 To calculate the flux linkage in a plain, circular loop of wire immersed in a homogenous magnetic field (figure RTZ) you take as the area a value of π times the square of the loop radius, r -

$\lambda =\pi {r}^{2}B$
Equation RTD

This result you may easily extend to the case of a 'single layer' solenoid coil by multiplying by N. For a 'multi layer' coil the situation is more complicated because turns near the outside of the winding have a greater area and will link more flux than turns on the inside. A little thought suggests that the geometry is closely represented by an Archimedean spiral, in which the radius, $r$, increases linearly as a function of azimuthal angle, $φ$, -

$r=mφ+ r1$
Equation RTA

where $r 1$ is the radius of the inside of the coil, and

$m= r 2 - r 1 2π$
Equation RTB

where $r 2$ is the radius of the outside of the coil.

You could analyze a 100 turn coil by integrating the area, A, swept out by the radius vector over the interval φ = 0 to φ = 200 π. However, it is simpler to consider the area of a one turn (0 to 2π) spiral and then multiply by N afterwards.

$ⅆA= 1 2 r( r⋅ⅆφ )= 1 2 r 2 ⅆφ$
Equation RTC

Substituting equation RTA -

$ⅆA= 1 2 ( mφ+ r 1 ) 2 ⅆφ$
Equation RTE

$= 1 2 ⁢ m 2 φ 2 +2m r 1 φ+ r 1 2 ⅆφ$
Equation RTF

$A= 1 2 ∫ 0 2π m 2 φ 2 +2m r 1 φ+ r 1 2 ⅆφ$
Equation RTG

$= 1 2 [ m 2 φ 3 3 +m r 1 φ 2 + r 1 2 φ ] 0 2π$
Equation RTH

$= 4 π 3 m 2 3 +2 π 2 m r 1 +π r 1 2$
Equation RTI

Substituting equation RTB -

$= π ( r 2 - r 1 ) 2 3 +π( r 2 - r 1 ) r 1 +π r 1 2$
Equation RTJ

$A= π 3 ( r 2 2 + r 2 r 1 + r 1 2 )$
Equation RTK

For the degenerate case of $r 1 = r 2$ this gives $A=π r 2$

## Example

Taking R1 = 5 mm and R2 = 11 mm (as figure RTX) you have

$A= π 3 ( 11 2 +11×5+ 5 2 )=210$ mm2
Equation RTL

Had you taken the arithmetic mean radius ((5+11)/2 = 8 mm) then π×82 gives 201 mm2 : an error of -4.3%.

[ Producing wound components] [Air coils index]

E-mail:R.Clarke@surrey.ac.uk