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Publication Number: FHWAHRT06139
Date: October 2006


Traffic Detector Handbook:Third Edition—Volume II
APPENDIX B.CURRENT SHEET FORMULA FOR CALCULATION OF LOOP INDUCTANCE
A formula for calculating the inductance produced by a rectangular current sheet is given by^{(1)}

(B1) 
where

(B2) 

(B3) 
N = number of turns and 
α_{1} = length of shorter side of loop. 
Inductive loops are modeled as very short solenoids. 
The term, “length of current sheet,” is defined as the axial length of a coil or solenoid. For an inductive loop, the length of the current sheet is equivalent to the height of the wires in the slot, as described in the sample calculation that follows. Since the length of the current sheet is very small compared to the longer side of the rectangle, γ is also very small. The factor F’ adjusts for the fact that inductive loops perform as very short solenoids (i.e., the loop area is much greater than the turn spacing).
The values of β are obtained from Table B1. The factor K that appears in the table is given by

(B4) 
Table B1. Values of β coefficients for short rectangular solenoids.^{(2)}
K  β_{1}  β_{1}’  β_{2}  β_{3}  β_{5}  β_{7} 
1.00 
0.4622  0.6366  0.2122  −0.0046  0.0046  −0.0382 
0.95 
0.4574  0.6534  0.2234  −0.0046  0.0053  
0.90 
0.4512  0.6720  0.2358  −0.0046  0.0064  −0.0525 
0.85 
0.4448  0.6928  0.2496  −0.0042  0.0080  
0.80 
0.4364  0.7162  0.2653  −0.0031  0.0103  −0.0831 
0.75 
0.4260  0.7427  0.2829  −0.0010  0.0141  
0.70 
0.4132  0.7730  0.3032  0.0026  0.0198  −0.1564 
0.65 
0.3971  0.8080  0.3265  0.0085  0.0291  
0.60 
0.3767  0.8488  0.3537  0.0179  0.0432  −0.3372 
0.55 
0.3500  0.8970  0.3858  0.0331  0.0711  
0.50 
0.3151  0.9549  0.4244  0.0578  0.1183  −0.7855 
0.40 
0.1836  1.1141  0.5305  0.1697  0.3898  −2.403 
0.30 
0.0314  1.3359  0.7074  0.5433  2.0517  −7.85 
0.20 
−0.6409  1.9099  1.0610  2.3230  14.507  15.51 
0.10 
−3.2309  3.5014  2.1220  22.5480  497.36  14280.00 
SAMPLE INDUCTANCE CALCULATION
An example from reference 1 is used to demonstrate the calculation of the self inductance for a short, threeturn, 6 by 6foot (1.8 by 1.8m) loop, using the current sheet model described above.
The following parameters are used: 
a_{1}  = 6 ft = 182.88 cm 
a  = 6 ft = 182.88 cm 
N  = 3 turns 
P  = turn spacing = 150 mils = 0.381 cm 
b  = N P = (3)(0.381) = 1.143 cm 
γ = (b/a)  = 1.143/182.88 = 0.00625 cm 
K  = (a_{1} / a) = 1. 
From Table B1,
 β_{1}  = 0.4622 
 β_{1}’  = 0.6366 
 β_{2}  = 0.2122 
 β_{3}  = 0.0046 
 β_{5}  = 0.0046 
 β_{7}  = 0.0382. 
Solving Equation B2 for F’ yields
 (B5) 
 (B6) 
where
 (B7) 
Using Equation B1, the inductance of the loop is found as
 (B8) 
or
 (B9) 
REFERENCES
 Grover, F.W. Inductance Calculations. Dover Publications, Inc. 1962.
 Y. Niwa, “A Study of Coils Wound on Rectangular Frames with Special Reference to the Calculation of Inductance,” Research of the Electrotechnical Laboratory, No. 141, Tokyo, Japan, 1924.
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