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{{About|Ampère's force law|the law relating the integrated magnetic field around a closed loop to the electric current through the loop|Ampère's circuital law}}
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{{Use dmy dates|date=September 2011}}
{{Electromagnetism|cTopic=[[Magnetostatics]]}}
 
[[File:MagneticWireAttraction-2nd.png|thumb|The top wire with current ''I''<sub>1</sub> experiences a [[Lorentz force]] F<sub>12</sub> due to magnetic field ''B''<sub>2</sub> created by the bottom wire. (Not shown is the simultaneous process where the bottom wire ''I''<sub>2</sub> experiences a magnetic force F<sub>21</sub> due to magnetic field ''B''<sub>1</sub> 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
 
::<math>  F_m = 2 k_A \frac {I_1 I_2 } {r}</math>,
 
where ''k''<sub>A</sub> is the magnetic force constant, ''r'' is the separation of the wires, and ''I''<sub>1</sub>, ''I''<sub>2</sub> are the [[direct current]]s carried by the wires. This is a good approximation for finite lengths if the distance between the wires is small compared to their lengths, but large compared to their diameters. The value of ''k''<sub>A</sub> depends upon the system of units chosen, and the value of ''k''<sub>A</sub> decides how large the unit of current will be. In the [[SI]] system,<ref name=Serway>{{cite book
|author=Raymond A Serway & Jewett JW
|title=Serway's principles of physics: a calculus based text
|url=http://books.google.com/books?id=1DZz341Pp50C&pg=RA1-PA746&dq=wire+%22magnetic+force%22&lr=&as_brr=0&sig=4vMV_CH6Nm8ZkgjtDJFlupekYoA#PRA1-PA746,M1 |publisher=Thompson Brooks/Cole
|edition=Fourth Edition
|location=Belmont, CA
|year=2006
|page=746
|isbn=0-534-49143-X}}</ref>
<ref name=Monk>{{cite book
|author=Paul M. S. Monk
|title=Physical chemistry: understanding our chemical world
|url=http://books.google.com/books?vid=ISBN0471491802&id=LupAi35QjhoC&pg=PA16&lpg=PA16&ots=IMiGyIL-67&dq=ampere+definition+si&sig=9Y0k0wgvymmLNYFMcXodwJZwvAM |publisher=Chichester: Wiley
|location=New York
|year=2004
|page=16
|isbn=0-471-49181-0}}</ref>
 
::<math>  k_A \ \overset{\underset{\mathrm{def}}{}}{=}\  \frac {\mu_0}{ 4\pi} \ </math>
 
with μ<sub>0</sub> the [[magnetic constant]], ''defined'' in SI units as<ref>[http://www.bipm.org/en/si/si_brochure/chapter2/2-1/ampere.html ''BIPM definition'']</ref><ref name="NIST">{{cite web |url=http://physics.nist.gov/cgi-bin/cuu/Value?mu0 |title=Magnetic constant |accessdate=8 August 2007 |work=2006 [[CODATA]] recommended values |publisher=[[NIST]] | archiveurl= http://web.archive.org/web/20070820144036/http://physics.nist.gov/cgi-bin/cuu/Value?mu0| archivedate= 20 August 2007 <!--DASHBot-->| deadurl= no}}</ref> 
 
::<math>  \mu_0  \ \overset{\underset{\mathrm{def}}{}}{=}\  4 \pi \times 10^{-7} \  </math> [[Newton (unit)|newtons]] / ([[ampere]])<sup>2</sup>.
 
Thus, in vacuum,
 
:''the force per meter of length between two parallel conductors – spaced apart by 1&nbsp;m and ''each'' carrying a current of 1&nbsp;[[Ampere|A]] – is exactly''
 
:: <math> \displaystyle 2 \times 10^{-7} </math> [[Newton (unit)|N]]/m.
 
The general formulation of the magnetic force for arbitrary geometries is based on [[line integral]]s and combines the [[Biot-Savart law]] and [[Lorentz force]] in one equation as shown below.
<ref>The integrand of this expression appears in the official documentation regarding definition of the ampere [http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf BIPM SI Units brochure, 8th Edition, p. 105]</ref>
:<ref name=Chow>{{cite book
|author=Tai L. Chow
|title=Introduction to electromagnetic theory: a modern perspective
|url=http://books.google.com/books?id=dpnpMhw1zo8C&pg=PA153&lpg=PA153&dq=%22ampere's+law+of+force%22&source=web&ots=uZOFz9dWv7&sig=NJp3UQvbCOvcVm7eJN4IUdlC9bs |publisher=Jones and Bartlett
|location=Boston
|year=2006
|page=153
|isbn=0-7637-3827-1}}</ref><ref>[http://info.ee.surrey.ac.uk/Workshop/advice/coils/unit_systems/ampereForce.html Ampère's Force Law] '' Scroll to section "Integral Equation" for formula. ''</ref>
 
:<math>  \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}</math>,
 
where
*<math>\vec{F}_{12}</math> is the total force felt by wire 1 due to wire 2 (usually measured in [[Newton (unit)|newtons]]),
*''I''<sub>1</sub> and ''I''<sub>2</sub> are the currents running through wires 1 and 2, respectively (usually measured in [[ampere]]s),
*The double line integration sums the force upon each element of wire 1 due to the magnetic field of each element of wire 2,
*<math>d \vec{\ell}_1</math> and <math>d \vec{\ell}_2</math> are infinitesimal vectors associated with wire 1 and wire 2 respectively (usually measured in metres); see [[line integral]] for a detailed definition,
*The vector <math>\hat{\mathbf{r}}_{21}</math> 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 [[Cross product|vector cross product]],
*The sign of ''I''<sub>n</sub> is relative to the orientation <math>d \vec{\ell}_n</math> (for example, if <math>d \vec{\ell}_1</math> points in the direction of [[conventional current]], then ''I''<sub>1</sub>>0).
 
To determine the force between wires in a material medium, the [[magnetic constant]] is replaced by the actual [[Permeability (electromagnetism)|permeability]] of the medium.
 
==Historical Background==
 
[[Image:AmpereDiagram.jpg|thumbnail|500px|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 [[André-Marie Ampère|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:<ref>{{cite book|last=O'Rahilly|first=Alfred|title=Electromagnetic Theory|year=1965|publisher=Dover|pages=104}}</ref>
 
:<math>  dF_x = k I I' ds' \int ds \frac {\cos(xds) \cos(rds') - \cos(rx)\cos(dsds')} {r^2}. </math>
 
Following Ampere, a number of scientists, including [[Wilhelm Eduard Weber|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:
 
:<math>  \frac{\cos(xds) \cos(rds')} {r^2} = -\cos(rx) \frac{(\cos\epsilon - 3 \cos\phi \cos\phi')} {r^2} </math>.
 
and also the identity:
 
:<math>  \frac{\cos(rx) \cos(dsds')} {r^2} = \frac{\cos(rx) \cos\epsilon} {r^2} </math>.
 
With these expressions, Ampere's force law can be expressed as:
 
:<math>  dF_x = k I I' ds'\int ds' \cos(rx) \frac{2\cos\epsilon - 3\cos\phi \cos\phi'} {r^2} </math>.
 
Using the identities:
 
:<math>  \frac{\partial r} {\partial s} = \cos\phi,  \frac{\partial r} {\partial s'} = -\cos\phi'    </math>.
 
and
 
:<math>  \frac{\partial^2 r} {\partial s \partial s'} = \frac{-\cos\epsilon + \cos\phi \cos\phi'} {r} </math>.
 
Ampere's results can be expressed in the form:
 
:<math> 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)  </math>.
 
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':<ref>{{cite book|last=Maxwell|first=James Clerk|title=Treatise on Electricity and Magnetism|year=1904|publisher=Oxford|pages=173}}</ref>
 
:<math>  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]  </math>.
 
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:
 
:<math>  Q = - \frac{(1+k)} {2r} </math>
 
We obtain the general expression for the force exerted on ds by ds:
 
:<math>  \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] </math>.
 
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:
 
:<math>  \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] </math>
 
If we take k=+1, we obtain
 
:<math>  \mathbf{d^2F} = -\frac{k I I'} {r^3} \left[ \mathbf{r} (\mathbf{dsds'}) - \mathbf{ds (r ds')} -\mathbf{ds'(r ds)} \right] </math>
 
Using the vector identity for the triple cross product, we may express this result as
 
:<math>  \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] </math>
 
When integrated around ds' the second term is zero, and thus we find the form of Ampere's force law given by Maxwell:
 
:<math>  \mathbf{F} = k I I' \int \int \frac{\mathbf{ds}\times(\mathbf{ds'}\times\mathbf{r})} {|r|^3} </math>
 
==Derivation of parallel straight wire case from general formula==
 
[[Image:Ampere Force-2nd.PNG|thumbnail|300px|Another depiction of the Lorentz force law, showing both the force F<sub>12</sub> on wire 1 due to magnetic field of wire 2, and the equal and opposite force F<sub>21</sub> on wire 2 due to magnetic field of wire 1.]]
 
Start from the general formula:
:<math>  \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}</math>,
Assume wire 2 is along the x-axis, and wire 1 is at y=D, z=0, parallel to the x-axis. Let <math>x_1,x_2</math> 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 <math>(x_1,D,0)</math> and the differential element of wire 2 is at <math>(x_2,0,0)</math>. By properties of line integrals, <math>d\vec{\ell}_1=(dx_1,0,0)</math> and <math>d\vec{\ell}_2=(dx_2,0,0)</math>. Also,
:<math> \hat{\mathbf{r}}_{21} = \frac{1}{\sqrt{(x_1-x_2)^2+D^2}}(x_1-x_2,D,0)</math>
and
:<math> |r| = \sqrt{(x_1-x_2)^2+D^2}</math>
Therefore the integral is
:<math>  \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}}</math>.
Evaluating the cross-product:
:<math>  \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}}</math>.
Next, we integrate <math>x_2</math> from <math>-\infty</math> to <math>+\infty</math>:
:<math>  \vec{F}_{12} = \frac {\mu_0 I_1 I_2} {4 \pi}\frac{2}{D}(0,-1,0) \int_{L_1} dx_1 </math>.
If wire 1 is also infinite, the integral diverges, because the ''total'' attractive force between two infinite parallel wires is infinity. In fact, we want to know the attractive force ''per unit length'' of wire 1. Therefore, assume wire 1 has a large but finite length <math>L_1</math>. Then the force vector felt by wire 1 is:
:<math>  \vec{F}_{12} = \frac {\mu_0 I_1 I_2} {4 \pi}\frac{2}{D}(0,-1,0) L_1 </math>.
As expected, the force that the wire feels is proportional to its length. The force per unit length is:
:<math>  \frac{\vec{F}_{12}}{L_1} = \frac {\mu_0 I_1 I_2} {2 \pi D}(0,-1,0) </math>.
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 <math>F_m</math> shown above.
 
==See also==
* [[Ampere]]
* [[Magnetic constant]]
* [[Lorentz force]]
* [[Ampère's circuital law]]
* [[Free space]]
 
==References and notes==
<references/>
 
==External links==
*[http://info.ee.surrey.ac.uk/Workshop/advice/coils/unit_systems/ampereForce.html Ampère's force law] Includes animated graphic of the force vectors.
 
{{DEFAULTSORT:Ampere's Force Law}}
[[Category:Electrostatics|Ampere's law]]
[[Category:Magnetostatics|Ampere's law]]
[[Category:Concepts in physics]]
 
[[de:Lorentzkraft#Definition_des_Amperes]]

Latest revision as of 08:28, 12 January 2015

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