# Why can't we remove Ampere from being base unit?

1) Analogy in F = ma

From experiment, we find that

$F = k m a$

and we deliberately choose $k = 1$ (dimensionaless) and thus we define the unit Newton, a derived unit.​

2) In the definition of ampere, it is in terms of the force per unit length between two straight parallel conductors having equal current.

$f = K I^2 / r$

One ampere is defined for

$I = K' \sqrt{f r}$

As we know today, the constant $K'$, has dimensions ($A/ \sqrt{N}$).

What if we instead take K' as dimensionless, just like the constant in Newton's 1st law?

In this case, the new Ampere 1 A would have the same dimension as $\sqrt{N}$.

In other words, this new ampere is a derived unit instead?

I'm quite sure someone must have thought about this, or I must have made some terrible mistake. But as I can't find any references, here's what I can come up with so far:

a) $\mu_0$ becomes dimensionless. So the unit of $\epsilon_0$ has to be changed too (related to speed of light).
b) can't think of it yet..

Shouldn't there be a more fundamental reason for this?​

Born2bwire
Gold Member
Ampere is just a name, a word. Current is the flow of charge over time. So whatever unit we choose to use for current, it must still retain the same physical meaning of C/s. We cannot assume that the proportionality constant between current and the square root of force to be dimensionless because then we would be unable to have a unit that relates the amount of charge per unit time. Now the value of that proportionality constant is something that we have the freedom of choosing, but the dimensionality has to be fixed to retain physical meaning.

Ampere is just a name, a word. Current is the flow of charge over time. So whatever unit we choose to use for current, it must still retain the same physical meaning of C/s. We cannot assume that the proportionality constant between current and the square root of the product of distance and force to be dimensionless because then we would be unable to have a unit that relates the amount of charge per unit time. Now the value of that proportionality constant is something that we have the freedom of choosing, but the dimensionality has to be fixed to retain physical meaning.

In my equation, the physical meaning of ampere is not altered. It is still having exactly the same definition as now:
the current which two parallel conductor would generate a force of 1N on each other.

As long as there is no inconsistency in the definition, there shouldn't be problem for its physical relation with charge. After all, charge is just a derived quantity of current, which is a base quantity.

PS Thanks for your input. I'm defending my argument not because I believe it's correct; rather, I do believe it's wrong, just that I have yet found out why...

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Born2bwire
Gold Member
How can you say that the physical meaning is not altered?

Originally we have something like this:

$$A = K*\sqrt{F}$$

where F is units of N. Now, whatever value we choose for K, K MUST have units of $\frac{C}{s\sqrt{N}}$. The definition of an Ampere being the X Newtons of force between two infinite wires that are Y meters apart only determines the magnitude of the constant $K$ above. It does not determine the units of $K$, that has to be fixed so that the units of A agrees with the physical definition of A.

So I don't know what you are arguing here. If you mean that we can change the value of K, then sure, we can redefine the definition of Ampere so that the K is different (right now we define K such that the distance and forces involved are 1 m and 1 N respectively) but the units of K are fixed.

Now, whatever value we choose for K, K MUST have units of $\frac{C}{s\sqrt{N}}$.

Thanks. Of course here I'm not talking about the value, but the dimension.

Let me put it this way.

Why can't we define $A = \frac{C}{s} = {\sqrt{N}}$ ?

What kind of inconsistency would occur?

rcgldr
Homework Helper
Currently, a newton is a unit derived from the kilogram, which is based on an international prototype kilogram (and some backup copies).

The ampere could be a derived unit from coulumb, based on the standard

1 elementary charge = 1 / (1.602176565(35) x 10-19) coulomb

1 coulomb = 6.2415093 ×1018 elemetary charges

This could make the kilogram a derived unit, but I doubt there's any existing method to accurately measure a newton based on the current definition of an ampere, which is the current flowing in each of two parallel wires in a vacuum spaced 1 meter apart producing a force per meter of exactly

2 x 10-7 netwons / meter

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Born2bwire
Gold Member
I would say that the inconsistency is that the Ampere is a base unit. It is not derived from other units like the Newton. So to me, it does not make sense for someone to say that an Ampere is $\sqrt{kg-m}/s$. The kg, m and s are base units as is the Ampere.

I would say that the inconsistency is that the Ampere is a base unit. It is not derived from other units like the Newton. So to me, it does not make sense for someone to say that an Ampere is $\sqrt{kg-m}/s$. The kg, m and s are base units as is the Ampere.

Thanks! This goes back to my original question in the title:

"Why is Ampere a base unit, rather than a derived unit?"

I thought about this because I noticed that while other base units' definition are quite self explanatory, the definition of Ampere seems to be derived from Newton.

Dale
Mentor
2020 Award
Hi tsoits,

It sounds like you are thinking along the lines of what is done in the Gaussian cgs system of units:
http://en.wikipedia.org/wiki/Gaussian_units

You certainly can set up a system of units (like the Gaussian cgs units) where the electrical quantities are expressed purely in terms of the mechanical quantities. There is some appeal to doing that since it reduces your required number of fundamental dimensionful parameters and it does not unncecssarily add new base units.

However, Born2bwire is 100% correct in his assertion that the physical meaning is altered. You can set up a consistent set of units in this manner but the equations for the physical laws are different in this system of units:
http://en.wikipedia.org/wiki/Gaussian_units#List_of_equations

Note that there are no factors of vacuum permittivity or permeability in Gaussian cgs units. Also, there are additional factors of c in some places, and the factors of pi are moved around.

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Thanks DaleSpam, this is exactly the answer I want!

It's a pity that the copy of Jackson's I got is no longer using cgs; so I just barely heard of the name cgs before without knowing such a consequence.

Thanks also to Born2bwire =)