Electromagnetic Induction

Step 1: EMF as work per unit charge

The EMF around a loop is $\mathcal{E} = \oint(\mathbf{f}/q)\cdot d\boldsymbol{\ell}$, where $\mathbf{f}$ is the force on a charge. For a conductor moving with velocity $\mathbf{v}$ in field $\mathbf{B}$:

$$\mathcal{E} = \oint(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$$

Step 2: Relate to flux change

The change in flux through the loop during time $dt$ arises from the area swept by each element $d\boldsymbol{\ell}$ moving with velocity $\mathbf{v}$: $d(\Phi_B) = \oint\mathbf{B}\cdot(\mathbf{v}\times d\boldsymbol{\ell})\,dt$.

Step 3: Use the scalar triple product identity

$\mathbf{B}\cdot(\mathbf{v}\times d\boldsymbol{\ell}) = (\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell} = -d\boldsymbol{\ell}\cdot(\mathbf{B}\times\mathbf{v})$. Therefore:

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

Step 4: Recover Faraday's law

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

Step 5: Generalization

When $\mathbf{B}$ itself is time-dependent, the electric field acquires a non-conservative part from $\nabla\times\mathbf{E} = -\partial\mathbf{B}/\partial t$. Both contributions combine into the universal flux rule.

Step 1: Vector potential of loop 1

The vector potential due to current $I_1$ in loop $\mathcal{C}_1$ is:

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

Step 2: Flux through loop 2

By Stokes' theorem, $\Phi_{21} = \int_{\mathcal{S}_2}\mathbf{B}_1\cdot d\mathbf{a}_2 = \oint_{\mathcal{C}_2}\mathbf{A}_1\cdot d\boldsymbol{\ell}_2$.

Step 3: Substitute the vector potential

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

Step 4: Extract the mutual inductance

Since $\Phi_{21} = M_{21}I_1$, we read off: $$M_{21} = \frac{\mu_0}{4\pi}\oint_{\mathcal{C}_1}\oint_{\mathcal{C}_2}\frac{d\boldsymbol{\ell}_1\cdot d\boldsymbol{\ell}_2}{|\mathbf{r}_1-\mathbf{r}_2|}$$

Step 5: Symmetry

The double integral is symmetric under $1\leftrightarrow 2$, proving $M_{12} = M_{21} \equiv M$. This is the Neumann formula, establishing that mutual inductance is a purely geometric quantity independent of which loop carries the current.

Step 1: Power delivered to an inductor

The back-EMF is $\mathcal{E} = -L\,dI/dt$. The power required to drive current against this EMF is $P = -\mathcal{E}\,I = LI\,dI/dt$.

Step 2: Integrate to find total energy

$$W = \int_0^T P\,dt = L\int_0^{I_f}I\,dI = \frac{1}{2}LI_f^2$$

Step 3: Express in terms of B-field

For a solenoid: $L = \mu_0 n^2 V$ and $B = \mu_0 nI$, so $W = \frac{1}{2}\mu_0 n^2 V I^2 = \frac{B^2}{2\mu_0}V$.

Step 4: Generalize to arbitrary fields

$$W = \frac{1}{2\mu_0}\int|\mathbf{B}|^2\,d^3r$$

The energy density is $u_B = B^2/(2\mu_0)$ J/m$^3$.

Step 5: Alternative form using A and J

Using $\mathbf{B} = \nabla\times\mathbf{A}$ and the vector identity $\mathbf{B}\cdot(\nabla\times\mathbf{A}) = \nabla\cdot(\mathbf{A}\times\mathbf{B}) + \mathbf{A}\cdot(\nabla\times\mathbf{B})$, with the surface term vanishing at infinity:

$$\boxed{W = \frac{1}{2}\int\mathbf{A}\cdot\mathbf{J}\,d^3r}$$

Step 1: Start from Maxwell's equations in a conductor

In a conductor with conductivity $\sigma$, Ohm's law gives $\mathbf{J} = \sigma\mathbf{E}$. For a good conductor, the displacement current is negligible compared to the conduction current, so $\nabla\times\mathbf{B} \approx \mu_0\sigma\mathbf{E}$.

Step 2: Derive the diffusion equation

Taking the curl of Faraday's law and substituting: $$\nabla^2\mathbf{B} = \mu_0\sigma\frac{\partial\mathbf{B}}{\partial t}$$

This is a diffusion equation with diffusivity $D = 1/(\mu_0\sigma)$.

Step 3: Assume harmonic time dependence

For $\mathbf{B} \propto e^{-i\omega t}$: $\nabla^2\mathbf{B} = -i\omega\mu_0\sigma\mathbf{B}$. For a planar geometry with $B = B_0 e^{ikz}$:

$$k^2 = i\omega\mu_0\sigma, \quad k = \frac{1+i}{\delta}$$

Step 4: Identify the skin depth

$$\boxed{\delta = \sqrt{\frac{2}{\omega\mu_0\sigma}}}$$

Step 5: Physical interpretation

The field decays as $B_0 e^{-z/\delta}\cos(z/\delta - \omega t)$. For copper at 60 Hz, $\delta \approx 8.5$ mm. At 1 GHz, $\delta \approx 2$ $\mu$m. This is why RF currents flow in a thin surface layer.

Step 1: Build up current in loop 1 first

With $I_2 = 0$, bring $I_1$ to its final value. The work is $W_1 = \frac{1}{2}L_1 I_1^2$.

Step 2: Then build up current in loop 2

Holding $I_1$ fixed, bring $I_2$ to its final value. The work against loop 2's self-inductance is $\frac{1}{2}L_2 I_2^2$. Additionally, the changing $I_2$ induces an EMF in loop 1, requiring extra work $MI_1 I_2$.

Step 3: Total energy

$$\boxed{W = \frac{1}{2}L_1 I_1^2 + MI_1 I_2 + \frac{1}{2}L_2 I_2^2}$$

Step 4: Energy must be non-negative

Since $W \geq 0$ for all $I_1, I_2$, the quadratic form must be positive semi-definite. This requires the discriminant to be non-positive:

$$M^2 \leq L_1 L_2$$

Step 5: Coupling coefficient bound

Defining $k = M/\sqrt{L_1 L_2}$, the constraint becomes $k^2 \leq 1$. Equality holds only when all flux from one loop links the other (perfect coupling). In practice, $k \approx 0.95$-$0.99$ for iron-core transformers.