Introduction to Mathematical Physics/Electromagnetism/Electromagnetic induction

Introduction
Electromagnetic induction refers to the induction of an electric motive force (emf) in a closed loop $$C_2$$ via Faraday's law from the magnetic field generated by current in a closed loop $$C_1$$.

The two laws involved in electromagnetic induction are:

Ampere's Law (static version): $$\nabla \times B = \mu_0 j$$

Faraday's Law: $$\nabla \times E = -\frac{\partial B}{\partial t}$$

where $$E$$ and $$B$$ are the electric and magnetic fields respectively, $$j$$ is the current density and $$\mu_0$$ is the magnetic permeability.

Loops, multi-loops, and divergence-free vector fields
The relationship between paths, loops, and divergence free vector fields is an important mathematical preliminary that merits a brief introduction.

Given any oriented path $$C$$, $$C$$ can be characterized by a vector field $$\delta(r;C)$$. $$\delta(r;C) = 0$$ for all positions $$r \notin C$$. For all positions $$r \in C$$, $$\delta(r;C)$$ is infinite in the direction of $$C$$ in a manner similar to the Dirac delta function. The integral property that must be satisfied by $$\delta(r;C)$$ is that for any oriented surface $$\sigma$$, if $$C$$ passes through $$\sigma$$ in the preferred direction a net total of $$N$$ times, then

$$ \iint_{r \in \sigma} \delta(r;C) \bullet dA = N $$ ($$ dA $$ is a vector that denotes an infinitesimal oriented surface segment)

($$C$$ passing through $$\sigma$$ in the reverse direction decreases $$N$$ by 1.)

Given any vector field $$F(r)$$, $$\int_{r \in C} F(r) \bullet dr = \iiint_{r \in \mathbb{R}^3} (F(r) \bullet \delta(r;C))d\tau$$ ($$ dr $$ is a vector that denotes an infinitesimal oriented path segment, and $$d\tau$$ is an infinitesimal volume segment)

It is easy to verify that if $$C$$ is a closed loop, then $$\nabla \bullet \delta(r;C) = 0$$

Given any sequence of closed loops $$C_1, C_2, \dots, C_k$$, these loops can be added in a linear fashion to get a "multi-loop" denoted by the vector field $$\delta(r;C_1) + \delta(r;C_2) + \dots + \delta(r;C_k)$$. The multi-loop is denoted by: $$C_1+C_2+\dots+C_k$$.

Most importantly, given any divergence-free vector field $$F$$ that decreases faster than $$o(1/|r|^2)$$ as $$|r| \rightarrow +\infty$$, then there exists a family $$C[\xi]$$ of closed loops where $$\xi \in D_C$$ is an arbitrary continuous indexing parameter such that $$F(r) = \iint_{\xi \in D_C} \delta(r;C[\xi])d\xi$$. In simpler terms, any divergence free vector field can be expressed as a linear combination of closed loops.

Surfaces, multi-surfaces, and irrotational vector fields
The relationship between surfaces, closed surfaces, and irrotational vector fields is also an important mathematical preliminary that merits a brief introduction.

Given any oriented surface $$\sigma$$, $$\sigma$$ can be characterized by a vector field $$\delta(r;\sigma)$$. $$\delta(r;\sigma) = 0$$ for all positions $$r \notin \sigma$$. For all positions $$r \in \sigma$$, $$\delta(r;\sigma)$$ is infinite in the direction of the outwards normal direction to $$\sigma$$ in a manner similar to the Dirac delta function. The integral property that must be satisfied by $$\delta(r;\sigma)$$ is that for any oriented path $$C$$, if $$C$$ passes through $$\sigma$$ in the preferred direction a net total of $$N$$ times, then

$$ \int_{r \in C} \delta(r;\sigma) \bullet dr = N $$

($$C$$ passing through $$\sigma$$ in the reverse direction decreases $$N$$ by 1.)

Given any vector field $$F(r)$$, $$\int_{r \in \sigma} F(r) \bullet dA = \iiint_{r \in \mathbb{R}^3} (F(r) \bullet \delta(r;\sigma))d\tau$$

It is easy to verify that if $$\sigma$$ is a closed surface, then $$\delta(r;\sigma)$$ is irrotational.

Given any sequence of surfaces $$\sigma_1, \sigma_2, \dots, \sigma_k$$, these surfaces can be added in a linear fashion to get a "multi-surface" denoted by the vector field $$\delta(r;\sigma_1) + \delta(r;\sigma_2) + \dots + \delta(r;\sigma_k)$$. The multi-surface is denoted by: $$\sigma_1+\sigma_2+\dots+\sigma_k$$.

Most importantly, given any irrotational vector field $$F$$ that decreases faster than $$o(1/|r|^2)$$ as $$|r| \rightarrow +\infty$$, then there exists a family $$\sigma[\xi]$$ of closed surfaces where $$\xi \in D_\sigma$$ is an arbitrary continuous indexing parameter such that $$F(r) = \iint_{\xi \in D_\sigma} \delta(r;\sigma[\xi])d\xi$$. In simpler terms, any irrotational vector field can be expressed as a linear combination of closed surfaces.

Given an oriented surface $$\sigma$$ with a counter-clockwise oriented boundary $$C$$, it is then the case that $$\nabla \times \delta(r;\sigma) = \delta(r;C)$$. Given any vector field $$F$$ that denotes a multi-surface, then $$\nabla \times F$$ is a vector field that denotes the counter-clockwise oriented boundary of the multi-surface denoted by $$F$$. This property is important as it enables a magnetic field to denote a multi-surface interior for the closed loop of current that generates it.

Definition of Mutual Inductance
Let $$C_1$$ and $$C_2$$ be two oriented closed loops, and let $$\sigma_1$$ and $$\sigma_2$$ be oriented surfaces whose counter-clockwise boundaries are respectively $$C_1$$ and $$C_2$$.

Given a current of $$I$$ flowing around $$C_1$$, let $$B_1$$ be the magnetic field induced via Ampere's law. Note that $$B_1(r) \propto I$$. The magnetic flux through surface $$\sigma_2$$ is

$$\Phi_{B,2} = \iint_{r \in \sigma_2} B_1(r) \bullet dA$$ where $$dA$$ is the vector representation of an infinitesimal surface element of $$\sigma_2$$.

Note that also, $$\Phi_{B,2} \propto I$$. This constant of proportionality, $$M_{1,2} = \frac{\Phi_{B,2}}{I}$$, is the mutual electromagnetic induction from $$C_1$$ to $$C_2$$.

The mutual electromagnetic induction from $$C_1$$ to $$C_2$$ will be denoted with $$M(C_1,C_2)$$

Self Inductance
When $$C_2 = C_1$$, the inductance $$L(C_1) = M(C_1,C_1)$$ is referred to as the "self inductance".

Linearity of Mutual Inductance
Given loops $$C_1$$, $$C_2$$, and $$C_3$$, it is relatively simple to demonstrate that $$M(C_1+C_3,C_2) = M(C_1,C_2)+M(C_3,C_2)$$ and $$M(C_1,C_2+C_3) = M(C_1,C_2)+M(C_1,C_3)$$.

Let $$B_1(r)$$, $$B_2(r)$$, and $$B_3(r)$$ be the magnetic fields generated when a current of $$I$$ flows through $$C_1$$, $$C_2$$, or $$C_3$$ respectively.

The magnetic field generated by $$C_1$$ and $$C_3$$ together is $$B_1(r)+B_3(r)$$ due to the linearity of Maxwell's equations. This leads to $$M(C_1+C_3,C_2) = M(C_1,C_2)+M(C_3,C_2)$$.

The flux through $$C_2+C_3$$ is the sum of the flux through $$C_2$$ and $$C_3$$ separately. This leads to $$M(C_1,C_2+C_3) = M(C_1,C_2)+M(C_1,C_3)$$.

Symmetry of Mutual Inductance
It is the case that given loops $$C_1$$ and $$C_2$$, that $$M(C_1,C_2) = M(C_2,C_1)$$. This symmetry, while apparent from explicit formulas for the mutual inductance, is far from obvious however. To make this fact more intuitive, the magnetic fields that are generated by $$C_1$$ and $$C_2$$ will be interpreted as multi-surfaces whose boundaries are respectively $$C_1$$ and $$C_2$$.

Let there exist a current of $$I$$ in loop $$C_1$$, and let $$B_1(r)$$ denote the resultant magnetic field. Ampere's law requires that $$\nabla \times B_1(r) = \mu_0 I\delta(r;C_1) \implies \nabla \times \frac{1}{\mu_0}\frac{B_1(r)}{I} = \delta(r;C_1)$$, and therefore $$\frac{1}{\mu_0}\frac{B_1(r)}{I}$$ is a multi-surface whose boundary is $$C_1$$. Since $$B_1(r) \propto I$$, let $$b_1(r) = \frac{B_1(r)}{I}$$.

Given a divergence free vector field $$F$$, the flux of $$F$$ through $$\sigma_1$$ is:

$$ \iint_{r \in \sigma_1} F(r) \bullet dA = \iiint_{r \in \mathbb{R}^3} F(r) \bullet \delta(r;\sigma_1)d\tau = \iiint_{r \in \mathbb{R}^3} F(r) \bullet \frac{b_1(r)}{\mu_0} d\tau$$

The final equality holds due to the fact that $$F$$ is divergence free and that $$\delta(r;\sigma_1)$$ and $$\frac{b_1(r)}{\mu_0}$$ are multi-surfaces with a common boundary of $$C_1$$.

$$B_1(r)$$ is divergence free. The flux of $$B_1(r)$$ through $$\sigma_2$$ is:

$$\Phi_{B,2} = \iiint_{r \in \mathbb{R}^3} I \frac{b_1(r) \bullet b_2(r)}{\mu_0} d\tau$$

Therefore: $$M(C_1,C_2) = \iiint_{r \in \mathbb{R}^3} \frac{b_1(r) \bullet b_2(r)}{\mu_0} d\tau$$ from which the symmetry $$M(C_1,C_2) = M(C_2,C_1)$$ is now apparent.

Approach #1 (use the vector potential)
Gauss' law of magnetism requires that $$\nabla \bullet B = 0$$. This makes possible a "vector potential" for $$B$$: a vector field $$A$$ which satisfies $$\nabla \times A = B$$. The condition $$\nabla \bullet A = 0$$ can also be enforced.

Using the vector identity:

For any vector field $$F$$: $$\nabla \times (\nabla \times F) = \nabla(\nabla \bullet F) - \nabla^2 F$$

Ampere's law becomes:

$$\nabla \times B = \mu_0 j \iff \nabla \times (\nabla \times A) = \mu_0 j \iff \nabla(\nabla \bullet A) - \nabla^2 A = \mu_0 j \iff \nabla^2 A = -\mu_0 j$$

$$\nabla^2 A = -\mu_0 j$$ is an instance of Poisson's equation which has the solution: $$ A(r) = \frac{\mu_0}{4\pi}\iiint_{r' \in \mathbb{R}^3} \frac{j(r')}{|r-r'|}d\tau' $$

It can be checked that for this solution, since $$\nabla \bullet j = 0$$, that $$\nabla \bullet A = 0$$.

The vector potential generated by a current of $$I$$ flowing through closed loop $$C_1$$ is: $$A_1(r) = \frac{\mu_0}{4\pi}\iiint_{r_1 \in \mathbb{R}^3} \frac{I\delta(r_1;C_1)}{|r-r_1|}d\tau_1 = \frac{\mu_0}{4\pi}\int_{r_1 \in C_1} \frac{I}{|r-r_1|}dr_1$$

The magnetic field generated by a current of $$I$$ flowing through closed loop $$C_1$$ is: $$B_1 = \nabla \times A_1$$. The flux through surface $$\sigma_2$$ (which is counter-clockwise bounded by $$C_2$$), is

$$ \Phi_{B,2} = \iint_{r_2 \in \sigma_2} B_1(r_2) \bullet dA_2 = \iint_{r_2 \in \sigma_2} (\nabla \times A_1)(r_2) \bullet dA_2 = \int_{r_2 \in C_2} A_1(r_2) \bullet dr_2 $$ via Stoke's theorem.

$$ \Phi_{B,2} = \frac{\mu_0}{4\pi}\int_{r_2 \in C_2}\int_{r_1 \in C_1} \frac{I}{|r_2-r_1|}(dr_1 \bullet dr_2) $$ so the mutual inductance is: $$ M(C_1,C_2) = \frac{\Phi_{B,2}}{I} = \frac{\mu_0}{4\pi}\int_{r_2 \in C_2}\int_{r_1 \in C_1} \frac{dr_1 \bullet dr_2}{|r_2-r_1|} $$

This equation is known as "Neumann formula".

It can also be seen from this expression that the mutual inductance is symmetric: $$M(C_1,C_2) = M(C_2,C_1)$$.

Approach #2 (use linearity and loop dipoles)
Given any closed loop $$C$$, let $$\sigma$$ be an oriented surface that has $$C$$ as its counterclockwise boundary. For each infinitesimal area vector element $$dA$$ of $$\sigma$$, let the infinitesimal $$\partial(dA)$$ be an infinitesimal closed loop that is the counterclockwise boundary of $$dA$$. It is then the case that $$C = \int_{r \in \sigma} \partial(dA)$$.

The linearity of the mutual inductance gives:

$$M(C_1, C_2) = M\left(\iint_{r_1 \in \sigma_1} \partial(dA_1), \iint_{r_2 \in \sigma_2} \partial(dA_2)\right) = \iint_{r_1 \in \sigma_1}\iint_{r_2 \in \sigma_2} M(\partial(dA_1),\partial(dA_2))$$

In other words, the mutual inductance between two large loops can be expressed as the sum of mutual inductances between several mini loops.

Given an area vector $$A$$, and a current $$I$$ that flows around the boundary of $$A$$ in a counterclockwise manner, then the magnetic dipole (vector) formed is $$P = IA$$. If the area shrinks, then the current increases proportionally if the magnetic dipole is to remain constant.

Given a magnetic dipole $$P$$ with an infinitesimal area at position $$0$$, the magnetic field produced by $$P$$ is:

$$B(r) = \frac{\mu_0}{4\pi}\left(\frac{3(P \bullet r)r}{|r|^5} - \frac{P}{|r|^3}\right)$$

Let $$a_1$$ and $$a_2$$ be area vectors of the interiors of two infinitesimal loops, with the second loop displaced from the first by $$r$$. Let a current $$I$$ flow around the boundary of $$a_1$$ in a counter clockwise manner forming the dipole $$Ia_1$$. The flux of the magnetic field generated by $$Ia_1$$ through $$a_2$$ is:

$$\Phi_{B,2} = \frac{\mu_0I}{4\pi}\left(\frac{3(a_1 \bullet r)(a_2 \bullet r)}{|r|^5} - \frac{a_1 \bullet a_2}{|r|^3}\right) $$

Therefore if $$c_1$$ and $$c_2$$ are the counter clockwise boundaries of $$a_1$$ and $$a_2$$:

$$M(c_1,c_2) = \frac{\mu_0}{4\pi}\left(\frac{3(a_1 \bullet r)(a_2 \bullet r)}{|r|^5} - \frac{a_1 \bullet a_2}{|r|^3}\right) $$

Returning to computing the mutual inductance between $$C_1$$ and $$C_2$$ gives:

$$M(C_1,C_2) = \frac{\mu_0}{4\pi}\iint_{r_1 \in \sigma_1}\iint_{r_2 \in \sigma_2} \left(\frac{3(dA_1 \bullet (r_2-r_1))(dA_2 \bullet (r_2-r_1))}{|r_2-r_1|^5} - \frac{dA_1 \bullet dA_2}{|r_2-r_1|^3}\right) $$

This formula is centered around surface integrals as opposed to loop integrals.