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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,
$\phi $, -

$r=m\phi +{r}_{1}$

Equation RTA

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

$m=\frac{{r}_{2}-{r}_{1}}{2\pi}$

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=\frac{1}{2}r\left(r\cdot d\phi \right)=\frac{1}{2}{r}^{2}d\phi $

Equation RTC

Substituting equation RTA -

$dA=\frac{1}{2}{\left(m\phi +{r}_{1}\right)}^{2}d\phi $

Equation RTE

$=\frac{1}{2}\left({m}^{2}{\phi}^{2}+2m{r}_{1}\phi +{r}_{1}^{2}\right)d\phi $

Equation RTF

$A=\frac{1}{2}{\int}_{0}^{2\pi}{m}^{2}{\phi}^{2}+2m{r}_{1}\phi +{r}_{1}^{2}d\phi $

Equation RTG

$=\frac{1}{2}{\left[\frac{{m}^{2}{\phi}^{3}}{3}+m{r}_{1}{\phi}^{2}+{r}_{1}^{2}\phi \right]}_{0}^{2\pi}$

Equation RTH

$=\frac{4{\pi}^{3}{m}^{2}}{3}+2{\pi}^{2}m{r}_{1}+\pi {r}_{1}^{2}$

Equation RTI

Substituting equation RTB -

$=\frac{\pi {\left({r}_{2}-{r}_{1}\right)}^{2}}{3}+\pi \left({r}_{2}-{r}_{1}\right){r}_{1}+\pi {r}_{1}^{2}$

Equation RTJ

$A=\frac{\pi}{3}\left({r}_{2}^{2}+{r}_{2}{r}_{1}+{r}_{1}^{2}\right)$

Equation RTK

For the degenerate case of
${r}_{1}={r}_{2}$ this gives $A=\pi {r}^{2}$

Taking R_{1} = 5 mm and R_{2} = 11 mm (as figure RTX)
you have

$A=\frac{\pi}{3}\left({11}^{2}+11\times 5+{5}^{2}\right)=210$
mm^{2}

Equation RTL

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

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

Last modified: 2008 May 11th.