Discussions on the Original Form of Ampere’s Law

Ampere’s Law provides a method for calculating the force between two current sources. However, the modern form of Ampere’s Law is not the one originally proposed by Ampere. While they differ mathematically, there is no measurable physical distinction under steady-state conditions.

This article will introduce the derivation process of the original form of Ampere’s Law and discuss its relationship with the modern form.

I. Derivation of the Original Ampere’s Law #

Experimental Basis #

1. Force reverses when current reverses

   $$ d\vec{F}_{12} \propto i_1 i_2 \tag{1} $$

2. Current elements possess vector superposition property

   $$ d\vec{F}_{12} = d\vec{F}_{12} (i_1, i_2, d\vec{l}_1, d\vec{l}_2, \vec{r}_{12}) \tag{2} $$

3. Force on a current element is perpendicular to the element itself

   $$ \left( \oint_{L_1}{d\vec{F}_{12}} \right) \cdot d\vec{l}_2 = 0 \tag{3} $$

4. Force magnitude is independent of geometric scale

   $$ |d\vec{F}_{12}| \propto \frac{|d\vec{l}_1| |d\vec{l}_2|}{|\vec{r}_{12}|^2} \tag{4} $$

Assumption #

The force between two current elements \( d\vec{l}_1 \) and \( d\vec{l}_2 \) acts along the direction of \( \vec{r}_{12} \).

Derivation Process #

Based on the assumption and considering experimental basis (1) and (2), we can write

$$ d\vec{F}_{12} = k i_1 i_2 \varphi(d\vec{l}_1, d\vec{l}_2, \vec{r}_{12}) \hat{r}_{12} \tag{5} $$

where \( \varphi \) is a scalar-valued function.

Considering experiment (2), the path element \( d\vec{l} \) characterizing the current element exhibits vector properties.

A simple consideration is that \( \varphi \) depends linearly on \( d\vec{l}_1 \) and \( d\vec{l}_2 \).

Thus, we assume

$$ \varphi = \varphi_1(\vec{r}_{12})(d\vec{l}_1 \cdot d\vec{l}_2) + \varphi_2(\vec{r}_{12})(d\vec{l}_1 \cdot \vec{r}_{12})(d\vec{l}_2 \cdot \vec{r}_{12}) + \varphi_3(\vec{r}_{12})(d\vec{l}_1 \times d\vec{l}_2) \cdot \vec{r}_{12} \quad \text{(6)} $$

The rationale for this assumption is that terms within \( \varphi \) must take forms such as

$$ (d{l}_1)_\alpha (d{l}_2)_\beta ({r}_{12})_\gamma ({r}_{12})_\mu \cdots \tag{7} $$

① \( \alpha = \beta \): This corresponds to the \( (d\vec{l}_1 \cdot d\vec{l}_2) \) term.

② \( \alpha \neq \beta \): Since \( \varphi \) is a scalar function, there are two possible forms

1. \( (d{l}_1)_\alpha (d{l}_2)_\beta ({r}_{12})_\alpha ({r}_{12})_\beta \cdots \)

i.e., the \( (d\vec{l}_1 \cdot \vec{r}_{12})(d\vec{l}_2 \cdot \vec{r}_{12}) \) term.

2. \( \varepsilon_{\gamma\alpha\beta} (d{l}_1)_\alpha (d{l}_2)_\beta ({r}_{12})_\gamma \cdots \)

i.e., the \( (d\vec{l}_1 \times d\vec{l}_2) \cdot \vec{r}_{12} \) term.

Considering (4), we have

$$ d\vec{F}_{12} = i_1 i_2 \vec{r}_{12} \left[ (d\vec{l}_1 \cdot d\vec{l}_2) \frac{A}{r_{12}^3} + (d\vec{l}_1 \cdot \vec{r}_{12})(d\vec{l}_2 \cdot \vec{r}_{12}) \frac{B}{r_{12}^5} + (d\vec{l}_1 \times d\vec{l}_2) \cdot \vec{r}_{12} \frac{C}{r_{12}^4} \right] \quad \text{(8)} $$

Considering (3), as shown in the figure,

where \(d\vec{l}_1 = - d\vec{r}_{12}\).

Note equation (3)

$$ \left( \oint_{L_1} d\vec{F}_{12} \right) \cdot d\vec{l}_{2} = \oint_{L_1} (d\vec{F}_{12} \cdot d\vec{l}_{2}) = 0 \tag{9} $$

This equality is independent of the path \(L_1\), implying that \(d\vec{F}_{12} \cdot d\vec{r}_{12}\) is the differential of some potential function \(\phi\),

i.e., \( d\phi = d\vec{F}_{12} \cdot d\vec{l}_{2} \).

For \(\phi\), considering the above, we have

$$ d\phi \propto \frac{|d\vec{r}_{12}| |d\vec{l}_2|^2}{|\vec{r}_{12}|^2} \tag{10} $$

$$ \phi \propto \frac{|d\vec{l}_2|^2}{|\vec{r}_{12}|} \tag{11} $$

Thus, we formally express \(\phi\) as

$$ \phi = (d\vec{l}_2 \cdot d\vec{l}_2) \frac{a}{r_{12}} + (d\vec{l}_2 \cdot \vec{r}_{12})^2 \frac{b}{r_{12}^3} + (d\vec{l}_2 \times \vec{r}_{12})^2 \frac{c}{r_{12}^3} \tag{12} $$

Taking the differential, we obtain

$$ \begin{align} d\phi = & - (d\vec{l}_2 \cdot d\vec{l}_2) \frac{a \vec{r}_{12} \cdot d\vec{r}_{12}}{r_{12}^3} + \notag \\ &\ 2(d\vec{l}_2 \cdot \vec{r}_{12})(d\vec{l}_2 \cdot d\vec{r}_{12}) \frac{b}{r_{12}^3} - 3(d\vec{l}_2 \cdot \vec{r}_{12})^2 \frac{b \vec{r}_{12} \cdot d\vec{r}_{12}}{r_{12}^5} + \notag \\ &\ 2(d\vec{l}_2 \times \vec{r}_{12}) \cdot (d\vec{l}_2 \times d\vec{r}_{12}) \frac{c}{r_{12}^3} - 3(d\vec{l}_2 \times \vec{r}_{12})^2 \frac{c \vec{r}_{12} \cdot d\vec{r}_{12}}{r_{12}^5} \tag{13} \end{align} $$

From (5), we also have

$$ d\vec{F}_{12} \cdot d\vec{l}_2 = -i_1 i_2 (\vec{r}_{12} \cdot d\vec{l}_2) \left[ (d\vec{r}_{12} \cdot d\vec{l}_2) \frac{A}{r_{12}^3} + (d\vec{r}_{12} \cdot \vec{r}_{12}) (d\vec{l}_2 \cdot \vec{r}_{12}) \frac{B}{r_{12}^5} + (d\vec{r}_{12} \times d\vec{l}_2) \cdot \vec{r}_{12} \frac{C}{r_{12}^4} \right] \quad \text{(14)} $$

And since \( d\phi = d\vec{F}_{12} \cdot d\vec{l}_2 \), comparing coefficients yields, we have

$$ \begin{cases} a = 0 \\ -A = 2b \\ -B = -3b \\ c = 0 \\ C = 0 \end{cases} \tag{15} $$

Thus, we can set \( A = -2k, B = 3k, C = 0\). Substituting into (9) gives

$$ d\vec{F}_{12} = -k i_1 i_2 \vec{r}_{12} \left[ (d\vec{l}_{1} \cdot d\vec{l}_2) \frac{2}{r_{12}^3} - (d\vec{l}_{1} \cdot \vec{r}_{12}) (d\vec{l}_2 \cdot \vec{r}_{12}) \frac{3}{r_{12}^5} \right] \tag{16} $$

This is the original form of Ampere’s Law.

II. Discussion on the Applicability of the Original Ampere’s Law #

Under steady-state conditions, the force between a pair of current elements cannot be measured in practice.

The physically measurable quantity is at least the force exerted by a closed loop on a current element.

$$ \begin{align} d\vec{F}_{L_1,2} =& \oint_{L_1} d\vec{F}_{12} \notag \\ =& \oint_{L_1} k i_1 i_2 \vec{r}_{12} \left[ \frac{2 (d\vec{r}_{12} \cdot d\vec{l}_2)}{r_{12}^3} - \frac{3 (d\vec{r}_{12} \cdot \vec{r}_{12}) (d\vec{l}_2 \cdot \vec{r}_{12})}{r_{12}^5} \right] \notag \\ =& \oint_{L_1} k i_1 i_2 \left[ d\left(\vec{r}_{12} \frac{(d\vec{l}_2 \cdot \vec{r}_{12})}{r_{12}^3} \right) + \frac{1}{r_{12}^3} \left[ \vec{r}_{12} (d\vec{r}_{12} \cdot d\vec{l}_2) - d\vec{r}_{12} (d\vec{r}_{12} \cdot d\vec{l}_2) \right] \right] \notag \quad \text{(17)} \end{align} $$

Here,

$$ \oint_{L_1} k i_1 i_2 d\left( \frac{\vec{r}_{12} (d\vec{l}_2 \cdot \vec{r}_{12})}{r_{12}^3} \right) = 0 \tag{18} $$$$ \vec{r}_{12} (d\vec{r}_{12} \cdot d\vec{l}_2) - d\vec{r}_{12} (d\vec{r}_{12} \cdot d\vec{l}_2) = d\vec{l}_2 \times (\vec{r}_{12} \times d\vec{r}_{12}) \tag{19} $$

This shows that the modern Ampere’s Law differs from the original Ampere’s Law by an exact differential term.

Therefore, for closed loops under steady-state conditions, there is no practical distinction between them.

Thus,

$$ d\vec{F}_{L_1,2} = \oint_{L_1} k i_1 i_2 \frac{1}{r_{12}^3} d\vec{l}_2 \times (\vec{r}_{12} \times d\vec{l}_1) \tag{20} $$

Considering that \( d\vec{r}_{12} = -d\vec{l}_1 \), we have

$$ d\vec{F}_{L_1,2} = \oint_{L_1} k i_1 i_2 \frac{1}{r_{12}^3} d\vec{l}_2 \times (d\vec{l}_1 \times \vec{r}_{12}) \tag{21} $$

This is consistent with the result obtained from the modern Ampere’s Law.

III. Why the Modern Form of Ampere’s Law is Used #

While stable, isolated current elements do not exist in practice,

they can exist under non-steady conditions.

For example, a moving point charge can be treated as an independent current element

$$ q \vec{v} = i d\vec{l} \tag{22} $$

Consider the scenario depicted.

Using the relativistic transformation of electromagnetic fields, the electromagnetic field produced by \( q_1 \) at position \( \vec{r}_{12} \) at that instant satisfies

$$ \begin{cases} \vec{E}(\vec{r}_{12}) = \left[ \gamma \vec{\vec{I}} - (\gamma - 1) \frac{\vec{v}_1 \vec{v}_1}{|\vec{v}_1|^2} \right] \cdot \frac{1}{4\pi\varepsilon_0} \frac{q_1 \vec{r}_{12}'}{|\vec{r}_{12}'|^3} \\\\ \vec{B}(\vec{r}_{12}) = \gamma \frac{\vec{v}_1}{c^2} \times \frac{1}{4\pi\varepsilon_0} \frac{q_1 \vec{r}_{12}'}{|\vec{r}_{12}'|^3} \end{cases} \tag{23} $$

where

$$ \vec{r}_{12}' = \left[ \vec{\vec{I}} + (\gamma - 1) \frac{\vec{v}_1 \vec{v}_1}{|\vec{v}_1|^2} \right] \vec{r}_{12} \tag{24} $$

Note that for a real current element, besides the moving charge carriers, there should be stationary charges maintaining overall electrical neutrality. Therefore, the actual electric field, after accounting for the stationary charges, should be negligible and is not considered here.

Thus, under the approximation \( v \ll c \), the net force on \( q_2 \) is

$$ \begin{align} \vec{F}_{12} =&\ q_2 \vec{v}_2 \times \vec{B} \\ =&\ q_2 \vec{v}_2 \times \left( \frac{\vec{v}_1}{c^2} \times \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 \vec{r}_{12}}{|\vec{r}_{12}|^3} \right) \\ =&\ \frac{\mu_0}{4\pi} \frac{q_2 \vec{v}_2 \times \left( q_1 \vec{v}_1 \times \vec{r}_{12} \right)}{|\vec{r}_{12}|^3} \end{align} \tag{25} $$

If we treat the moving charge \( q\vec{v} \) as a current element \( i d\vec{l} \), then

$$ \vec{F}_{12} = \frac{\mu_0}{4\pi} \frac{(i_2 d\vec{l}_2) \times \left( (i_1 d\vec{l}_1) \times \vec{r}_{12} \right)}{|\vec{r}_{12}|^3} \tag{26} $$

This is consistent with the form of the modern Ampere’s Law.