]> Flux linkage in a radially thick coil

Flux linkage in a radially thick coil

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Figure RTZ: A loop of wire, radius r, sitting in a homogenous magnetic field 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 -

λ = π r^{2} B

Equation RTD

This result you may easily extend to the case of a 'single layer' solenoid coil by multiplying by N.

Figure RTY: A radially thick multilayer solenoid 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.

Figure RTX: An Archimedean spiral 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 = m φ + r_{1}

Equation RTA

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

m = \frac{r_{2} - r_{1}}{2 π}

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.

ⅆ A = \frac{1}{2} r (r \cdot ⅆ φ) = \frac{1}{2} r^{2} ⅆ φ

Equation RTC

Substituting equation RTA -

ⅆ A = \frac{1}{2} {(m φ + r_{1})}^{2} ⅆ φ

Equation RTE

= \frac{1}{2} (m^{2} φ^{2} + 2 m r_{1} φ + r_{1}^{2}) ⅆ φ

Equation RTF

A = \frac{1}{2} \int_{0}^{2 π} m^{2} φ^{2} + 2 m r_{1} φ + r_{1}^{2} ⅆ φ

Equation RTG

= \frac{1}{2} {[\frac{m^{2} φ^{3}}{3} + m r_{1} φ^{2} + r_{1}^{2} φ]}_{0}^{2 π}

Equation RTH

= \frac{4 π^{3} m^{2}}{3} + 2 π^{2} m r_{1} + π r_{1}^{2}

Equation RTI

Substituting equation RTB -

= \frac{π {(r_{2} - r_{1})}^{2}}{3} + π (r_{2} - r_{1}) r_{1} + π r_{1}^{2}

Equation RTJ

A = \frac{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 = \frac{π}{3} (11^{2} + 11 \times 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%.

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E-mail:R.Clarke@surrey.ac.uk
Last modified: 2008 May 11th.