Chapter 6 • Theory & Derivations

Electromagnetic Induction and Inductance

Faraday's law of induction, Lenz's law and energy conservation, motional EMF, eddy currents and magnetic damping, self-inductance of solenoids and toroids, mutual inductance and coupling coefficient, LR circuit growth and decay transients, and magnetic field energy storage.

§6.1 Faraday's Law of Induction and Lenz's Law

Michael Faraday (1831) made the epochal discovery that a changing magnetic flux induces an electromotive force in an electric circuit.

1. Magnetic Flux

The magnetic flux $\Phi_B$ through an oriented surface $S$ is defined as: $$\Phi_B = \int_S \vec{B} \cdot d\vec{A}$$ SI Unit: **Weber (Wb)** ($1 \text{ Weber} = 1 \text{ T}\cdot\text{m}^2 = 1 \text{ Volt}\cdot\text{second}$).

2. Faraday's Law of Induction

The induced electromotive force $\mathcal{E}$ in a closed conducting loop is directly proportional to the negative time rate of change of magnetic flux through the loop: $$\mathcal{E} = -\frac{d\Phi_B}{dt}$$ For a closely wound coil of $N$ identical turns: $$\mathcal{E} = -N \frac{d\Phi_B}{dt}$$

3. Lenz's Law and Conservation of Energy

Heinrich Lenz (1834) established the physical origin of the negative sign:

The polarity of the induced electromotive force is always such that any induced current establishes a magnetic field that opposes the original change in magnetic flux that produced it.

Proof by Conservation of Energy: If the induced current reinforced the flux change instead of opposing it, a minuscule initial flux increase would induce current creating more flux, accelerating indefinitely without external energy input—a perpetual motion machine. Because Lenz's law dictates opposition, an external mechanical agent must perform work against magnetic retarding forces to move a magnet or conductor, and this mechanical work is transformed into electrical energy.

4. Differential Form of Faraday's Law

Since $\mathcal{E} = \oint_C \vec{E} \cdot d\vec{l}$: $$\oint_C \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int_S \vec{B} \cdot d\vec{A} = -\int_S \frac{\partial \vec{B}}{\partial t} \cdot d\vec{A}$$ Applying Stokes' Theorem: $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$ Revolutionary Consequence: A time-varying magnetic field creates a non-conservative, non-electrostatic **induced electric field** whose field lines form closed continuous loops ($\oint \vec{E} \cdot d\vec{l} e 0$)!

§6.2 Motional EMF, Eddy Currents, and Magnetic Braking

Electromotive force can also arise purely from the physical motion of a conductor through a static magnetic field.

1. Motional EMF

Consider a conducting rod of length $L$ sliding with velocity $\vec{v}$ along frictionless parallel rails in a uniform magnetic field $\vec{B}$ perpendicular to the rail plane. Free electrons in the rod experience magnetic Lorentz force: $$\vec{F}_m = -e (\vec{v} \times \vec{B})$$ This force pushes electrons to one end, creating an internal electrostatic separating field $\vec{E}_{\text{ind}}$: $$\mathcal{E} = \int_0^L (\vec{v} \times \vec{B}) \cdot d\vec{l} = v B L$$ Alternatively, via Faraday's flux rule: $$\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(B L x) = -B L \frac{dx}{dt} = -B L v$$ If the rails are connected to an external load resistor $R$, induced current is $I = \mathcal{E}/R = BLv/R$. The current-carrying rod experiences a retarding magnetic drag force: $$F_{\text{drag}} = I L B = \frac{B^2 L^2 v}{R}$$ The mechanical power required to pull the rod equals the electrical power dissipated as Joule heat: $$P_{\text{mech}} = F_{\text{drag}} v = \frac{B^2 L^2 v^2}{R} = I^2 R = P_{\text{elec}}$$

2. Eddy Currents and Induction Heating

When a solid metallic block moves through a localized magnetic field, circulating loops of induced current termed **eddy currents (Foucault currents)** are set up within the bulk metal:

Magnetic Braking: By Lenz's law, eddy currents oppose the relative motion, generating smooth, wear-free braking forces (used in high-speed bullet trains and rollercoasters).
Lamination of Transformer Cores: Eddy currents cause severe $I^2 R$ energy losses. Transformer cores are assembled from thin, insulated silicon-steel laminations to interrupt eddy current loops, slashing core losses by over 95%.

§6.3 Self-Inductance, Mutual Inductance, and Inductive Coupling

When the current through a circuit changes, its own magnetic flux varies, inducing a back EMF in the circuit itself—a phenomenon termed self-induction.

1. Self-Inductance ($L$)

The magnetic flux linkage $N\Phi_B$ through a circuit carrying current $I$ is directly proportional to $I$: $$N \Phi_B = L I \implies L = \frac{N \Phi_B}{I}$$ The constant of proportionality $L$ is the Self-Inductance. SI Unit: **Henry (H)** ($1 \text{ Henry} = 1 \text{ Wb/A} = 1 \text{ V}\cdot\text{s/A}$). By Faraday's law, the **self-induced back EMF** is: $$\mathcal{E}_L = -N \frac{d\Phi_B}{dt} = -L \frac{dI}{dt}$$ Inductance represents the electrical inertia of a circuit; it opposes any change in current.

2. Self-Inductance of an Ideal Solenoid

For a long solenoid of length $l$, cross-sectional area $A$, and $N$ total turns ($n = N/l$): $$B = \mu_0 n I = \mu_0 \left(\frac{N}{l}\right) I$$ $$\Phi_B = B A = \mu_0 \left(\frac{N}{l}\right) A I$$ $$L = \frac{N \Phi_B}{I} = \frac{N [\mu_0 (N/l) A I]}{I} = \mu_0 \frac{N^2 A}{l} = \mu_0 n^2 A l = \mu_0 n^2 \cdot (\text{Volume})$$ Inductance scales with the square of the turn count ($L \propto N^2$). If the core is filled with ferromagnetic material of relative permeability $\mu_r$: $L = \mu_0 \mu_r n^2 A l$.

3. Mutual Inductance ($M$) and Coupling Coefficient

When two coils 1 and 2 are in proximity, changing current $I_1$ in coil 1 produces changing flux $\Phi_{21}$ through coil 2: $$\mathcal{E}_2 = -M_{21} \frac{dI_1}{dt}, \quad \mathcal{E}_1 = -M_{12} \frac{dI_2}{dt}$$ By the Neumann Reciprocity Theorem: $$M_{12} = M_{21} = M$$ The magnetic coupling coefficient $k$ is: $$k = \frac{M}{\sqrt{L_1 L_2}}, \quad 0 \le k \le 1$$ $k = 1$ denotes ideal perfect magnetic flux linkage (toroidal transformers).

§6.4 LR Circuit Transients and Magnetic Field Energy Storage

In circuits containing resistors and inductors, the back EMF prevents instantaneous changes in current.

1. Growth of Current in a Series LR Circuit

A battery $\mathcal{E}$, resistor $R$, and inductor $L$ are connected in series; switch closed at $t = 0$. KVL: $$\mathcal{E} - i R - L \frac{di}{dt} = 0 \implies L \frac{di}{dt} + R i = \mathcal{E}$$ Solving with initial condition $i(0) = 0$: $$i(t) = \frac{\mathcal{E}}{R} \left( 1 - e^{-t/\tau_L} \right) = I_0 \left( 1 - e^{-t/\tau_L} \right)$$ where $\tau_L = \frac{L}{R}$ is the inductive time constant (Units: seconds, $\text{H}/\Omega = \text{s}$).

At $t = 0$: $i = 0$, back EMF is maximum ($\mathcal{E}_L = -\mathcal{E}$). The inductor acts as an open circuit.
At $t = \tau_L$: Current reaches $(1 - 1/e) \approx 63.2\%$ of $I_0$.
At $t \to \infty$: $di/dt \to 0$, $i \to I_0 = \mathcal{E}/R$. The inductor acts as an ideal zero-resistance short circuit.

2. Decay of Current in an LR Circuit

When the battery is switched out: $$L \frac{di}{dt} + R i = 0 \implies i(t) = I_0 e^{-t/\tau_L}$$

3. Energy Stored in a Magnetic Field

To establish current $I$ against the opposing back EMF, the source must perform work: $$dW = P dt = (-\mathcal{E}_L) i \, dt = \left( L \frac{di}{dt} \right) i \, dt = L i \, di$$ Integrating from $i = 0$ to $i = I$: $$U_B = \int_0^I L i \, di = \frac{1}{2} L I^2$$ For a long solenoid where $L = \mu_0 n^2 A l$ and $B = \mu_0 n I \implies I = B / (\mu_0 n)$: $$U_B = \frac{1}{2} (\mu_0 n^2 A l) \left( \frac{B}{\mu_0 n} \right)^2 = \frac{B^2}{2\mu_0} (A l)$$ Because $Al$ is the enclosed core volume, the magnetic energy density $u_B$ (J/m³) is: $$u_B = \frac{B^2}{2\mu_0} = \frac{1}{2} \vec{B} \cdot \vec{H}$$ Analogous to $u_E = \frac{1}{2}\epsilon_0 E^2$, magnetic energy is localized continuously throughout the space occupied by the magnetic field!

Standard University Exam Examination Problems

Rigorous Analytical & Numerical Solved Problems

Comprehensive step-by-step mathematical proofs, dimensional evaluations, and calculations matching B.Sc. Honors university examinations.

Undergraduate Standard Classical Exam Example 6.1: Motional EMF and Dynamic Terminal Velocity

A metal rod of mass $m = 40.0\text{ g}$ and length $L = 25.0\text{ cm}$ slides downward under gravity along two frictionless vertical conductive rails separated by distance $L$. The rails are connected at the top by a resistor $R = 2.50\ \Omega$. A uniform horizontal magnetic field $B = 1.20\text{ Tesla}$ is directed perpendicular to the rail plane. Acceleration due to gravity $g = 9.80\text{ m/s}^2$.\n(a) Derive an expression for the motional EMF $\mathcal{E}(v)$ and induced current $I(v)$ as a function of downward velocity $v$,\n(b) Calculate the steady-state terminal velocity $v_t$ attained by the falling rod, and\n(c) Verify that at terminal velocity, the rate of loss of gravitational potential energy equals the electrical power dissipated in resistor $R$.

Step 1: Express motional EMF, current, and magnetic drag

$$\mathcal{E} = B L v$$ $$I = \frac{\mathcal{E}}{R} = \frac{B L v}{R}$$ $$F_{\text{drag}} = I L B = \frac{B^2 L^2 v}{R} \quad (\text{directed upward by Lenz's law})$$

As velocity increases, upward magnetic drag force grows linearly with $v$.

Step 2: Terminal velocity equilibrium

$$m g - F_{\text{drag}} = 0 \implies m g = \frac{B^2 L^2 v_t}{R}$$ $$v_t = \frac{m g R}{B^2 L^2} = \frac{(0.0400 \text{ kg}) \times (9.80 \text{ m/s}^2) \times (2.50\ \Omega)}{(1.20 \text{ T})^2 \times (0.250 \text{ m})^2}$$ $$v_t = \frac{0.980}{1.44 \times 0.0625} = \frac{0.980}{0.0900} = 10.889 \text{ m/s} = 10.89 \text{ m/s}$$

The rod accelerates until drag balances gravity, capping terminal speed at 10.89 m/s.

Step 3: Power conservation verification

$$P_{\text{grav}} = m g v_t = (0.0400 \times 9.80) \times 10.889 = 0.3920 \times 10.889 = 4.268 \text{ Watts}$$ $$P_{\text{elec}} = I^2 R = \left( \frac{B L v_t}{R} \right)^2 R = \frac{B^2 L^2 v_t^2}{R} = \frac{0.0900 \times (10.889)^2}{2.50}$$ $$P_{\text{elec}} = \frac{0.0900 \times 118.57}{2.50} = \frac{10.671}{2.50} = 4.268 \text{ Watts}$$ $$P_{\text{grav}} = P_{\text{elec}} = 4.268 \text{ W} \implies \text{Energy strictly conserved!}$$

Gravitational potential energy is transformed directly into electrical Joule heating with 100% efficiency.

Honors Inductance Problem Example 6.2: Self-Inductance and Stored Energy of a Toroidal Inductor

A toroidal coil has a rectangular cross-section with inner radius $a = 8.00\text{ cm}$, outer radius $b = 14.0\text{ cm}$, and axial height $h = 5.00\text{ cm}$. It consists of $N = 1200$ closely wound turns of copper wire carrying steady current $I = 4.00\text{ A}$ in air ($\mu_r = 1.00$).\n(a) Derive the exact analytical formula for the self-inductance $L$ taking into account the radial variation of $B(r)$,\n(b) Calculate the numerical self-inductance in millihenries (mH), and\n(c) Find the total magnetic energy stored in the toroid.

Step 1: Integrate magnetic flux over cross-section

$$B(r) = \frac{\mu_0 N I}{2\pi r}$$ $$dA = h \, dr$$ $$\Phi_B = \int_a^b B(r) (h \, dr) = \frac{\mu_0 N I h}{2\pi} \int_a^b \frac{dr}{r} = \frac{\mu_0 N I h}{2\pi} \ln\left(\frac{b}{a}\right)$$ $$L = \frac{N \Phi_B}{I} = \frac{\mu_0 N^2 h}{2\pi} \ln\left(\frac{b}{a}\right)$$

Because the magnetic field varies inversely with radius across the core, flux integration is logarithmic.

Step 2: Numerical evaluation of self-inductance

$$\ln(b/a) = \ln(14.0 / 8.00) = \ln(1.75) = 0.55962$$ $$L = \frac{(4\pi \times 10^{-7}) \times (1200)^2 \times 0.0500}{2\pi} \times 0.55962$$ $$L = 2 \times 10^{-7} \times (1.44 \times 10^6) \times 0.0500 \times 0.55962$$ $$L = 0.0144 \times 0.55962 = 8.0585 \times 10^{-3} \text{ Henry} = 8.06 \text{ mH}$$

The toroid has a self-inductance of 8.06 millihenries.

Step 3: Stored magnetic energy

$$U_B = \frac{1}{2} L I^2 = \frac{1}{2} \times (8.0585 \times 10^{-3} \text{ H}) \times (4.00 \text{ A})^2$$ $$U_B = 0.5 \times 8.0585 \times 10^{-3} \times 16.0 = 6.447 \times 10^{-2} \text{ Joules} = 64.5 \text{ mJ}$$

The toroid stores 64.5 mJ of magnetic energy completely confined within its core.

Standard University Exam Problem Example 6.3: LR Circuit Transients and Inductive Back EMF

A series LR circuit has an inductor $L = 2.50\text{ H}$ and a resistor $R = 50.0\ \Omega$ connected across an ideal DC source $\mathcal{E} = 120.0\text{ V}$. The circuit switch is closed at $t = 0$.\n(a) Determine the inductive time constant $\tau_L$ and steady-state maximum current $I_0$,\n(b) Find the instantaneous current $i(t)$ and back EMF $\mathcal{E}_L(t)$ at $t = 0.0500\text{ s}$, and\n(c) At what time $t$ will the energy stored in the magnetic field reach $50.0\%$ of its final maximum value?

Step 1: Compute time constant and steady-state current

$$\tau_L = \frac{L}{R} = \frac{2.50 \text{ H}}{50.0\ \Omega} = 0.0500 \text{ seconds} = 50.0 \text{ ms}$$ $$I_0 = \frac{\mathcal{E}}{R} = \frac{120.0 \text{ V}}{50.0\ \Omega} = 2.400 \text{ Amperes}$$

The inductive time constant is exactly 50 ms and maximum current is 2.40 A.

Step 2: Current and back EMF at t = 50 ms (one time constant)

$$i(\tau_L) = I_0 (1 - e^{-1}) = 2.400 \times (1 - 0.36788) = 2.400 \times 0.63212 = 1.517 \text{ Amperes}$$ $$\mathcal{E}_L(t) = -\mathcal{E} e^{-t/\tau_L} \implies |\mathcal{E}_L(\tau_L)| = 120.0 \times e^{-1} = 120.0 \times 0.36788 = 44.15 \text{ Volts}$$

After one time constant, current reaches 1.52 A (63.2%) and back EMF drops to 44.1 V (36.8%).

Step 3: Time to achieve 50% stored magnetic energy

$$U_B(t) = \frac{1}{2} L [i(t)]^2 = 0.500 U_{B,\max} = 0.500 \left( \frac{1}{2} L I_0^2 \right)$$ $$[i(t)]^2 = 0.500 I_0^2 \implies i(t) = \frac{I_0}{\sqrt{2}} = 0.7071 I_0$$ $$I_0 (1 - e^{-t/\tau_L}) = 0.7071 I_0 \implies e^{-t/\tau_L} = 1 - 0.7071 = 0.2929$$ $$-\frac{t}{\tau_L} = \ln(0.2929) = -1.228 \implies t = 1.228 \tau_L$$ $$t = 1.228 \times 0.0500 \text{ s} = 0.0614 \text{ seconds} = 61.4 \text{ ms}$$

Because energy scales as $i^2$, current must reach $1/\sqrt{2} \approx 70.7\%$ of maximum, requiring 61.4 ms.