# 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.

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