omega C}$
LCR impedance$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Resonant frequency$f_r = \frac{1}{2\pi\sqrt{LC}}$
Quality factor$Q = \frac{1}{R}\sqrt{\frac{L}{C}}$
AC power$P = V_{rms}I_{rms}\cos\phi$
Transformer ratio$\frac{V_2}{V_1} = \frac{N_2}{N_1} = \frac{I_1}{I_2}$

Revise the full chapter: Faraday’s Law · Lenz’s Law · Motional EMF · Self Inductance · Mutual Inductance · AC Circuits · LCR Circuits · Transformers