Ampère's force law

Diagram of original Ampere experiment

The form of Ampere's force law commonly given was derived by Maxwell and is one of several expressions consistent with the original experiments of Ampère and Gauss. The x-component of the force between two linear currents I and I’, as depicted in the adjacent diagram, was given by Ampère in 1825 and Gauss in 1833 as follows:[10]

${\displaystyle dF_{x}=kII'ds'\int ds{\frac {\cos(xds)\cos(rds')-\cos(rx)\cos(dsds')}{r^{2}}}.}$

Following Ampère, a number of scientists, including Wilhelm Weber, Rudolf Clausius, James Clerk Maxwell, Bernhard Riemann, Hermann Grassmann,[11] and Walther Ritz, developed this expression to find a fundamental expression of the force. Through differentiation, it can be shown that:

${\displaystyle {\frac {\cos(xds)\cos(rds')}{r^{2}}}=-\cos(rx){\frac {(\cos \epsilon -3\cos \phi \cos \phi ')}{r^{2}}}}$.

and also the identity:

${\displaystyle {\frac {\cos(rx)\cos(dsds')}{r^{2}}}={\frac {\cos(rx)\cos \epsilon }{r^{2}}}}$.

With these expressions, Ampère's force law can be expressed as:

${\displaystyle dF_{x}=kII'ds'\int ds'\cos(rx){\frac {2\cos \epsilon -3\cos \phi \cos \phi '}{r^{2}}}}$.

Using the identities:

${\displaystyle {\frac {\partial r}{\partial s}}=\cos \phi ,{\frac {\partial r}{\partial s'}}=-\cos \phi '}$.

and

${\displaystyle {\frac {\partial ^{2}r}{\partial s\partial s'}}={\frac {-\cos \epsilon +\cos \phi \cos \phi '}{r}}}$.

Ampère's results can be expressed in the form:

${\displaystyle d^{2}F={\frac {kII'dsds'}{r^{2}}}\left({\frac {\partial r}{\partial s}}{\frac {\partial r}{\partial s'}}-2r{\frac {\partial ^{2}r}{\partial s\partial s'}}\right)}$.

As Maxwell noted, terms can be added to this expression, which are derivatives of a function Q(r) and, when integrated, cancel each other out. Thus, Maxwell gave "the most general form consistent with the experimental facts" for the force on ds arising from the action of ds':[12]

${\displaystyle d^{2}F_{x}=kII'dsds'{\frac {1}{r^{2}}}\left[\left(\left({\frac {\partial r}{\partial s}}{\frac {\partial r}{\partial s'}}-2r{\frac {\partial ^{2}r}{\partial s\partial s'}}\right)+r{\frac {\partial ^{2}Q}{\partial s\partial s'}}\right)\cos(rx)+{\frac {\partial Q}{\partial s'}}\cos(xds)-{\frac {\partial Q}{\partial s}}\cos(xds')\right]}$.

Q is a function of r, according to Maxwell, which "cannot be determined, without assumptions of some kind, from experiments in which the active current forms a closed circuit." Taking the function Q(r) to be of the form:

${\displaystyle Q=-{\frac {(1+k)}{2r}}}$

We obtain the general expression for the force exerted on ds by ds:

${\displaystyle \mathbf {d^{2}F} =-{\frac {kII'}{2r^{2}}}\left[(3-k){\hat {\mathbf {r_{1}} }}(\mathbf {dsds'} )-3(1-k){\hat {\mathbf {r_{1}} }}(\mathbf {{\hat {r_{1}}}ds} )(\mathbf {{\hat {r_{1}}}ds'} )-(1+k)\mathbf {ds} (\mathbf {{\hat {r_{1}}}ds'} )-(1+k)\mathbf {d's} (\mathbf {{\hat {r_{1}}}ds} )\right]}$.

Integrating around s' eliminates k and the original expression given by Ampère and Gauss is obtained. Thus, as far as the original Ampère experiments are concerned, the value of k has no significance. Ampère took k=-1; Gauss took k=+1, as did Grassmann and Clausius, although Clausius omitted the S component. In the non-ethereal electron theories, Weber took k=-1 and Riemann took k=+1. Ritz left k undetermined in his theory. If we take k = -1, we obtain the Ampère expression:

${\displaystyle \mathbf {d^{2}F} =-{\frac {kII'}{r^{3}}}\left[2\mathbf {r} (\mathbf {dsds'} )-3\mathbf {r} (\mathbf {rds} )(\mathbf {rds'} )\right]}$

If we take k=+1, we obtain

${\displaystyle \mathbf {d^{2}F} =-{\frac {kII'}{r^{3}}}\left[\mathbf {r} (\mathbf {dsds'} )-\mathbf {ds(rds')} -\mathbf {ds'(rds)} \right]}$

Using the vector identity for the triple cross product, we may express this result as

${\displaystyle \mathbf {d^{2}F} ={\frac {kII'}{r^{3}}}\left[\left(\mathbf {ds} \times \mathbf {ds'} \times \mathbf {r} \right)+\mathbf {ds'(rds)} \right]}$

When integrated around ds' the second term is zero, and thus we find the form of Ampère's force law given by Maxwell:

${\displaystyle \mathbf {F} =kII'\int \int {\frac {\mathbf {ds} \times (\mathbf {ds'} \times \mathbf {r} )}{|r|^{3}}}}$

Start from the general formula:

${\displaystyle {\vec {F}}_{12}={\frac {\mu _{0}}{4\pi }}\int _{L_{1}}\int _{L_{2}}{\frac {I_{1}d{\vec {\ell }}_{1}\ \mathbf {\times } \ (I_{2}d{\vec {\ell }}_{2}\ \mathbf {\times } \ {\hat {\mathbf {r} }}_{21})}{|r|^{2}}}}$,

Assume wire 2 is along the x-axis, and wire 1 is at y=D, z=0, parallel to the x-axis. Let ${\displaystyle x_{1},x_{2}}$ be the x-coordinate of the differential element of wire 1 and wire 2, respectively. In other words, the differential element of wire 1 is at ${\displaystyle (x_{1},D,0)}$ and the differential element of wire 2 is at ${\displaystyle (x_{2},0,0)}$. By properties of line integrals, ${\displaystyle d{\vec {\ell }}_{1}=(dx_{1},0,0)}$ and ${\displaystyle d{\vec {\ell }}_{2}=(dx_{2},0,0)}$. Also,

${\displaystyle {\hat {\mathbf {r} }}_{21}={\frac {1}{\sqrt {(x_{1}-x_{2})^{2}+D^{2}}}}(x_{1}-x_{2},D,0)}$

and

${\displaystyle |r|={\sqrt {(x_{1}-x_{2})^{2}+D^{2}}}}$

Therefore, the integral is

${\displaystyle {\vec {F}}_{12}={\frac {\mu _{0}I_{1}I_{2}}{4\pi }}\int _{L_{1}}\int _{L_{2}}{\frac {(dx_{1},0,0)\ \mathbf {\times } \ \left[(dx_{2},0,0)\ \mathbf {\times } \ (x_{1}-x_{2},D,0)\right]}{|(x_{1}-x_{2})^{2}+D^{2}|^{3/2}}}}$.

Evaluating the cross-product:

${\displaystyle {\vec {F}}_{12}={\frac {\mu _{0}I_{1}I_{2}}{4\pi }}\int _{L_{1}}\int _{L_{2}}dx_{1}dx_{2}{\frac {(0,-D,0)}{|(x_{1}-x_{2})^{2}+D^{2}|^{3/2}}}}$.

Next, we integrate ${\displaystyle x_{2}}$ from ${\displaystyle -\infty }$ to ${\displaystyle +\infty }$:

${\displaystyle {\vec {F}}_{12}={\frac {\mu _{0}I_{1}I_{2}}{4\pi }}{\frac {2}{D}}(0,-1,0)\int _{L_{1}}dx_{1}}$.

If wire 1 is also infinite, the integral diverges, because the total attractive force between two infinite parallel wires is infinity. In fact, what we really want to know is the attractive force per unit length of wire 1. Therefore, assume wire 1 has a large but finite length ${\displaystyle L_{1}}$. Then the force vector felt by wire 1 is:

${\displaystyle {\vec {F}}_{12}={\frac {\mu _{0}I_{1}I_{2}}{4\pi }}{\frac {2}{D}}(0,-1,0)L_{1}}$.

As expected, the force that the wire feels is proportional to its length. The force per unit length is:

${\displaystyle {\frac {{\vec {F}}_{12}}{L_{1}}}={\frac {\mu _{0}I_{1}I_{2}}{2\pi D}}(0,-1,0)}$.

The direction of the force is along the y-axis, representing wire 1 getting pulled towards wire 2 if the currents are parallel, as expected. The magnitude of the force per unit length agrees with the expression for ${\displaystyle {\frac {F_{m}}{L}}}$ shown above.

Chronologically ordered:

• Ampère's original 1823 derivation:
  • Assis, André Koch Torres; Chaib, J. P. M. C; Ampère, André-Marie (2015). Ampère's electrodynamics: analysis of the meaning and evolution of Ampère's force between current elements, together with a complete translation of his masterpiece: Theory of electrodynamic phenomena, uniquely deduced from experience (PDF). Montreal: Apeiron. ISBN 978-1-987980-03-5.
• Maxwell's 1873 derivation:
  • Treatise on Electricity and Magnetism vol. 2, part 4, ch. 2 (§§502–527)
• Pierre Duhem's 1892 derivation:
  • Duhem, Pierre Maurice Marie (9 September 2018). Ampère's Force Law: A Modern Introduction. Alan Aversa (trans.). doi:10.13140/RG.2.2.31100.03206/1. Retrieved 3 July 2019. (EPUB)
    • translation of: Leçons sur l'électricité et le magnétisme vol. 3, appendix to book 14, pp. 309-332 (in French)
• Alfred O'Rahilly's 1938 derivation:
  • Electromagnetic Theory: A Critical Examination of Fundamentals vol. 1, pp. 102–104 (cf. the following pages, too)

• Ampere
• Magnetic constant
• Lorentz force
• Ampère's circuital law
• Free space

• Ampère's force law Includes animated graphic of the force vectors.