JEE Main & Advanced

Electromagnetic Induction and Alternating Current

Electromagnetic Induction and Alternating Current for JEE Main & Advanced

1
Module 1

Electromagnetic Induction

Faraday's & Lenz's Laws, Motional EMFTopic 1

Magnetic Flux: $\Phi_B = \int \vec{B}\cdot d\vec{A} = BA\cos\theta$ for uniform field. SI unit: weber (Wb) = T·m².

Faraday's Law: Induced EMF equals negative rate of change of flux: $$\varepsilon = -\frac{d\Phi_B}{dt}$$ For $N$ turns: $\varepsilon = -N\,d\Phi_B/dt$.

Lenz's Law: Induced current opposes the change in flux (consequence of energy conservation). Sign of negative in Faraday's law.

Motional EMF: A conductor of length $L$ moving with velocity $v$ perpendicular to $\vec{B}$: $$\varepsilon = BLv$$ General form: $\varepsilon = \int(\vec{v}\times\vec{B})\cdot d\vec{l}$.

Induced electric field (non-conservative): $\oint \vec{E}\cdot d\vec{l} = -d\Phi_B/dt$.

Eddy Currents: Currents induced in bulk conductors by changing flux; dissipate as heat (used in induction heating, electromagnetic braking; reduced by lamination in transformer cores).

Rotating Coil in Magnetic Field: $N$ turns, area $A$, angular velocity $\omega$ in field $B$: $$\varepsilon = NAB\omega\sin\omega t, \quad \varepsilon_0 = NAB\omega$$ Basis of AC generator.

Worked Examples
1

A rod of length $0.5$ m moves at $4$ m/s perpendicular to a magnetic field of $0.2$ T. Find induced EMF.

Show solution

$$\varepsilon = BLv = 0.2 \times 0.5 \times 4 = 0.4 \text{ V}$$

Final Answer: $\varepsilon = 0.4$ V.

2

A coil of $200$ turns and area $0.01$ m² is rotated at $50$ Hz in a field of $0.1$ T about an axis perpendicular to field. Find peak EMF.

Show solution

$\omega = 2\pi f = 100\pi$ rad/s. $$\varepsilon_0 = NAB\omega = 200 \times 0.01 \times 0.1 \times 100\pi = 20\pi \approx 62.8 \text{ V}$$

Final Answer: $\varepsilon_0 \approx 62.8$ V.

✎ Self-Check — 5 questions0 / 5
Q1.

SI unit of magnetic flux:

Q2.

Lenz's law is consequence of:

Q3.

A magnet falls through a copper ring. The fall is:

Q4.

Motional EMF in a rod of length $L$, velocity $v$, perpendicular to $B$:

Q5.

Eddy currents are minimized in transformer cores by:

Self & Mutual Inductance, Energy in InductorTopic 2

Self-Inductance ($L$): EMF induced in a coil due to change in its own current. $$\varepsilon = -L\frac{dI}{dt}, \quad L = \frac{N\Phi}{I}$$ SI unit: henry (H) = Wb/A = V·s/A.

Self-Inductance of Common Configurations:

ConfigurationSelf-inductance
Long solenoid ($n$ turns/m, area $A$, length $\ell$)$L = \mu_0 n^2 A\ell$
Toroid (radius $r$, $N$ turns, area $A$)$L = \mu_0 N^2 A/(2\pi r)$
Circular loop (radius $R$, wire radius $a$)$L \approx \mu_0 R[\ln(8R/a) - 2]$

Mutual Inductance ($M$): EMF induced in one coil due to current change in another. $$\varepsilon_2 = -M\frac{dI_1}{dt}, \quad M = \frac{N_2\Phi_{21}}{I_1}$$

For two coils with self-inductances $L_1, L_2$ and coupling coefficient $k$: $$M = k\sqrt{L_1 L_2}, \quad 0 \leq k \leq 1$$

Series/Parallel Combination of Inductors (no mutual coupling):

  • Series: $L_{\text{eq}} = L_1 + L_2$
  • Parallel: $1/L_{\text{eq}} = 1/L_1 + 1/L_2$

With mutual: $L_{\text{series}} = L_1 + L_2 \pm 2M$.

Energy Stored in Inductor: $$U = \frac{1}{2}LI^2$$

Magnetic Energy Density: $u_B = B^2/(2\mu_0)$.

LR Circuit (growth): $I = (\varepsilon/R)(1 - e^{-Rt/L})$. Time constant: $\tau = L/R$.

LR Circuit (decay): $I = I_0 e^{-Rt/L}$.

Worked Examples
1

A solenoid has $1000$ turns over $50$ cm, area $10$ cm². Find self-inductance.

Show solution

$n = 1000/0.5 = 2000$ turns/m, $A = 10 \times 10^{-4}$ m², $\ell = 0.5$ m. $$L = \mu_0 n^2 A\ell = 4\pi \times 10^{-7} \times (2000)^2 \times 10^{-3} \times 0.5$$ $$= 4\pi \times 10^{-7} \times 4 \times 10^6 \times 5 \times 10^{-4} = 8\pi \times 10^{-4} \approx 2.51 \times 10^{-3} \text{ H}$$

Final Answer: $L \approx 2.51$ mH.

2

Current in an inductor of $L = 5$ mH changes from $1$ A to $4$ A in $0.1$ s. Find induced EMF and energy stored at $I = 4$ A.

Show solution

$$\varepsilon = -L\,\frac{dI}{dt} = -5 \times 10^{-3} \times \frac{3}{0.1} = -0.15 \text{ V (magnitude } 0.15\text{ V)}$$ $$U = \frac{1}{2}LI^2 = \frac{1}{2}(5 \times 10^{-3})(4)^2 = 0.04 \text{ J}$$

Final Answer: $|\varepsilon| = 0.15$ V; $U = 0.04$ J.

✎ Self-Check — 5 questions0 / 5
Q1.

SI unit of self-inductance:

Q2.

Energy stored in inductor of $L$ carrying current $I$:

Q3.

Time constant of an LR circuit:

Q4.

Mutual inductance depends on:

Q5.

Two coils $L_1$ and $L_2$ are coupled with coefficient $k = 1$. Then $M$ equals:

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Module 2

Alternating Current

AC Circuits, RMS, Reactance, ImpedanceTopic 1

AC Voltage/Current: $V = V_0\sin\omega t$, $I = I_0\sin(\omega t \pm \phi)$. Frequency $f = \omega/(2\pi)$; period $T = 1/f$.

Mean and RMS Values: For sinusoidal AC over one full cycle:

  • Average value over full cycle = $0$
  • Average over half-cycle: $\bar V = 2V_0/\pi \approx 0.637 V_0$
  • RMS value: $V_{\text{rms}} = V_0/\sqrt{2} \approx 0.707 V_0$

RMS value is the DC equivalent that produces same heating. Power: $P = V_{\text{rms}}I_{\text{rms}}\cos\phi$.

AC Through Single Elements:

ElementImpedancePhase ($V$ relative to $I$)
Resistor $R$$R$In phase
Inductor $L$$X_L = \omega L$$V$ leads $I$ by $\pi/2$
Capacitor $C$$X_C = 1/(\omega C)$$V$ lags $I$ by $\pi/2$

Reactance $X_L, X_C$ measured in ohms. $X_L$ increases with frequency; $X_C$ decreases.

Inductor blocks high $f$; capacitor blocks low $f$ (DC).

Power in Pure Reactive Element: $\langle P\rangle = 0$ (over full cycle). Pure $L$ or $C$ are "wattless".

Power Factor: $\cos\phi = R/Z$, where $Z$ = impedance. $\phi$ = phase angle.

Worked Examples
1

An AC source $V = 200\sin(100\pi t)$ V is connected across a $20$ Ω resistor. Find RMS current and average power.

Show solution

$V_0 = 200$ V, $V_{\text{rms}} = 200/\sqrt{2} \approx 141.4$ V. $I_{\text{rms}} = V_{\text{rms}}/R = 141.4/20 \approx 7.07$ A. $P = V_{\text{rms}}I_{\text{rms}} = 141.4 \times 7.07 = 1000$ W = $1$ kW.

Final Answer: $I_{\text{rms}} \approx 7.07$ A; $P = 1$ kW.

2

An inductor of $L = 100$ mH connected to AC of $V_{\text{rms}} = 220$ V at $50$ Hz. Find current and power.

Show solution

$X_L = \omega L = 2\pi \times 50 \times 0.1 = 10\pi \approx 31.4$ Ω. $I_{\text{rms}} = V_{\text{rms}}/X_L = 220/31.4 \approx 7$ A. Power $P = V_{\text{rms}}I_{\text{rms}}\cos\phi$. For pure inductor, $\phi = 90°$, $\cos\phi = 0$, so $P = 0$.

Final Answer: $I_{\text{rms}} \approx 7$ A; $P = 0$ (wattless current).

✎ Self-Check — 5 questions0 / 5
Q1.

RMS value of $I = 10\sin\omega t$ A:

Q2.

The reactance of a capacitor at high frequency:

Q3.

The current in a pure capacitor leads the voltage by:

Q4.

Average power in a pure inductor over one cycle:

Q5.

Power factor of a circuit having only resistance:

LCR Series Circuit, Resonance, Power, TransformerTopic 2

Series LCR Circuit with AC source: $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ $$\tan\phi = \frac{X_L - X_C}{R}$$

Phase angle $\phi$: positive if inductive ($X_L > X_C$), negative if capacitive.

Current: $I_{\text{rms}} = V_{\text{rms}}/Z$. Voltage drops: $V_R = IR$, $V_L = IX_L$, $V_C = IX_C$ (vector addition needed since not in phase).

Resonance occurs when $X_L = X_C$: $$\omega_0 = \frac{1}{\sqrt{LC}}, \quad f_0 = \frac{1}{2\pi\sqrt{LC}}$$

At resonance: $Z = R$ (minimum), $I$ = maximum, $\phi = 0$ (in phase), power factor = 1.

Quality Factor (Q): $$Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$$ High $Q$ → sharp resonance peak (narrow bandwidth). $Q$ = $\omega_0/\Delta\omega$ = $f_0$/bandwidth.

Average Power: $$P = V_{\text{rms}}I_{\text{rms}}\cos\phi = I_{\text{rms}}^2 R$$

At resonance, power is maximum ($\cos\phi = 1$).

Transformer: Two coils (primary, secondary) on common iron core. For ideal transformer: $$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$$

Step-up: $N_s > N_p$ → voltage increases, current decreases. Step-down: opposite.

Power conserved (ideal): $V_p I_p = V_s I_s$. Real transformer efficiency $\eta = P_{\text{out}}/P_{\text{in}}$ (typically $95-99\%$).

Losses in transformer: copper loss ($I^2R$), iron loss (eddy + hysteresis), flux leakage, humming.

Worked Examples
1

A series LCR circuit has $R = 30$ Ω, $L = 100$ mH, $C = 100\,\mu$F connected to $200$ V, $50$ Hz AC. Find impedance, current, and phase angle.

Show solution

$X_L = 2\pi(50)(0.1) = 10\pi \approx 31.4$ Ω. $X_C = 1/(2\pi \cdot 50 \cdot 100 \times 10^{-6}) = 1/(0.0314) \approx 31.8$ Ω. $Z = \sqrt{30^2 + (31.4 - 31.8)^2} = \sqrt{900 + 0.16} \approx 30 $ Ω. $I = V/Z = 200/30 \approx 6.67$ A. $\tan\phi = (31.4 - 31.8)/30 \approx -0.013$, so $\phi \approx -0.8°$ (capacitive, but nearly $0$ — near resonance).

Final Answer: $Z \approx 30$ Ω, $I \approx 6.67$ A, $\phi \approx 0°$.

2

A transformer steps $2200$ V to $220$ V. Primary current is $1$ A. Find secondary current (ideal).

Show solution

$$V_pI_p = V_sI_s \implies I_s = \frac{V_pI_p}{V_s} = \frac{2200 \times 1}{220} = 10 \text{ A}$$

Final Answer: $I_s = 10$ A.

✎ Self-Check — 5 questions0 / 5
Q1.

At resonance in series LCR, the impedance:

Q2.

The resonant frequency of LCR:

Q3.

Step-up transformer:

Q4.

Power dissipated in pure capacitor over a cycle:

Q5.

Quality factor $Q$ for high selectivity should be:

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