# UnitMath Examples: ( Inductance ) Last Modified: 1/24/2000

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I have tried to make sure the following formulas are correct, but I do not guarantee them.

### The following examples show how to find the inductances of the following shapes. ### Inductance of a Coaxial Cable TOP

The following equation gives the inductance of a coaxial cable.

L: mu/(2 pi) ln( b/a ) * H/m;

Where
a is the inner radius.
b is the outer radius.
mu is the permeability

### Following are example calculations. In the first two examples, I omit the internal inductance; which, is valid for high frequencies.

L: mu/(2 pi) ln( b/a );
mu: 1 Áo;
a: 1 mm;
b: 3 mm;
L as microH/m ~ 0.220 microH/m

L: mu/(2 pi) ln( b/a );
mu: 1 Áo;
a: 2.0 mm;
b: 9.0 mm;
L as microH/m ~ ( 0.295 to 0.301 to 0.307 ) microH/m

### Taking the internal inductances into account ( low frequencies ). Where f is the outer radius of the outer conductor.

L: Áo / ( 2 pi ) * ln( b/a ) + mu_core /(8 pi) + mu_outer /(2 pi) ( f^4 / (f^2 - b^2)^2 * ln (f/b) - (3 f^2 - b^2) / (4 (f^2 - b^2) ) )
mu_core: 1 Áo;
mu_outer: 1 Áo;
a: 2 mm;
b: 9 mm;
f: 10 mm;
L as microH/m ~ 0.358 microH/m ### Inductance of Round Parallel Wires TOP

The following formula gives the inductance per length of a pair of round parallel wires.

L: mu /pi acosh( d / (2a) );

Where
d is the distance between the centers of the wires.
a is the wires radius.
mu is the permeability

### Following are example calculations.

L: mu /pi acosh( d / (2a) );
mu: 1 Áo;
d: 25 ft;
a: ( 0.803 pm 0 ) in;
L as microH/m ~ 2.37 microH/m

L: mu /pi acosh( d / (2a) );
mu: 1 Áo;
d: 25. ft;
a: 0.80 in;
L as microH/m ~ ( 2.36 to 2.37 to 2.38 ) microH/m

L: mu /pi acosh( d / (2a) );
mu: 1 Áo;
d: 12 mm;
a: 1 mm;
L as microH/m ~ 0.99 microH/m ### Inductance of Round Wire over a Ground Plane TOP

The following formula gives inductance per length of a round wire over a ground plane.

L: mu /( 2 pi ) acosh( d / a );

Where
d is the distance between the centers of the wire and the ground plane.
a is the wire's radius.
mu is the permeability

### Following are example calculations.

L: mu /( 2 pi ) acosh( d / a );
mu: 1 Áo;
d: 25/2 ft;
a: ( 0.803 pm 0 ) in;
L as microH/m ~ 1.185 microH/m

L: mu /( 2 pi ) acosh( d / a );
mu: 1 Áo;
d: 12.5 ft pm 9 in;
a: 0.8 in;
L as microH/m ~ ( 1.16 to 1.19 to 1.21 ) microH/m ### Self Inductance of a Straight Round Wire TOP

Find its self inductance of a round wire. Where Lo is the self inductance, and L is the high frequency inductance.

Lo: 2 length ( ln( 4 length / diameter ) - 1 + mu/4 ) nH / cm;
L: 2 length ( ln( 4 length / diameter ) - 1 ) nH / cm;

Where
length is the wires length
diameter is the wires diameter
mur is the relative permeability

### Following are example calculations.

mur: 1;
length: 25 cm;
diameter: 1/10 cm;
Lo: 2 length ( ln( 4 length / diameter ) - 1 + mur/4 ) nH/cm;
L: 2 length ( ln( 4 length / diameter ) - 1 ) nH/cm;

Lo as microH ~ 0.308 microH
L as microH ~ 0.295 microH

mur: 1;
length: 5 in;
diameter: 0.02535 in;
Lo: 2 length ( ln( 4 length / diameter ) - 1 + mur/4 ) nH/cm;
L: 2 length ( ln( 4 length / diameter ) - 1 ) nH/cm;

Lo as microH ~ 0.150 microH
L as microH ~ 0.144 microH ### Inductance of a Single-Layered Air Core Coil TOP

The following empirical formula called Wheeler's Formula can be used to find the inductance of a single-layer air coil. These equations are said to be accurate to about 1% when 2 R / len > 3.

L: sq ( N IR ) / ( 9 IR + N 10 D ) * microH/inch;
or
L: sq ( N IR ) / ( 9 IR +10 length ) * microH/inch;

Where
N is the number of turns.
IR is the inner radius.
D is the diameter of the wire.
length is the length of the coil.

### Following are example calculations.

L: sq ( N IR ) / ( 9 IR + N 10 D ) * microH/inch;

IR: 1 inch;
D: (0.086 pm 0) inch;
N:30;
L as microH ~ 25.862 microH

L: sq ( N IR ) / ( 9 IR + 10 length ) * microH/inch;

IR: 2.00 cm;
length: 50. cm;
N:300;
L as microH ~ ( 270. to 274. to 278. ) microH ### Inductance of a Multi-Layered Air Core Coil TOP

The following empirical formula can be used to find the approximate inductance of a multi-layered air coil.

L: 316/10 * sq ( N r1 ) / ( 6 r1 + 9 length + 10 ( r2 - r1 ) ) * microH;

Where
N is the number of turns.
r1 is the inner radius.
r2 is the outer radius.
length is the length of the coil.

### Following are example calculations.

L: 316/10 * sq ( N r1 ) / ( 6 r1 + 9 length + 10 ( r2 - r1 ) ) * microH/m;
N:300;
r1: 9 mm;
r2: 25 mm;
length: 20 mm;
L as microH ~ 584.68 microH

L: 316/10 * sq ( N r1 ) / ( 6 r1 + 9 length + 10 ( r2 - r1 ) ) * microH/m;
N:800;
r1: 1 cm;
r2: 2 cm;
length: 3 cm;
L as mH ~ 4.70 mH ### Inductance of a Toroid with a Square Cross SectionTOP

The following equation gives the inductance of a toroid coil with a square cross section.

L: mu N^2 a/( 2 pi) * ln( r2/r1 ) ;

Where
r2 is the outer radius.
21 is the inner radius.
mu is the permeability
N is the number of turns
a is the height

### Following are example calculations.

L: mu N^2 a/( 2 pi) * ln( r2/r1 );
N:700;
mu: 1 Áo;
r1: 1 cm;
r2: 2 cm;
a: 3/2 cm;
L as mH ~ 1.019 mH

L: mu N^2 a/( 2 pi) * ln( r2/r1 );
N:700;
mu: 1 Áo;
r1: 80.0 cm;
r2: 82.0 cm;
a: 3/2 cm;
L as microH ~ ( 34.5 to 36.3 to 38.1 ) microH