So, here is what the BARD chat bot had to say (Some engineering book!). This took 10 seconds.

?

It has been said that a chat bot will never replace an engineer, but an engineer with a chat bot WILL replace you!

?

My initial question:

“Given two toroid coils with the same number of turns, wound with the same wire but of two different gauges, what would be the difference in inductive reactance?”

?

BARD answer:

View other drafts (there where several other answers available!)

The inductive reactance of a toroid coil is directly proportional to the number of turns wound on the core, and inversely proportional to the cross-sectional area of the wire. Therefore, if two toroid coils have the same number of turns, but the wire used to wind them has different gauges, the inductive reactance of the coils will be different.

The gauge of a wire is a measure of its thickness. A lower gauge number corresponds to a thicker wire, while a higher gauge number corresponds to a thinner wire. Therefore, a toroid coil wound with a thicker wire will have a lower inductive reactance than a toroid coil wound with a thinner wire.

The following formula can be used to calculate the inductive reactance of a toroid coil:

XL = 2πfL / N

where:

• XL is the inductive reactance in ohms
• f is the frequency of the AC current in hertz
• L is the inductance of the coil in henries
• N is the number of turns of wire on the core

The inductance of a toroid coil is given by the following formula:

L = μ0μrN?a? / (2πr)

where:

• L is the inductance in henries
• μ0 is the permeability of free space, which is 4π × 10?? henries per meter
• μr is the relative permeability of the core material
• N is the number of turns of wire on the core
• a is the mean radius of the toroid
• r is the minor radius of the toroid

From these formulas, it can be seen that the inductive reactance of a toroid coil is directly proportional to the square of the number of turns, and inversely proportional to the mean radius and the minor radius of the toroid.

Therefore, if two toroid coils have the same number of turns, but the mean radius or minor radius of the coils are different, the inductive reactance of the coils will also be different.

In conclusion, the inductive reactance of a toroid coil is affected by a number of factors, including the number of turns of wire on the core, the gauge of the wire, the mean radius of the toroid, and the minor radius of the toroid.

?

My follow up question (this took another 7 seconds!):

How much of a difference would one gauge number, from 18ga to 20ga make at: 1kHz, at 1mHz, at 30mHz?

View other drafts

To determine the difference in inductive reactance between a toroid coil wound with 18 AWG wire and a toroid coil wound with 20 AWG wire, let's assume the following parameters:

• Mean radius (a) = 0.1 meters
• Minor radius (r) = 0.05 meters
• Relative permeability (μr) of the core material = 1
• Number of turns (N) = 100

Using the formula for inductive reactance:

XL = 2πfL / N

First, calculate the inductance for each gauge of wire:

For 18 AWG wire:

L = (4π × 10?? H/m) × 1 × (100 turns)? × (0.1 m)? / (2π × 0.05 m)

L ≈ 0.00502 H

For 20 AWG wire:

L = (4π × 10?? H/m) × 1 × (100 turns)? × (0.1 m)? / (2π × 0.05 m)

L ≈ 0.00393 H

Now, calculate the inductive reactance for each frequency:

At 1 kHz:

XL_18 = 2π × 1 kHz × 0.00502 H / 100 turns

XL_18 ≈ 0.0315 Ω

XL_20 = 2π × 1 kHz × 0.00393 H / 100 turns

XL_20 ≈ 0.0248 Ω

At 1 mHz:

XL_18 = 2π × 1 mHz × 0.00502 H / 100 turns

XL_18 ≈ 31.5 ?Ω

XL_20 = 2π × 1 mHz × 0.00393 H / 100 turns

XL_20 ≈ 24.8 ?Ω

At 30 mHz:

XL_18 = 2π × 30 mHz × 0.00502 H / 100 turns

XL_18 ≈ 0.945 ?Ω

XL_20 = 2π × 30 mHz × 0.00393 H / 100 turns

XL_20 ≈ 0.744 ?Ω

Comparing the inductive reactance values for the two gauges:

As you can see, the difference in inductive reactance between the two gauges decreases as the frequency decreases. At 1 kHz, the difference is about 6.7%. At 1 mHz, the difference is about 27%. At 30 mHz, the difference is about 27%.

This is because the inductive reactance of a coil is directly proportional to the frequency of the AC current. At lower frequencies, the inductive reactance is smaller, and the difference between the two gauges is less noticeable.

Best 73,

Bill - WA2WIO

?