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# Ampère's force law

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### Ampère's force law

The top wire with current I1 experiences a Lorentz force F12 due to magnetic field B2 created by the bottom wire. (Not shown is the simultaneous process where the bottom wire I2 experiences a magnetic force F21 due to magnetic field B1 created by the top wire.)

In magnetostatics, the force of attraction or repulsion between two current-carrying wires (see first figure below) is often called Ampère's force law. The physical origin of this force is that each wire generates a magnetic field, as defined by the Biot–Savart law, and the other wire experiences a magnetic force as a consequence, as defined by the Lorentz force.

## Equation

The best-known and simplest example of Ampère's force law, which underlies the definition of the ampere, the SI unit of current, states that the force per unit length between two straight parallel conductors is

\frac {F_m} {L} = 2 k_A \frac {I_1 I_2 } {r},

where kA is the magnetic force constant from the Biot–Savart law, Fm/L is the total force on either wire per unit length of the shorter (the longer is approximated as infinitely long relative to the shorter), r is the distance between the two wires, and I1, I2 are the direct currents carried by the wires.

This is a good approximation if one wire is sufficiently longer than the other that it can be approximated as infinitely long, and if the distance between the wires is small compared to their lengths (so that the one infinite-wire approximation holds), but large compared to their diameters (so that they may also be approximated as infinitely thin lines). The value of kA depends upon the system of units chosen, and the value of kA decides how large the unit of current will be. In the SI system,[1] [2]

k_A \ \overset{\underset{\mathrm{def}}{}}{=}\ \frac {\mu_0}{ 4\pi}

with μ0 the magnetic constant, defined in SI units as[3][4]

\mu_0 \ \overset{\underset{\mathrm{def}}{}}{=} \ 4 \pi \times 10^{-7} N / A2.

Thus, in vacuum,

the force per meter of length between two parallel conductors – spaced apart by 1 m and each carrying a current of 1 A – is exactly
\displaystyle 2 \times 10^{-7} N / m.

The general formulation of the magnetic force for arbitrary geometries is based on iterated line integrals and combines the Biot–Savart law and Lorentz force in one equation as shown below. [5]

[6][7]
\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},

where

• \vec{F}_{12} is the total force felt by wire 1 due to wire 2 (usually measured in newtons),
• I1 and I2 are the currents running through wires 1 and 2, respectively (usually measured in amperes),
• The double line integration sums the force upon each element of wire 1 due to the magnetic field of each element of wire 2,
• d \vec{\ell}_1 and d \vec{\ell}_2 are infinitesimal vectors associated with wire 1 and wire 2 respectively (usually measured in metres); see line integral for a detailed definition,
• The vector \hat{\mathbf{r}}_{21} is the unit vector pointing from the differential element on wire 2 towards the differential element on wire 1, and |r| is the distance separating these elements,
• The multiplication × is a vector cross product,
• The sign of In is relative to the orientation d \vec{\ell}_n (for example, if d \vec{\ell}_1 points in the direction of conventional current, then I1>0).

To determine the force between wires in a material medium, the magnetic constant is replaced by the actual permeability of the medium.

## Historical Background

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 Ampere and Gauss. The x-component of the force between two linear currents I and I’, as depicted in the diagram to the right, was given by Ampere in 1825 and Gauss in 1833 as follows:[8]

dF_x = k I I' ds' \int ds \frac {\cos(xds) \cos(rds') - \cos(rx)\cos(dsds')} {r^2}.

Following Ampere, a number of scientists, including Wilhelm Weber, Rudolf Clausius, James Clerk Maxwell, Bernhard Riemann and Walter Ritz, developed this expression to find a fundamental expression of the force. Through differentiation, it can be shown that:

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

and also the identity:

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

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

dF_x = k I I' ds'\int ds' \cos(rx) \frac{2\cos\epsilon - 3\cos\phi \cos\phi'} {r^2} .

Using the identities:

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

and

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

Ampere's results can be expressed in the form:

d^2 F = \frac{k I I' ds ds'} {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':[9]

d^2 F_x = k I I' ds ds' \left[ \left( \frac{1} {r^2} \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:

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

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

\mathbf{d^2F} = -\frac{k I I'} {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 Ampere and Gauss is obtained. Thus, as far as the original Ampere experiments are concerned, the value of k has no significance. Ampere 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 Ampere expression:

\mathbf{d^2F} = -\frac{k I I'} {r^3} \left[ 2 \mathbf{r} (\mathbf{dsds'}) - 3\mathbf{r} (\mathbf{r ds}) (\mathbf{r ds'}) \right]

If we take k=+1, we obtain

\mathbf{d^2F} = -\frac{k I I'} {r^3} \left[ \mathbf{r} (\mathbf{dsds'}) - \mathbf{ds (r ds')} -\mathbf{ds'(r ds)} \right]

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

\mathbf{d^2F} = \frac{k I I'} {r^3} \left[ \left(\mathbf{ds}\times\mathbf{ds'}\times\mathbf{r}\right)+ \mathbf{ds'(r ds)} \right]

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

\mathbf{F} = k I I' \int \int \frac{\mathbf{ds}\times(\mathbf{ds'}\times\mathbf{r})} {|r|^3}

## Derivation of parallel straight wire case from general formula

Start from the general formula:

\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 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 (x_1,D,0) and the differential element of wire 2 is at (x_2,0,0). By properties of line integrals, d\vec{\ell}_1=(dx_1,0,0) and d\vec{\ell}_2=(dx_2,0,0). Also,

\hat{\mathbf{r}}_{21} = \frac{1}{\sqrt{(x_1-x_2)^2+D^2}}(x_1-x_2,D,0)

and

|r| = \sqrt{(x_1-x_2)^2+D^2}

Therefore the integral is

\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:

\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 x_2 from -\infty to +\infty:

\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 L_1. Then the force vector felt by wire 1 is:

\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:

\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 \frac {F_m} {L} shown above.

## References and notes

1. ^
2. ^
3. ^ BIPM definition
4. ^
5. ^ The integrand of this expression appears in the official documentation regarding definition of the ampere BIPM SI Units brochure, 8th Edition, p. 105
6. ^
7. ^ Ampère's Force Law Scroll to section "Integral Equation" for formula.
8. ^ (cf. , which appears in )
9. ^