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.

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

$\iint _{r\in \sigma }\delta (r;C)\bullet dA=N$ ($dA$ 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.

Calculating the Mutual Inductance

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:

$\Phi _{B,2}={\frac {\mu _{0}I}{4\pi }}\left({\frac {3(a_{1}\bullet r)(a_{2}\bullet r)}{|r|^{5}}}-{\frac {a_{1}\bullet a_{2}}{|r|^{3}}}\right)$

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.