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): $\mathrm{\nabla }\times B={\mu }_{0}j$

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

where $E$ and $B$ are the electric and magnetic fields respectively, $j$ is the current density and ${\mu }_0$ is the magnetic permeability.

Mathematical Preliminaries

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

$\underset{r\in \sigma }{\iint }\delta (r;C)\bullet dA=N$ ($dA$ 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 $F(r)$, $\underset{r\in C}{\int }F(r)\bullet dr=\underset{r\in {ℝ}^3}{\iiint }(F(r)\bullet \delta (r;C))d\tau$ ($dr$ 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 $\mathrm{\nabla }\bullet \delta (r;C)=0$

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 $|r|\to +\mathrm{\infty }$, then there exists a family $C[\xi ]$ of closed loops where $\xi \in D_C$ is an arbitrary continuous indexing parameter such that $F(r)=\underset{\xi \in D_C}{\iint }\delta (r;C[\xi ])d\xi$. 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

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

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

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

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 $|r|\to +\mathrm{\infty }$, then there exists a family $\sigma [\xi ]$ of closed surfaces where $\xi \in D_{\sigma }$ is an arbitrary continuous indexing parameter such that $F(r)=\underset{\xi \in D_{\sigma }}{\iint }\delta (r;\sigma [\xi ])d\xi$. 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 $\mathrm{\nabla }\times \delta (r;\sigma )=\delta (r;C)$. Given any vector field $F$ that denotes a multi-surface, then $\mathrm{\nabla }\times F$ 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

${\mathrm{\Phi }}_{B,2}=\underset{r\in {\sigma }_2}{\iint }{B}_1(r)\bullet dA$ where $dA$ is the vector representation of an infinitesimal surface element of ${\sigma }_2$.

Note that also, ${\mathrm{\Phi }}_{B,2}\propto I$. This constant of proportionality, $M_{1,2}=\frac{{\mathrm{\Phi }}_{B,2}}{I}$, 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 $\mathrm{\nabla }\times B_1(r)={\mu }_{0}I\delta (r;C_1)⟹\mathrm{\nabla }\times \frac{1}{{\mu }_0}\frac{B_1(r)}{I}=\delta (r;C_1)$, 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:

$\underset{r\in {\sigma }_1}{\iint }F(r)\bullet dA=\underset{r\in {ℝ}^3}{\iiint }F(r)\bullet \delta (r;{\sigma }_1)d\tau =\underset{r\in {ℝ}^3}{\iiint }F(r)\bullet \frac{b_1(r)}{{\mu }_0}d\tau$

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:

${\mathrm{\Phi }}_{B,2}=\underset{r\in {ℝ}^3}{\iiint }I\frac{b_1(r)\bullet b_2(r)}{{\mu }_0}d\tau$

Therefore: $M(C_1,C_2)=\underset{r\in {ℝ}^3}{\iiint }\frac{b_1(r)\bullet b_2(r)}{{\mu }_0}d\tau$ from which the symmetry $M(C_1,C_2)=M(C_2,C_1)$ is now apparent.

Calculating the Mutual Inductance

Approach #1 (use the vector potential)

Gauss' law of magnetism requires that $\mathrm{\nabla }\bullet B=0$. This makes possible a "vector potential" for $B$: a vector field $A$ which satisfies $\mathrm{\nabla }\times A=B$. The condition $\mathrm{\nabla }\bullet A=0$ can also be enforced.

Using the vector identity:

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

Ampere's law becomes:

$\mathrm{\nabla }\times B={\mu }_{0}j⟺\mathrm{\nabla }\times (\mathrm{\nabla }\times A)={\mu }_{0}j⟺\mathrm{\nabla }(\mathrm{\nabla }\bullet A)-{\mathrm{\nabla }}^{2}A={\mu }_{0}j⟺{\mathrm{\nabla }}^{2}A=-{\mu }_{0}j$

${\mathrm{\nabla }}^{2}A=-{\mu }_{0}j$ is an instance of Poisson's equation which has the solution: $A(r)=\frac{{\mu }_0}{4\pi }\underset{r^{\prime }\in {ℝ}^3}{\iiint }\frac{j(r^{\prime })}{|r-r^{\prime }|}d{\tau }^{\prime }$

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

The vector potential generated by a current of $I$ flowing through closed loop $C_1$ is: $A_1(r)=\frac{{\mu }_0}{4\pi }\underset{r_1\in {ℝ}^3}{\iiint }\frac{I\delta (r_1;C_1)}{|r-r_1|}d{\tau }_1=\frac{{\mu }_0}{4\pi }\underset{r_1\in C_1}{\int }\frac{I}{|r-r_1|}d{r}_1$

The magnetic field generated by a current of $I$ flowing through closed loop $C_1$ is: $B_1=\mathrm{\nabla }\times A_1$. The flux through surface ${\sigma }_2$ (which is counter-clockwise bounded by $C_2$), is

${\mathrm{\Phi }}_{B,2}=\underset{r_2\in {\sigma }_2}{\iint }{B}_1(r_2)\bullet d{A}_2=\underset{r_2\in {\sigma }_2}{\iint }(\mathrm{\nabla }\times A_1)(r_2)\bullet d{A}_2=\underset{r_2\in C_2}{\int }{A}_1(r_2)\bullet d{r}_2$ via Stoke's theorem.

${\mathrm{\Phi }}_{B,2}=\frac{{\mu }_0}{4\pi }\underset{r_2\in C_2}{\int }\underset{r_1\in C_1}{\int }\frac{I}{|r_2-r_1|}(d{r}_1\bullet d{r}_2)$ so the mutual inductance is: $M(C_1,C_2)=\frac{{\mathrm{\Phi }}_{B,2}}{I}=\frac{{\mu }_0}{4\pi }\underset{r_2\in C_2}{\int }\underset{r_1\in C_1}{\int }\frac{d{r}_1\bullet d{r}_2}{|r_2-r_1|}$

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 $C=\underset{r\in \sigma }{\int }\partial (dA)$.

The linearity of the mutual inductance gives:

$M(C_1,C_2)=M(\underset{r_1\in {\sigma }_1}{\iint }\partial (d{A}_1),\underset{r_2\in {\sigma }_2}{\iint }\partial (d{A}_2))=\underset{r_1\in {\sigma }_1}{\iint }\underset{r_2\in {\sigma }_2}{\iint }M(\partial (d{A}_1),\partial (d{A}_2))$

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:

$B(r)=\frac{{\mu }_0}{4\pi }(\frac{3(P\bullet r)r}{|r{|}^5}-\frac{P}{|r{|}^3})$

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 $I{a}_1$. The flux of the magnetic field generated by $I{a}_1$ through $a_2$ is:

${\mathrm{\Phi }}_{B,2}=\frac{{\mu }_{0}I}{4\pi }(\frac{3(a_1\bullet r)(a_2\bullet r)}{|r{|}^5}-\frac{a_1\bullet a_2}{|r{|}^3})$

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

$M(c_1,c_2)=\frac{{\mu }_0}{4\pi }(\frac{3(a_1\bullet r)(a_2\bullet r)}{|r{|}^5}-\frac{a_1\bullet a_2}{|r{|}^3})$

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

$M(C_1,C_2)=\frac{{\mu }_0}{4\pi }\underset{r_1\in {\sigma }_1}{\iint }\underset{r_2\in {\sigma }_2}{\iint }(\frac{3(d{A}_1\bullet (r_2-r_1))(d{A}_2\bullet (r_2-r_1))}{|r_2-r_1{|}^5}-\frac{d{A}_1\bullet d{A}_2}{|r_2-r_1{|}^3})$

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

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