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AXIOM PHYSICS Class 12 • Open Access

CBSE Class XII • Unit 04 (Electromagnetic Induction)

Chapter 06: Electromagnetic Induction

Master Revision Dossier • Faraday-Lenz Laws, Motional Induction, Energy Balance, Solenoid Inductances & AC Generator

1. Definitive Formula & Induction Parameter Matrix

Concept Mathematical Formulation SI Unit / Dim Boundary Conditions & Notes
Magnetic Flux ($\Phi_B$) $$\Phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta$$ $\text{Weber (Wb)} = \text{T}\cdot\text{m}^2$
$[\text{M}^1\text{L}^2\text{T}^{-2}\text{A}^{-1}]$		 Scalar quantity. $\theta$ is angle between $\vec{B}$ and normal area vector $\hat{n}$. $\Phi = 0$ when plane of loop is parallel to field.
Faraday-Lenz Law $$\mathcal{E} = -N\frac{d\Phi_B}{dt}, \quad I = -\frac{N}{R}\frac{d\Phi_B}{dt}$$ $\text{Volt (V)}$ Negative sign represents Lenz's law (opposition to flux change). Grounded in Law of Conservation of Energy.
Induced Charge ($\Delta q$) $$\Delta q = \frac{N|\Delta\Phi_B|}{R}$$ $\text{Coulomb (C)}$ Strictly independent of time $\Delta t$ or speed of withdrawal! Depends only on net flux change and loop resistance.
Translational Motional EMF $$\mathcal{E} = Bvl \quad (\vec{v}\perp\vec{B}\perp\vec{l})$$ $\text{Volt (V)}$ Arises from magnetic Lorentz force on free electrons. Polarity determined by Fleming's Right-Hand Rule.
Rotating Conductor EMF $$\mathcal{E} = \frac{1}{2}B\omega l^2 = B f (\pi l^2) = B f A$$ $\text{Volt (V)}$ Between center axle and outer rim. For a wheel with $N$ spokes, EMF remains unchanged (spokes in parallel).
Mechanical-Electrical Power $$P_{\text{mech}} = F_{\text{ext}}v = \frac{B^2 l^2 v^2}{R} = I^2 R = P_{\text{Joule}}$$ $\text{Watt (W)}$ Retarding magnetic force $F_m = \frac{B^2 l^2 v}{R}$ opposes velocity; external work completely converts to Joule heat.
Self-Inductance of Solenoid $$L = \frac{\mu_0 N^2 A}{l} = \mu_0 n^2 A l = \mu_0 n^2 V$$ $\text{Henry (H)} = \text{V}\cdot\text{s/A}$
$[\text{M}^1\text{L}^2\text{T}^{-2}\text{A}^{-2}]$		 Self-induced back-EMF $\mathcal{E} = -L\frac{dI}{dt}$. With magnetic core: $L = \mu_r \mu_0 n^2 A l$. Note $L \propto N^2$.
Coaxial Mutual Inductance $$M = \mu_0 n_1 n_2 (\pi r_1^2) l = \frac{\mu_0 N_1 N_2 A_1}{l}$$ $\text{Henry (H)}$ Reciprocity theorem holds: $M_{12} = M_{21}$. Always uses the cross-sectional area of the inner solenoid $A_1$.
Magnetic Energy & Density $$U = \frac{1}{2}LI^2, \quad u_B = \frac{B^2}{2\mu_0}$$ $U: \text{Joule (J)}$
$u_B: \text{J/m}^3$		 Energy is stored in the magnetic field volume of the inductor, analogous to capacitor electrostatic energy $u_E = \frac{1}{2}\varepsilon_0 E^2$.
AC Generator Equation $$\mathcal{E} = \mathcal{E}_0\sin\omega t, \quad \mathcal{E}_0 = NBA\omega$$ $\text{Volt (V)}$ Flux is $\Phi = NBA\cos\omega t$. When $\Phi$ is maximum ($\theta=0^\circ$), induced EMF is zero ($\sin 0^\circ = 0$). $\mathcal{E}$ leads $\Phi$ by $90^\circ$.

2. Core Derivations Blueprint (Step-by-Step Board Proofs)

Blueprint 6.1: Motional EMF on Sliding Conductor & Mechanical-Electrical Energy Equivalence

1. A rod of length $l$ moves at constant velocity $v = \frac{dx}{dt}$ along parallel rails connected across resistance $R$ in uniform perpendicular field $B$.

2. Flux enclosed: $\Phi_B = Blx$. By Faraday's Law, induced EMF:

$$\mathcal{E} = -\frac{d\Phi_B}{dt} = -Bl\frac{dx}{dt} = -Bvl \implies |\mathcal{E}| = Bvl$$

3. Induced current: $I = \frac{\mathcal{E}}{R} = \frac{Bvl}{R}$. Conductor experiences opposing magnetic retarding force: $F_m = IlB = \frac{B^2 l^2 v}{R}$.

4. Mechanical power required to pull the rod at constant velocity $v$:

$$P_{\text{mech}} = F_{\text{ext}} \cdot v = \left(\frac{B^2 l^2 v}{R}\right)v = \frac{B^2 l^2 v^2}{R}$$

5. Electrical Joule heat dissipation rate in resistor $R$:

$$P_{\text{Joule}} = I^2 R = \left(\frac{Bvl}{R}\right)^2 R = \frac{B^2 l^2 v^2}{R} \equiv P_{\text{mech}} \quad (\text{Energy is Strictly Conserved})$$

Blueprint 6.2: Motional EMF in a Rotating Conducting Rod ($\mathcal{E} = \frac{1}{2}B\omega l^2$)

1. Consider a rod of length $l$ rotating with angular velocity $\omega$ about one end in a plane perpendicular to magnetic field $B$.

2. Radial element $dr$ at distance $r$ from pivot moves with linear velocity $v = \omega r$. Elementary EMF across $dr$:

$$d\mathcal{E} = B v \, dr = B (\omega r) dr = B\omega r \, dr$$

3. Total EMF induced across the full length between center pivot and outer rim:

$$\mathcal{E} = \int_0^l B\omega r \, dr = B\omega \left[\frac{r^2}{2}\right]_0^l = \frac{1}{2}B\omega l^2 = B f (\pi l^2) = B f A$$

Blueprint 6.3: Self-Inductance of a Long Uniform Solenoid ($L = \mu_0 n^2 A l$)

1. For a solenoid of length $l$, area $A$, total turns $N$ ($n = N/l$) carrying current $I$: Field inside is $B = \mu_0 n I = \mu_0 \left(\frac{N}{l}\right)I$.

2. Flux linked per turn: $\Phi_1 = BA = \mu_0 \left(\frac{N}{l}\right) I A$. Total flux linkage for all $N$ turns:

$$N\Phi_B = N \left[\mu_0 \frac{N}{l} I A\right] = \frac{\mu_0 N^2 A}{l} I$$

3. Comparing with $N\Phi_B = LI$:

$$L = \frac{\mu_0 N^2 A}{l} = \mu_0 n^2 A l = \mu_0 n^2 V$$

Blueprint 6.4: Alternating EMF Generated in a Rotating Armature Coil

1. Coil of $N$ turns and area $A$ rotates at angular velocity $\omega$ in field $B$. At time $t$, angle between $\vec{A}$ and $\vec{B}$ is $\theta = \omega t$.

2. Instantaneous magnetic flux: $\Phi_B = N (\vec{B}\cdot\vec{A}) = NBA\cos\omega t$.

3. Differentiating with respect to time according to Faraday's Law:

$$\mathcal{E} = -\frac{d\Phi_B}{dt} = -NBA(-\omega\sin\omega t) = NBA\omega\sin\omega t = \mathcal{E}_0\sin\omega t \quad (\mathcal{E}_0 = NBA\omega)$$

3. High-Yield "Red Alert" Exam Trap Matrix

Master Trap 1: Induced Effects across Standard Physical Configurations

Physical Scenario Induced EMF ($\mathcal{E}$) Induced Current ($I$) Kinematics / Board Verdict
Magnet falling in long copper pipe $\mathcal{E} \ne 0$ $I_{\text{eddy}} \ne 0$ Upward Lenz drag causes $a < g$; reaches steady terminal velocity ($a = 0$).
Magnet falling through cut ring $\mathcal{E} \ne 0$ $I = 0$ (Broken loop) No retarding magnetic force acts; falls freely with $a = g$.
Fast vs Slow magnet withdrawal $\mathcal{E}_{\text{fast}} > \mathcal{E}_{\text{slow}}$ $I_{\text{fast}} > I_{\text{slow}}$ Total charge $\Delta q = \Delta\Phi/R$ is identical in both cases!
Airplane flying horizontally $\mathcal{E} = B_V v l$ $I = 0$ (Open wings) Horizontal component $B_H \parallel \vec{v} \implies$ Only vertical component $B_V$ induces EMF.
Bicycle wheel with $N$ metallic spokes $\mathcal{E} = \frac{1}{2}B\omega l^2$ Parallel network All spokes are in parallel between axle and rim; EMF is independent of spoke count $N$.

⚠️ Trap 2: Mutual Inductance Area Selection

For two coaxial solenoids where outer radius is $r_2$ and inner radius is $r_1$ ($r_2 > r_1$): $$M = \mu_0 n_1 n_2 (\pi r_1^2) l = \frac{\mu_0 N_1 N_2 A_1}{l}$$ Never use $\pi r_2^2$! The magnetic field of the outer solenoid threads only through the physical cross-section of the inner solenoid. Area is strictly $A_{\text{inner}} = \pi r_1^2$.

⚠️ Trap 3: AC Generator Quadrature Phase Shift

$$\Phi_B = NBA\cos\omega t \quad \text{vs} \quad \mathcal{E} = NBA\omega\sin\omega t$$ When coil plane $\perp \vec{B}$ ($\omega t = 0^\circ$): Flux is maximum ($\Phi_{\max} = NBA$), but induced EMF is ZERO ($d\Phi/dt = 0$). When coil plane $\parallel \vec{B}$ ($\omega t = 90^\circ$): Flux is zero, but induced EMF is MAXIMUM ($\mathcal{E}_0 = NBA\omega$).

4. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Rotating Conductor Wheel Across Earth's Magnetic Field Exemplar 01

A wheel with 10 metallic spokes each $0.5\text{ m}$ long is rotated at $120\text{ rev/min}$ in a plane normal to the horizontal component of Earth’s magnetic field $B_H = 0.40\text{ G}$. What is the induced EMF between the axle and the rim of the wheel?

Solution: Length $l = 0.5\text{ m}$, frequency $f = \frac{120}{60} = 2\text{ rev/s}$, $\omega = 2\pi f = 4\pi\text{ rad/s}$. Field $B = 0.40 \times 10^{-4}\text{ T}$.

$$\mathcal{E} = \frac{1}{2} B \omega l^2 = \frac{1}{2} \times (0.40 \times 10^{-4}) \times (4\pi) \times (0.5)^2$$
$$\mathcal{E} = 0.20 \times 10^{-4} \times 4\pi \times 0.25 = 2.0 \times 10^{-5}\pi\text{ V} \approx 6.28 \times 10^{-5}\text{ Volts} = 62.8\text{ }\mu\text{V}$$

Note: The number of spokes (10) does not change the result because all spokes are in parallel.

Problem 2: Solenoid Self-Inductance & Current Reversal Back-EMF Exemplar 02

A solenoid of length $50\text{ cm}$ and area $4.0\text{ cm}^2$ has $500\text{ turns}$. If a current of $2.0\text{ A}$ flowing through it is reversed in $0.10\text{ s}$, calculate (i) self-inductance $L$, (ii) induced back-EMF, and (iii) stored magnetic energy.

Solution: $l = 0.5\text{ m}$, $A = 4 \times 10^{-4}\text{ m}^2$, $N = 500$:

$$L = \frac{\mu_0 N^2 A}{l} = \frac{(4\pi \times 10^{-7}) \times (500)^2 \times (4 \times 10^{-4})}{0.50} = 8\pi \times 10^{-5}\text{ H} \approx 2.51 \times 10^{-4}\text{ H}$$

Current change: $\Delta I = (-2) - (+2) = -4\text{ A} \implies \frac{dI}{dt} = \frac{-4}{0.10} = -40\text{ A/s}$.

$$\mathcal{E} = -L \frac{dI}{dt} = -(2.51 \times 10^{-4}) \times (-40) \approx +10.05\text{ mV}$$ $$U = \frac{1}{2} L I^2 = \frac{1}{2} \times (2.51 \times 10^{-4}) \times (2.0)^2 \approx 5.03 \times 10^{-4}\text{ J} = 0.503\text{ mJ}$$

5. Printable Pocket Flashcard Grid

Induced Charge

$\Delta q = \frac{\Delta\Phi_B}{R}$ (Time Free)

Motional EMF

$\mathcal{E} = Bvl$

Solenoid $L$

$L = \mu_0 n^2 A l$ ($L \propto N^2$)

Peak AC Voltage

$\mathcal{E}_0 = NBA\omega$