Faraday's Law & Electromagnetic Induction

Step 1: Define the magnetic flux through a surface S bounded by loop C

$$\Phi_B = \int_S \mathbf{B} \cdot d\mathbf{a}$$

Step 2: Faraday's experimental result — the induced EMF is proportional to the rate of change of flux

$$\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}\int_S \mathbf{B} \cdot d\mathbf{a}$$

Step 3: The EMF around the loop is the line integral of the electric field

$$\mathcal{E} = \oint_C \mathbf{E} \cdot d\boldsymbol{\ell}$$

Step 4: Equate the two expressions (integral form of Faraday's law)

$$\oint_C \mathbf{E} \cdot d\boldsymbol{\ell} = -\frac{d}{dt}\int_S \mathbf{B} \cdot d\mathbf{a}$$

Step 5: For a stationary loop, move the time derivative inside the integral

$$\oint_C \mathbf{E} \cdot d\boldsymbol{\ell} = -\int_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{a}$$

Step 6: Apply Stokes' theorem to the left side

$$\int_S (\nabla \times \mathbf{E}) \cdot d\mathbf{a} = -\int_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{a}$$

Step 7: Since this holds for any surface S, the integrands must be equal

$$\boxed{\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}}$$

This is the differential form of Faraday's law — the third Maxwell equation.

Step 1: Consider a charge q in a wire segment moving with velocity v through field B

$$\mathbf{F}_{\text{mag}} = q(\mathbf{v} \times \mathbf{B})$$

Step 2: The force per unit charge (effective electric field) driving current along the wire

$$\mathbf{f}_{\text{mag}} = \frac{\mathbf{F}_{\text{mag}}}{q} = \mathbf{v} \times \mathbf{B}$$

Step 3: The motional EMF is the line integral of this force per unit charge around the loop

$$\mathcal{E}_{\text{motional}} = \oint_C \mathbf{f}_{\text{mag}} \cdot d\boldsymbol{\ell} = \oint_C (\mathbf{v} \times \mathbf{B}) \cdot d\boldsymbol{\ell}$$

Step 4: For a rectangular loop with one side (length l) sliding with velocity v in a uniform B

$$\mathcal{E} = vBl$$

Only the moving segment contributes since $\mathbf{v} \times \mathbf{B}$ is perpendicular to $d\boldsymbol{\ell}$ on the stationary segments.

Step 5: Verify consistency — the flux through the loop changes as the area changes

$$\Phi_B = B \cdot A = B \cdot l \cdot x(t), \quad \frac{d\Phi_B}{dt} = Blv$$

Step 6: Confirm Faraday's law gives the same result

$$\boxed{\mathcal{E} = -\frac{d\Phi_B}{dt} = -Blv = \oint_C (\mathbf{v} \times \mathbf{B}) \cdot d\boldsymbol{\ell}}$$

Step 1: The total flux change has two contributions

$$\frac{d\Phi_B}{dt} = \frac{d}{dt}\int_S \mathbf{B} \cdot d\mathbf{a}$$

The flux can change because B changes (transformer EMF) or because S changes (motional EMF).

Step 2: Separate the two contributions using the Leibniz integral rule

$$\frac{d}{dt}\int_S \mathbf{B} \cdot d\mathbf{a} = \int_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{a} + \oint_C (\mathbf{B} \times \mathbf{v}) \cdot d\boldsymbol{\ell}$$

Step 3: The transformer EMF (from time-varying B at fixed loop)

$$\mathcal{E}_{\text{transformer}} = -\int_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{a} = \oint_C \mathbf{E} \cdot d\boldsymbol{\ell}$$

Step 4: The motional EMF (from moving loop in static B)

$$\mathcal{E}_{\text{motional}} = -\oint_C (\mathbf{B} \times \mathbf{v}) \cdot d\boldsymbol{\ell} = \oint_C (\mathbf{v} \times \mathbf{B}) \cdot d\boldsymbol{\ell}$$

Using the cyclic property of the scalar triple product.

Step 5: The total EMF is the sum of both contributions

$$\mathcal{E}_{\text{total}} = \oint_C \mathbf{E} \cdot d\boldsymbol{\ell} + \oint_C (\mathbf{v} \times \mathbf{B}) \cdot d\boldsymbol{\ell}$$

Step 6: The universal Faraday's law — valid for any combination of changing B and moving loop

$$\boxed{\mathcal{E}_{\text{total}} = \oint_C (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot d\boldsymbol{\ell} = -\frac{d\Phi_B}{dt}}$$

The integrand is exactly the Lorentz force per unit charge — the total force driving current.

Step 1: The vector potential due to current loop 1 at a field point r

$$\mathbf{A}_1(\mathbf{r}) = \frac{\mu_0 I_1}{4\pi}\oint_{C_1} \frac{d\boldsymbol{\ell}_1}{|\mathbf{r} - \mathbf{r}_1|}$$

Step 2: The magnetic flux through loop 2 due to loop 1

$$\Phi_{21} = \int_{S_2} \mathbf{B}_1 \cdot d\mathbf{a}_2 = \oint_{C_2} \mathbf{A}_1 \cdot d\boldsymbol{\ell}_2$$

Using Stokes' theorem: $\int_S (\nabla \times \mathbf{A}) \cdot d\mathbf{a} = \oint_C \mathbf{A} \cdot d\boldsymbol{\ell}$

Step 3: Substitute the expression for A₁ into the flux integral

$$\Phi_{21} = \frac{\mu_0 I_1}{4\pi}\oint_{C_2}\oint_{C_1} \frac{d\boldsymbol{\ell}_1 \cdot d\boldsymbol{\ell}_2}{|\mathbf{r}_1 - \mathbf{r}_2|}$$

Step 4: Define the mutual inductance M₂₁ via Φ₂₁ = M₂₁ I₁

$$M_{21} = \frac{\Phi_{21}}{I_1} = \frac{\mu_0}{4\pi}\oint_{C_1}\oint_{C_2} \frac{d\boldsymbol{\ell}_1 \cdot d\boldsymbol{\ell}_2}{|\mathbf{r}_1 - \mathbf{r}_2|}$$

Step 5: Note the symmetry — the dot product and denominator are symmetric in 1↔2

$$d\boldsymbol{\ell}_1 \cdot d\boldsymbol{\ell}_2 = d\boldsymbol{\ell}_2 \cdot d\boldsymbol{\ell}_1, \quad |\mathbf{r}_1 - \mathbf{r}_2| = |\mathbf{r}_2 - \mathbf{r}_1|$$

Step 6: Therefore M₁₂ = M₂₁ = M (the Neumann formula)

$$\boxed{M = \frac{\mu_0}{4\pi}\oint_{C_1}\oint_{C_2} \frac{d\boldsymbol{\ell}_1 \cdot d\boldsymbol{\ell}_2}{|\mathbf{r}_1 - \mathbf{r}_2|}}$$

This is a purely geometric quantity depending only on the shapes and relative positions of the two loops.

Step 1: Apply Ampere's law to a rectangular path enclosing n turns per unit length

$$\oint \mathbf{B} \cdot d\boldsymbol{\ell} = \mu_0 I_{\text{enc}}$$

The rectangular Amperian loop has one side of length $h$ inside the solenoid, one outside.

Step 2: The enclosed current for a path of height h is n·h turns, each carrying current I

$$Bh = \mu_0 (n h) I \implies B = \mu_0 n I$$

The field outside an ideal solenoid is zero; only the interior segment contributes.

Step 3: The total flux through one turn of the solenoid (cross-sectional area A)

$$\Phi_{\text{one turn}} = B \cdot A = \mu_0 n I A$$

Step 4: The total flux linkage through all N = nℓ turns

$$\Phi_{\text{total}} = N \cdot \Phi_{\text{one turn}} = (n\ell)(\mu_0 n I A) = \mu_0 n^2 A \ell \cdot I$$

Step 5: The self-inductance is defined by Φ_total = LI

$$\boxed{L = \mu_0 n^2 A \ell = \mu_0 n^2 \pi r^2 \ell = \frac{\mu_0 N^2 A}{\ell}}$$

Note: $n = N/\ell$, so $n^2 \ell = N^2/\ell$. The inductance scales as the square of the number of turns.

Step 1: When current I flows through an inductor, the back-EMF is

$$\mathcal{E}_{\text{back}} = -L\frac{dI}{dt}$$

Step 2: The power delivered to the inductor (work per unit time against back-EMF)

$$P = -\mathcal{E}_{\text{back}} \cdot I = LI\frac{dI}{dt}$$

Step 3: Integrate from I = 0 to I = I_f to find total energy stored

$$W = \int_0^{I_f} P\,dt = \int_0^{I_f} LI\,dI = L\left[\frac{I^2}{2}\right]_0^{I_f} = \frac{1}{2}LI_f^2$$

Step 4: For a solenoid, substitute L = μ₀n²Aℓ and B = μ₀nI

$$W = \frac{1}{2}(\mu_0 n^2 A\ell)I^2 = \frac{1}{2\mu_0}(\mu_0 nI)^2 \cdot (A\ell) = \frac{1}{2\mu_0}B^2 \cdot V$$

where $V = A\ell$ is the volume of the solenoid.

Step 5: Identify the magnetic energy density

$$u_B = \frac{W}{V} = \frac{B^2}{2\mu_0}$$

Step 6: Generalize — for any magnetic field configuration, the total energy is the volume integral

$$\boxed{W = \frac{1}{2}LI^2 = \frac{1}{2\mu_0}\int_{\text{all space}} B^2\,d\tau}$$

The energy is stored in the magnetic field itself, distributed throughout space with density $u_B = B^2/(2\mu_0)$.