Introduction to Mathematical Physics/Electromagnetism/Electromagnetic induction

Introduction

Electromagnetic induction refers to the induction of an electric motive force (emf) in a closed loop ${\displaystyle C_{2}}$ via Faraday's law from the magnetic field generated by current in a closed loop ${\displaystyle C_{1}}$.

The two laws involved in electromagnetic induction are:

Ampere's Law (static version): ${\displaystyle \nabla \times B=\mu _{0}j}$

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

where ${\displaystyle E}$ and ${\displaystyle B}$ are the electric and magnetic fields respectively, ${\displaystyle j}$ is the current density and ${\displaystyle \mu _{0}}$ 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 ${\displaystyle C}$, ${\displaystyle C}$ can be characterized by a vector field ${\displaystyle \delta (r;C)}$. ${\displaystyle \delta (r;C)=0}$ for all positions ${\displaystyle r\notin C}$. For all positions ${\displaystyle r\in C}$, ${\displaystyle \delta (r;C)}$ is infinite in the direction of ${\displaystyle C}$ in a manner similar to the Dirac delta function. The integral property that must be satisfied by ${\displaystyle \delta (r;C)}$ is that for any oriented surface ${\displaystyle \sigma }$, if ${\displaystyle C}$ passes through ${\displaystyle \sigma }$ in the preferred direction a net total of ${\displaystyle N}$ times, then

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

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

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

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

Given any sequence of closed loops ${\displaystyle C_{1},C_{2},\dots ,C_{k}}$, these loops can be added in a linear fashion to get a "multi-loop" denoted by the vector field ${\displaystyle \delta (r;C_{1})+\delta (r;C_{2})+\dots +\delta (r;C_{k})}$. The multi-loop is denoted by: ${\displaystyle C_{1}+C_{2}+\dots +C_{k}}$.

Most importantly, given any divergence-free vector field ${\displaystyle F}$ that decreases faster than ${\displaystyle o(1/|r|^{2})}$ as ${\displaystyle |r|\rightarrow +\infty }$, then there exists a family ${\displaystyle C[\xi ]}$ of closed loops where ${\displaystyle \xi \in D_{C}}$ is an arbitrary continuous indexing parameter such that ${\displaystyle F(r)=\iint _{\xi \in D_{C}}\delta (r;C[\xi ])d\xi }$. 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 ${\displaystyle \sigma }$, ${\displaystyle \sigma }$ can be characterized by a vector field ${\displaystyle \delta (r;\sigma )}$. ${\displaystyle \delta (r;\sigma )=0}$ for all positions ${\displaystyle r\notin \sigma }$. For all positions ${\displaystyle r\in \sigma }$, ${\displaystyle \delta (r;\sigma )}$ is infinite in the direction of the outwards normal direction to ${\displaystyle \sigma }$ in a manner similar to the Dirac delta function. The integral property that must be satisfied by ${\displaystyle \delta (r;\sigma )}$ is that for any oriented path ${\displaystyle C}$, if ${\displaystyle C}$ passes through ${\displaystyle \sigma }$ in the preferred direction a net total of ${\displaystyle N}$ times, then

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

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

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

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

Given any sequence of surfaces ${\displaystyle \sigma _{1},\sigma _{2},\dots ,\sigma _{k}}$, these surfaces can be added in a linear fashion to get a "multi-surface" denoted by the vector field ${\displaystyle \delta (r;\sigma _{1})+\delta (r;\sigma _{2})+\dots +\delta (r;\sigma _{k})}$. The multi-surface is denoted by: ${\displaystyle \sigma _{1}+\sigma _{2}+\dots +\sigma _{k}}$.

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

Given an oriented surface ${\displaystyle \sigma }$ with a counter-clockwise oriented boundary ${\displaystyle C}$, it is then the case that ${\displaystyle \nabla \times \delta (r;\sigma )=\delta (r;C)}$. Given any vector field ${\displaystyle F}$ that denotes a multi-surface, then ${\displaystyle \nabla \times F}$ is a vector field that denotes the counter-clockwise oriented boundary of the multi-surface denoted by ${\displaystyle F}$. 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 ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ be two oriented closed loops, and let ${\displaystyle \sigma _{1}}$ and ${\displaystyle \sigma _{2}}$ be oriented surfaces whose counter-clockwise boundaries are respectively ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$.

Given a current of ${\displaystyle I}$ flowing around ${\displaystyle C_{1}}$, let ${\displaystyle B_{1}}$ be the magnetic field induced via Ampere's law. Note that ${\displaystyle B_{1}(r)\propto I}$. The magnetic flux through surface ${\displaystyle \sigma _{2}}$ is

${\displaystyle \Phi _{B,2}=\iint _{r\in \sigma _{2}}B_{1}(r)\bullet dA}$ where ${\displaystyle dA}$ is the vector representation of an infinitesimal surface element of ${\displaystyle \sigma _{2}}$.

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

The mutual electromagnetic induction from ${\displaystyle C_{1}}$ to ${\displaystyle C_{2}}$ will be denoted with ${\displaystyle M(C_{1},C_{2})}$

Self Inductance

When ${\displaystyle C_{2}=C_{1}}$, the inductance ${\displaystyle L(C_{1})=M(C_{1},C_{1})}$ is referred to as the "self inductance".

Linearity of Mutual Inductance

Given loops ${\displaystyle C_{1}}$, ${\displaystyle C_{2}}$, and ${\displaystyle C_{3}}$, it is relatively simple to demonstrate that ${\displaystyle M(C_{1}+C_{3},C_{2})=M(C_{1},C_{2})+M(C_{3},C_{2})}$ and ${\displaystyle M(C_{1},C_{2}+C_{3})=M(C_{1},C_{2})+M(C_{1},C_{3})}$.

Let ${\displaystyle B_{1}(r)}$, ${\displaystyle B_{2}(r)}$, and ${\displaystyle B_{3}(r)}$ be the magnetic fields generated when a current of ${\displaystyle I}$ flows through ${\displaystyle C_{1}}$, ${\displaystyle C_{2}}$, or ${\displaystyle C_{3}}$ respectively.

The magnetic field generated by ${\displaystyle C_{1}}$ and ${\displaystyle C_{3}}$ together is ${\displaystyle B_{1}(r)+B_{3}(r)}$ due to the linearity of Maxwell's equations. This leads to ${\displaystyle M(C_{1}+C_{3},C_{2})=M(C_{1},C_{2})+M(C_{3},C_{2})}$.

The flux through ${\displaystyle C_{2}+C_{3}}$ is the sum of the flux through ${\displaystyle C_{2}}$ and ${\displaystyle C_{3}}$ separately. This leads to ${\displaystyle M(C_{1},C_{2}+C_{3})=M(C_{1},C_{2})+M(C_{1},C_{3})}$.

Symmetry of Mutual Inductance

It is the case that given loops ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$, that ${\displaystyle M(C_{1},C_{2})=M(C_{2},C_{1})}$. 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 ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ will be interpreted as multi-surfaces whose boundaries are respectively ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$.

Let there exist a current of ${\displaystyle I}$ in loop ${\displaystyle C_{1}}$, and let ${\displaystyle B_{1}(r)}$ denote the resultant magnetic field. Ampere's law requires that ${\displaystyle \nabla \times B_{1}(r)=\mu _{0}I\delta (r;C_{1})\implies \nabla \times {\frac {1}{\mu _{0}}}{\frac {B_{1}(r)}{I}}=\delta (r;C_{1})}$, and therefore ${\displaystyle {\frac {1}{\mu _{0}}}{\frac {B_{1}(r)}{I}}}$ is a multi-surface whose boundary is ${\displaystyle C_{1}}$. Since ${\displaystyle B_{1}(r)\propto I}$, let ${\displaystyle b_{1}(r)={\frac {B_{1}(r)}{I}}}$.

Given a divergence free vector field ${\displaystyle F}$, the flux of ${\displaystyle F}$ through ${\displaystyle \sigma _{1}}$ is:

${\displaystyle \iint _{r\in \sigma _{1}}F(r)\bullet dA=\iiint _{r\in \mathbb {R} ^{3}}F(r)\bullet \delta (r;\sigma _{1})d\tau =\iiint _{r\in \mathbb {R} ^{3}}F(r)\bullet {\frac {b_{1}(r)}{\mu _{0}}}d\tau }$

The final equality holds due to the fact that ${\displaystyle F}$ is divergence free and that ${\displaystyle \delta (r;\sigma _{1})}$ and ${\displaystyle {\frac {b_{1}(r)}{\mu _{0}}}}$ are multi-surfaces with a common boundary of ${\displaystyle C_{1}}$.

${\displaystyle B_{1}(r)}$ is divergence free. The flux of ${\displaystyle B_{1}(r)}$ through ${\displaystyle \sigma _{2}}$ is:

${\displaystyle \Phi _{B,2}=\iiint _{r\in \mathbb {R} ^{3}}I{\frac {b_{1}(r)\bullet b_{2}(r)}{\mu _{0}}}d\tau }$

Therefore: ${\displaystyle M(C_{1},C_{2})=\iiint _{r\in \mathbb {R} ^{3}}{\frac {b_{1}(r)\bullet b_{2}(r)}{\mu _{0}}}d\tau }$ from which the symmetry ${\displaystyle M(C_{1},C_{2})=M(C_{2},C_{1})}$ is now apparent.

Calculating the Mutual Inductance

Approach #1 (use the vector potential)

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

Using the vector identity:

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

Ampere's law becomes:

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

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

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

The vector potential generated by a current of ${\displaystyle I}$ flowing through closed loop ${\displaystyle C_{1}}$ is: ${\displaystyle A_{1}(r)={\frac {\mu _{0}}{4\pi }}\iiint _{r_{1}\in \mathbb {R} ^{3}}{\frac {I\delta (r_{1};C_{1})}{|r-r_{1}|}}d\tau _{1}={\frac {\mu _{0}}{4\pi }}\int _{r_{1}\in C_{1}}{\frac {I}{|r-r_{1}|}}dr_{1}}$

The magnetic field generated by a current of ${\displaystyle I}$ flowing through closed loop ${\displaystyle C_{1}}$ is: ${\displaystyle B_{1}=\nabla \times A_{1}}$. The flux through surface ${\displaystyle \sigma _{2}}$ (which is counter-clockwise bounded by ${\displaystyle C_{2}}$), is

${\displaystyle \Phi _{B,2}=\iint _{r_{2}\in \sigma _{2}}B_{1}(r_{2})\bullet dA_{2}=\iint _{r_{2}\in \sigma _{2}}(\nabla \times A_{1})(r_{2})\bullet dA_{2}=\int _{r_{2}\in C_{2}}A_{1}(r_{2})\bullet dr_{2}}$ via Stoke's theorem.

${\displaystyle \Phi _{B,2}={\frac {\mu _{0}}{4\pi }}\int _{r_{2}\in C_{2}}\int _{r_{1}\in C_{1}}{\frac {I}{|r_{2}-r_{1}|}}(dr_{1}\bullet dr_{2})}$ so the mutual inductance is: ${\displaystyle M(C_{1},C_{2})={\frac {\Phi _{B,2}}{I}}={\frac {\mu _{0}}{4\pi }}\int _{r_{2}\in C_{2}}\int _{r_{1}\in C_{1}}{\frac {dr_{1}\bullet dr_{2}}{|r_{2}-r_{1}|}}}$

This equation is known as "Neumann formula" [1].

It can also be seen from this expression that the mutual inductance is symmetric: ${\displaystyle M(C_{1},C_{2})=M(C_{2},C_{1})}$.

Approach #2 (use linearity and loop dipoles)

Given any closed loop ${\displaystyle C}$, let ${\displaystyle \sigma }$ be an oriented surface that has ${\displaystyle C}$ as its counterclockwise boundary. For each infinitesimal area vector element ${\displaystyle dA}$ of ${\displaystyle \sigma }$, let the infinitesimal ${\displaystyle \partial (dA)}$ be an infinitesimal closed loop that is the counterclockwise boundary of ${\displaystyle dA}$. It is then the case that ${\displaystyle C=\int _{r\in \sigma }\partial (dA)}$.

The linearity of the mutual inductance gives:

${\displaystyle M(C_{1},C_{2})=M\left(\iint _{r_{1}\in \sigma _{1}}\partial (dA_{1}),\iint _{r_{2}\in \sigma _{2}}\partial (dA_{2})\right)=\iint _{r_{1}\in \sigma _{1}}\iint _{r_{2}\in \sigma _{2}}M(\partial (dA_{1}),\partial (dA_{2}))}$

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

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

${\displaystyle B(r)={\frac {\mu _{0}}{4\pi }}\left({\frac {3(P\bullet r)r}{|r|^{5}}}-{\frac {P}{|r|^{3}}}\right)}$

Let ${\displaystyle a_{1}}$ and ${\displaystyle a_{2}}$ be area vectors of the interiors of two infinitesimal loops, with the second loop displaced from the first by ${\displaystyle r}$. Let a current ${\displaystyle I}$ flow around the boundary of ${\displaystyle a_{1}}$ in a counter clockwise manner forming the dipole ${\displaystyle Ia_{1}}$. The flux of the magnetic field generated by ${\displaystyle Ia_{1}}$ through ${\displaystyle a_{2}}$ is:

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

Therefore if ${\displaystyle c_{1}}$ and ${\displaystyle c_{2}}$ are the counter clockwise boundaries of ${\displaystyle a_{1}}$ and ${\displaystyle a_{2}}$:

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

Returning to computing the mutual inductance between ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ gives:

${\displaystyle M(C_{1},C_{2})={\frac {\mu _{0}}{4\pi }}\iint _{r_{1}\in \sigma _{1}}\iint _{r_{2}\in \sigma _{2}}\left({\frac {3(dA_{1}\bullet (r_{2}-r_{1}))(dA_{2}\bullet (r_{2}-r_{1}))}{|r_{2}-r_{1}|^{5}}}-{\frac {dA_{1}\bullet dA_{2}}{|r_{2}-r_{1}|^{3}}}\right)}$

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

1. Griffiths, D. J., Introduction to Electrodynamics, 3rd edition, Prentice Hall, 1999.