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# With the expression of a magnetic flux, given by the surface integral equation

=
B
dS

where dS is the infinitesimal element of the surface. The magnetic flux can also be
described with self inductance, L, or mutual inductance, M, with these equations.
=LI
=MI

where ,I, is the magnitude of the electric current in the problem. So, the total magnetic

fluxes, a and b can be related to the induction coefficients through the matrix
equation.
a = Laa M I a
M Lbb I b
b

()(
Within this equation,

Laa

and

Lbb

)( )

## are the self inductions of the solenoids A and B,

with M being their mutual inductance. Each solenoid will receive two kinds of
contributions to the total magnetic flux. One will be from to its own magnetic field, and the
other will be from the other solenoid's magnetic flux. While this would raise some
problems that would cause problems in common textbooks, because they use the central
axis, even though the cross section is circular, this is no problem when expressed in this
manner. These irregularities when calculating the inductance matrix can be explained that
the magnetic field is even throughout the solenoid and not just the central axis, which
textbooks do not take into account. Other studies have shown that with the help of other
equations (the surface integral equation, the two magnetic flux equations and the matrix
equation) and their manipulation will lead to the coupling coefficient. However, the
coefficient can only be described as

a
b

## only when two cross sections are circular.

ab ba
M
a
=
=
a b
L
L
aa bb b
k =

This equation is gained when common assumptions are used. Some of those assumptions
are the shape of the cross-section which has been proven not to matter as the magnetic field
does not depend on the cross section shape. The calculus included in these equations are not
hard to follow, and can be included in lower course levels textbooks. To properly define the
problem, we consider two ideal solenoids where the only requirement is for one to be
Sa
Sb
S >S
Na
Nb
smaller than the other, specifically
and
( b b ,
and
, the
Ia

magnetic fluxes,

and

and

Ib

## currents through the matrix equation. They can be solved as:

L
( aa) ( I a ) + M ( I b )
a =
L
( bb) ( I b ) + M ( I a )
b =

If

I a=0

Lbb

## , since the electric

current only flows through the outer coil, and the inner zones of the solenoids, the modulus
of the magnetic field is B=

0 N b I b
h

where

a=N a Sa

n=N b /h

0 N b I b
=M I b
h

b=N b Sb

0 N b I b
=Lbb I b
h
Lbb

, can be

## rearranged and solved and turned into:

M=

0 Sa N a I b
h

0 S b N 2b
Lbb=
h
The equation is the same if one wanted to find

Laa

I b=0.

The final

## results through the inductance matrix that

L=

0 N 2b S a N a S a
h N a S a N 2b S a

k=

Sa
M
=
Laa Lbb S a S b

Sa

and

Sb

respectively.
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