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Help: From Ampere to Coulomb in SI units |
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| Jan18-13, 02:10 PM | #1 |
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Help: From Ampere to Coulomb in SI units
Hello,
Could someone help me finish this train of thought? This is how we think in SI units: First, just because we like the value of 1 Ampere as it is now, we choose the force between two parallel conductors to be exactly 2×10[itex]^{-7}[/itex]N= const ×[itex]\frac{1A×1A}{1m}[/itex] Then, purely as choice, we decide to formulate a constant from the above relation μ[itex]_{0}[/itex]=4π×10[itex]^{-7}[/itex] N/A^2 Next comes the Coulomb's law. Why do we choose the Coulomb constant to involve the [itex]\epsilon_{0}[/itex] from the expression c[itex]^{2}[/itex]=[itex]\frac{1}{\epsilon_{0}\mu_{0}}[/itex]? Why not choose something else? Is this tied to the definition of Ampere? Thanks. |
| Jan18-13, 06:10 PM | #2 |
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Recognitions:
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These days, the Ampere is defned by the force law - so it is the amount of current that gets you 2x10-7N of force between wires 1m apart.
Normally we'd want to make the force 1N for the definition, that would make the Ampere very large indeed. The funny number was selected for historical reasons. http://en.wikipedia.org/wiki/Ampere The Coulomb is defined in terms of the Amp - yes. This is likely to get reversed in 2014... with the Coulomb getting defined as a specific number of elementary charges and the Amp being defined as 1 C/s. Then people will ask - "why that exact number" and the reason will be "historical" ... i.e. so that values don't get thrown too far off what people expect. [I mean - the kinds of values that people feel are comfortable to use.] |
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