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Effective radius of a single-layer solenoid.
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1. Introduction 2. Homogeneous rectangular wire. 3. Strained rectangular wire. |
4 Homogeneous round wire. 5. Strained round wire. 99. Discussion. |
| Abstract: Due to path-length variation and strain, the current distribution in the wire of a solenoid inductor is not uniform at low frequencies. This causes the effective solenoid radius to differ from the physical average, leading to a small systematic error in inductance calculations. Formulae for the correct effective radius, for coils wound with solid rectangular and round wire, are derived. |
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1. Introduction. As has been pointed out by Fraga et al. [15]; the current distribution in the wire of a solenoid inductor is not uniform at low frequencies, because the length of the conduction path is greater on the outside of the coil than it is on the inside. To this observation we can also add; that the act of winding the wire into a helix causes the metal on the outside to stretch, thereby giving rise to a resistivity gradient. These effects are obviously minor when the diameter of the coil is large relative to the diameter of the wire, but can become significant as the curvature of the wire increases. The overall effect of strain and curvature is to concentrate the current in that part of the wire closest to the inside of the coil. This is a purely resistive phenomenon, as distinct from the complex skin and proximity effects which occur at high frequencies; but the consequence, as in the high-frequency case, is that the diameter of the equivalent current sheet is smaller than the average diameter of the coil. This causes a systematic error when inductance calculations are based on the diameter taken from wire-centre to wire-centre (exacerbated because inductance is proportional to the solenoid radius squared), and may explain why such calculations tend to come out a little too high. In this article we will derive the equivalent current-sheet radius (r0) for solenoids wound with rectangular and round wire. In both instances, we will consider the case when the wire is homogeneous (i.e., of uniform resistivity throughout its cross-section) and the case when the wire is strained (i.e., has resistivity which increases with distance from the coil axis). The only mathematical axiom for this study is that r0 is defined as: 'that distance, measured perpendicular to the coil axis, at which the current flowing on the outside is equal to the current flowing within'. |
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2. Homogeneous rectangular conductor. The diagram below represents the cross-section of a rectangular wire forming part of a solenoid of average conduction radius ra. The radial thickness of the wire is defined as 2rw (to be consistent with later analysis of round wire coils) and its axial thickness is w. The radius co-ordinate is r, and the current flowing in the solenoid is distributed over the range: ra-rw £ r £ ra+rw . |
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Now consider a slice of conductor,
of infinitesimal thickness dr, parallel
to the coil axis. The resistance per turn of this slice is: Rt(slice) = r  t(slice)
/ ( w dr )where r is the bulk resistivity, and t(slice)
is the length per turn. For coils of small pitch angle, the turn
length is approximately equal to the solenoid circumference at
radius r, i.e.:t(slice)
= 2prHence: Rt(slice) = r 2pr / ( w dr ) The current flowing in the slice is inversely proportional to its resistance, i.e., after dropping all constants from the proportionality relationship: I(slice) µ (1/r)dr |
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| The total current flowing in the solenoid is the sum of the currents flowing in the set of all infinitesimal slices of equal thickness which can exist in the range ra-rw £ r £ ra+rw . Thus: |
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ra+rw |
r |
| Now to find the equivalent current sheet radius r0; we note that the current flowing outside r0 is equal to the current flowing inside r0. Hence: |
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r0 |
r |
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ra+rw |
r |
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ò |
r |
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i.e.: ln(r0) - ln(ra-rw) = ln(ra+rw) - ln(r0) Combining the logarithms: ln[ r0/(ra-rw) ] = ln[ (ra+rw)/r0 ] and taking the antilog. of both sides: r0/(ra-rw) = (ra+rw)/r0 i.e.: r0² = (ra+rw)(ra-rw) = ra² - rw² Rearranging this to express the difference between r0 and ra as a correction factor gives: |
| r0 = ra Ö[1 - (rw/ra)²] |
Homogeneous rectangular wire. 2pra >> p |
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Note that in deriving the expression above, the length of the
conductor at radius r was approximated as the circumference of
the solenoid at that radius. Such an approximation is justifiable
in most inductor modeling situations, but it is interesting to
consider its effect when rw = ra.
It is impossible to wind wire on a coil-former in such a way
that the average solenoid radius is equal to half the radial
wire thickness, because to do so would require a former of zero
radius. Such structures can however be obtained by twisting two
wires together and then removing one of the wires, and they can
be machined out of solid rod; i.e, the situation ra
= rw represents the transition from coils
to screws. That being the case, we may observe that equation
(2.1) has an incorrect boundary
condition; because it predicts r0 = 0
when ra = rw, whereas
(say) a machine screw, due to its surface convolution, must have
greater inductance than a straight rod. The false boundary condition arises because the model allows the conduction path, and hence the resistance, to go to zero at the coil axis when ra=rw. This does not happen in reality, because a solenoid has finite length. For every turn around the former, the wire advances along the axis by a distance p, where p is known as the 'pitch distance'. Hence an exact expression for the length per turn of the conduction path at radius r is: t(slice)
= Ö[(2pr)²
+ p²]The resistance per turn of an infinitesimal slice parallel to the axis is now: Rt(slice) = r {Ö[(2pr)² + p²] } / ( w dr ) To simplify the working from now on, let us define a new pitch parameter: q = p / 2p Thus: Rt(slice) = r 2p[Ö(r² + q²)] / ( w dr ) The current in the slice is inversely proportional to the resistance; and so, dropping all constant factors: I(slice) µ dr / Ö(r² + q²) The equivalent current sheet radius is now defined by the relationship: |
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r0 |
Ö(r² + q²) |
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ra+rw |
Ö(r² + q²) |
where |
ò |
Ö(r² + q²) |
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ln[r + Ö(r² + q²)] + c |
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Ryshik & Gradstein [17]
sect. 2.27 (p83), 2.271, 4 (p84) also: sect. 2.26 (p79), 2.261 |
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Hence: ln[r0+Ö(r0²+q²)] - ln[ra-rw+Ö{(ra-rw)²+q²}] = ln[ra+rw+Ö{(ra+rw)²+q²}] - ln[r0+Ö(r0²+q²)] Rearranging and combining logarithms (bearing in mind that doubling a logarithm is equivalent to squaring its argument), then taking the antilog. of both sides gives: [r0+Ö(r0²+q²)]² = [ra+rw+Ö{(ra+rw)²+q²}] [ra-rw+Ö{(ra-rw)²+q²}] i.e.: r0+Ö(r0²+q²) = Ö[(ra+rw+Ö{(ra+rw)²+q²}) (ra-rw+Ö{(ra-rw)²+q²})] The right hand side of this expression is composed entirely of constants (i.e., input parameters). Thus, if we define: a² = [ra+rw+Ö{(ra+rw)²+q²}] [ra-rw+Ö{(ra-rw)²+q²}] then r0+Ö(r0²+q²) = a Subtracting r0 from each side and then squaring gives: r0²+q² = (a - r0)² Expanding the right hand side gives: r0²+q² = a² -2a r0 +r0² i.e.: r0 = (a² - q²) / ( 2a ) Hence: |
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| The effect of pitch on r0 is shown in the graph below, where coil parameters are expressed in units of rw [Spreadsheet calculation: r0_hrectw.ods]. Note that the data have been plotted on a logarithmic ra/rw scale to expand the region where pitch is significant. |
| For pitch values up to 8rw, equation (2.2) is convergent with (2.1) to within 0.07% for ra>6rw. In practical terms, ra=6rw might correspond to (say) a wire of 2mm radial thickness wound onto a 10mm diameter former, i.e., a tightly wound coil. The actual error in r0 incurred by using the approximation (2.1) instead the exact expression (2.2) is plotted below. |
| Note that, unless machined out of solid bar, the coils for which equation (2.1) is inadequate cannot be constructed without deforming the wire considerably. It follows that, for conventional solenoids, neither model is strictly valid; and accuracy in calculating the inductance of tightly wound coils will be improved by taking the effect of wire strain into account. |
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3. Strained rectangular conductor. When wire is wound into a coil, the strain induced by so doing will give rise to a resistivity gradient in the conductor cross-section. That this is realistic is evident from the resistivity data for copper in the CRC Handbook [16], which gives a figure of 17.241 nWm for annealed Cu and 17.71 nWm for hard-drawn Cu (both measurements at 20°C). In order to model the strain effect, we will assume that the increase in resistivity is proportional to the relative elongation. Referring again to the diagram in section 2; recall that when 2pra >> p, the resistance per turn of an infinitesimal slice is given by: Rt(slice) = r(slice) 2pr / ( w dr ) If we take the resistivity at the average radius ra to be r; then, bearing in mind that the elongation is proportional to the circumference, the resistivity at radius r is: r(slice) = r r / ra The choice of the average radius as the resistivity reference point is, incidentally, arbitrary and inconsequential. In reality, if we assume that the metal is effectively incompressible, then the material closest to the inside of the coil will have resistivity unchanged from its original value, and everything outside that radius will have been stretched. Whether we use ra or ra-rw as the reference radius however makes no difference, because either choice produces a constant parameter which will cancel from the integral equation which defines r0. Hence, sticking with ra as the reference: Rt(slice) = r 2pr² / ( ra w dr ) As before, the current in the slice is inversely proportional to its resistance; and so, dropping constants: I(slice) µ (1/r²)dr Hence, for p=0 or 2pra>>p, the equivalent current sheet radius can be obtained from the relationship: |
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r0 |
r² |
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ra+rw |
r² |
| In this case, integration is simply the reverse of differentiation, and so: |
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r0 |
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( ra-rw ) |
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( ra+rw ) |
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r0 |
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r0 |
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( ra+rw ) |
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( ra-rw ) |
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( ra+rw )( ra-rw ) |
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( ra² - rw² ) |
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Taking the reciprocal of both sides gives: r0 = ( ra² - rw² ) / ra And by factoring ra² from the numerator we get: |
| r0 = ra [1 - (rw/ra)²] |
Strained rectangular wire. 2pra >> p |
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As expected, the effective current sheet radius is reduced as
a consequence of strain, the difference between equation (3.1) and equation (2.1)
being the loss of a square-root symbol operating on the radius
correction factor. Equation (3.1), of course, also has a false boundary condition at ra = rw; and we can correct that, as before, by taking the length per turn at radius r as Ö[(2pr)² + p²]. Thus: Rt(slice) = r 2p {Ö[r² + q²]} r / ( ra w dr ) where q= p/2p. Hence: I(slice) µ dr / { r Ö[r² + q²] } and |
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r0 |
r Ö(r² + q²) |
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ra+rw |
r Ö(r² + q²) |
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where |
ò |
r Ö(r² + q²) |
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-(1/q) ln[{q + Ö(r² + q²)}/r ] + c |
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Ryshik & Gradstein. [17]
sect. 2.26 (p79), 2.266 (p81) also: sect. 2.27 (p83), 2.275, 4 (p85) |
| In this case, there is no solution for q=0 (because 1/q®¥ and ln[1]=0 ) but we already have that from equation (3.1). Hence, for q>0; |
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r0 |
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ra-rw |
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ra+rw |
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| Now let us define: |
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ra-rw |
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ra+rw |
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Which gives: q + Ö(r0² + q²) = b r0 Moving q to the right-hand side and squaring gives: r0² + q² = (b r0 - q)² and expanding gives: r0² + q² = ( b r0 )² -2q b r0 + q² i.e.: r0² (1 - b²) + 2q b r0 = 0 and hence: r0 [ r0 + 2q b / (1 - b²) ] = 0 The solutions are r0 =0 (but, as we have already deduced, not when q>0) or: r0 = -2q b / (1 - b²) which, after multiplying numerator and denominator by -1, gives: |
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| The effect of pitch on r0 for the strained wire case is shown below, with coil parameters expressed in units of rw as before [Spreadsheet calculation: r0_strectw.ods]. |
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In the previous section (homogeneous rectangular wire), a graph
showing the error incurred by ignoring pitch was shown. The analogous
graph for the present case turns out to be identical; an observation
which leads to a formula which is still defined when ra=rw. Using the following nomenclature: r0(h) = Effective radius, homogeneous wire, neglecting pitch (equation 2.1) r0(h, p) = Effective radius, homogeneous wire, including pitch (equation 2.2) r0(s) = Effective radius, strained wire, neglecting pitch (equation 3.1) r0(s, p) = Effective radius, strained wire, including pitch (equation 3.2) The identical error relationships for the strained and unstrained cases impy that: r0(s, p) / r0(s) = r0(h, p) / r0(h) i.e.: |
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r0(h) |
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ra Ö[1 - (rw/ra)²] |
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which, bearing in mind that any quantity is the square of its
own square root, gives: r0(s, p) = r0(h, p) Ö[1 - (rw/ra)²] Hence: |
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Algebraic proof of the relationship (3.3)
is likely to be laborious and has not been attempted, but the
formula is verified numerically in the spreadsheet r0_strectw2.ods.
The only difference between (3.2)
and (3.3) is that the latter
returns a valid result when ra=rw. What the formula returns when ra=rw is, incidentally, that r0=0. Such a result was noted earlier to constitute a false boundary condition for the homogeneous wire case, but it is mathematically valid when the wire is strained. The reason is that, in order for the inside surface of the conductor to lie on the axis, any point within the conductor at finite r must lie in a region which has undergone infinite extension. In other words, according to the model; the effective solenoid radius is zero when ra=rw, because all parts of the conductor not lying on the axis must have been stretched from zero to finite length (i.e., destroyed) in the act of winding the coil. The model is, of course, ideosyncratic. It assumes that stress can only be relieved by strain in a direction exactly perpendicular to the coil axis, and that infinitesimal slices of conductor parallel to the axis always have constant resistivity. This is a reasonable assumption when ra>>rw, but when the solenoid diameter is small, the strain will also be relieved by twisting. Consider, for example, the act of winding wire around a pin, with a pitch considerably greater than the radial wire thickness. Once the pin is removed, an additional twist can be applied to bring ra=rw, but the additional strain is centred (i.e., zero) within body of the wire rather than at the coil axis. Hence the model, despite having the correct boundary condition, assumes an unrealistic pattern of deformation at ra=rw. For coils of realtively small diameter, the actual effective radius will lie somewhere between the values given by equations (2.2) and (3.3). For practical radio coils however, where the objective is (or should be) to obtain the maximum inductance with the minimum amount of wire, equation (3.3) is a reasonable compromise between accuracy and mathematical tractability. |
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4. Homogeneous round conductor. Deriving expressions for the effective radius of a round wire solenoid is more difficult than it is for the rectangular wire case. The resulting integrals do however have analytical solutions; and although those solutions do not lead to closed-form expressions for r0, they nevertheless allow r0 to be evaluated to arbitrary precision. A little numerical investigation then leads to surprisingly simple and accurate approximations. Starting with the homogeneous case; consider (as before) an infinitesimal slice of conductor parallel to the solenoid axis. The resistance per turn of this slice, for the case when 2pra >> p, is : |
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Rt(slice) = r
2pr / ( w dr
) For a round wire however, the relationship is complicated by the fact that the width of the slice (w) varies as a function of r. Using Pythagoras' theorem: (w/2)² = rw² - (r-ra)² i.e.: w = 2 Ö[rw² - (r-ra)²] Hence: Rt(slice) = rpr / (Ö[rw² - (r-ra)²] dr ) The current in the slice is inversely proportional to the resistance; and so, dropping all constant factors: I(slice) µ (1/r) Ö[rw² - (r-ra)²] dr The equivalent current sheet radius is now defined in the relationship: |
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r0 |
r |
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ra+rw |
r |
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The piecewise solution for this integral is given by Ryshik and
Gradstein [17] (section
2.26); but in order to apply it, we need to put the problem into
standard form according to the conventions of that source. Hence
we start by defining a quadratic polynomial: s(r) = a + br + cr² = rw² - (r-ra)² = rw² - ra² + 2ra r - r² Thus: a = -(ra²-rw²) < 0 b = 2ra c = -1 < 0 and the determinant is: -D = b² - 4ac = 4ra² - 4(ra²-rw²) = 4rw² > 0 The solution is as follows: |
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| Now restoring the original parameters: |
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r |
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rw r |
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rw |
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This solution can now be applied to equation (4.1),
but it will save a considerable amount of confusion of we draw
up a table showing the values of the three tems at the upper
and lower limits r=ra+rw
and r=ra-rw. Simplification
of the middle term requires the substitution: ra²-rw² = (ra+rw)(ra-rw) |
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| Using the table; it can be seen that the integral evaluated at the lower limit on the left hand side of (4.1) is equal to the integral evaluated at the upper limit on the right, and complete cancellation occurs. All that remains is the integral evaluated at r0 on the left, and the negative of the same quantity on the right. Hence, moving everything to the left side and dividing by 2: |
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rw r0 |
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rw |
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There is no closed-form analytical solution for r0
in this case, because instances of r0
occur both within and outside the arguments of the arcsines.
Solutions can nevertheless be obtained to arbitrary precision
using an iterative method. A suitable technique in that case
is to select a particular instance of r0
as the definitive one and rearrange the equation to separate
it. All other instances are then replaced with an approximation
r0(prev) (i.e., 'previous'). Assuming
that recalculation is automatic after a parameter change (as
in suitably configured spreadsheets) r0(prev)
is adjusted until r0=r0(prev)
to the required degree of precision. When the iteration is incorporated
into a computer program, it is necessary to calculate the derivative
¶r0/¶r0(prev) in
order to estimate the amount by which r0(prev)
should be shifted to achieve convergence. There are also issues
regarding whether or not convergence will actually occur. When
using a spreadsheet however, it is sufficient to alter r0(prev) manually until it agrees with r0 to the required number of significant figures. Before anything can be done with the equation above however, it must be renormalised. The reason is that, in separating out the argument of an arcsine function, a sine function with everything else inside it will be obtained. On examining the expression it will be noted that the arcsine arguments are dimensionless, and so too must be the arguments of any sine functions produced (neglecting radians, which are ratiometric rather than physical). As it stands, the equation has dimensions of length, and so it must be divided throughout by a length. In the previous sections, it was found useful to express all parameters in units of rw, and the same choice is suitable here. Hence, dividing throughout by rw and moving the last term to the right hand side, we get: |
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rw² |
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rw² |
-1}]Arcsin |
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rw r0 |
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rw |
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rw |
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Now any one of the three major terms can be transposed in order
to extract an instance of r0, and it remains
to determine the best choice. The solution, in this case, was
found by the time-honoured method of trial and error; and it
transpires that the term on the right is the important one. Thus,
with that foreknowledge, we can define two quasi-constants: u(r0(prev)) = Ö[1 - (r0(prev)-ra)²/rw²] and v(r0(prev)) = [Ö{ (ra²/rw²)-1] Arcsin[ { ra r0(prev)-(ra²-rw²) }/(rw r0(prev) ) ] so that: |
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rw |
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rw |
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Hence: (ra - r0) / rw = Sin[ (rw/ra) ( u + v) ] the argument of the sine function being in radians. The best iteration formula for r0 is therefore: r0 = ra - rw Sin[ (rw/ra) ( u + v) ] This formula was used in the spreadsheet calculation: r0_hrndw.ods , not only to obtain a set of solutions for r0/ ra vs ra/rw, but also to evaluate the three major terms. The latter operation produces the most important result of all, which is that: (ra/rw) Arcsin[ (ra - r0) / rw ] » 1 i.e: u + v » 1 Hence: |
| r0 = ra - rw Sin(rw/ra) |
Homogeneous round wire. ra/rw >4 , 2pra >> p , rw/ra in radians |
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| The exact and the approximate solutions are shown plotted below, with a graph of the error incurred by using equation (4.2) below that. The two calculation methods are in agreement within 0.07% for ra>4rw; a convenience which makes the iterative method redundant for most purposes. |
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5. Strained round conductor. The principal effect of straining a round wire to make it follow a coil former, is of course, to produce a resistivity gradient as it did for the rectangular wire case. This introduces an r² dependence in the denominator of the current integral as before. Hence, the effective solenoid radius for the strained round wire case is captured in the relationship: |
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r0 |
r² |
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ra+rw |
r² |
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Once again, the solution is given by Ryshik and Gradstein, and
the polynomial inside the square-root bracket is put into standard
form as before: s(r) = a + br + cr² = rw² - (r-ra)² = rw² - ra² + 2ra r - r² a = -(ra²-rw²) < 0 b = 2ra c = -1 < 0 -D = b² - 4ac = 4rw² > 0 The solution is: |
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| Restoring the original parameters gives: |
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r² |
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r |
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Ö(ra²-rw²) |
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rw r |
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rw |
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| In applying this solution to equation (5.1), we may observe that all of the constant terms cancel, as they did in the homogeneous wire case discussed previously. That leaves the sum of terms containing r0 to be equal to zero, as before. Thus: |
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r0 |
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Ö(ra²-rw²) |
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rw r0 |
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rw |
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Renormalisation is not necessary because all of the terms are
dimensionless. Iteration for exact solutions for r0
is however still required. In this case, it transpires that the
middle term is the most important one; and so, defining: y(r0(prev)) = {Ö[rw² - (r0(prev)-ra)²]}/ r0(prev) and z(r0(prev)) = Arcsin[ (ra - r0(prev)) / rw ] we obtain the preferred iteration formula: |
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Ö(ra²-rw²) |
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rw r0 |
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This formula is evaluated numerically in the spreadsheet: r0_strndw.ods . The calculation
reveals that: z - y » 0 i.e.: |
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Ö(ra²-rw²) |
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rw r0 |
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» 0 |
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Now notice that the factor before the arcsine is always >1,
so it is the arcsine itself which is small. We are therefore
led to the remarkable conclusion that: ra r0-(ra²-rw²) » 0 i.e., to a good approximation: r0 = ra - rw²/ra Hence, the effective current sheet radius (neglecting pitch) is approximately the same for strained round and rectangular wires (cf. eqation (3.1) ); i.e.: |
| r0 = ra [1 - (rw/ra)²] |
Strained round or rectangular wire. ra/rw >4 , 2pra >> p |
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| Comparison between the exact round wire calculation and the approximation is given in the graphs below. The error in equation (5.2) is less than 0.07% for ra>4rw. The error is also positive, providing some compensation for the neglect of pitch. |
