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Ampère's force law

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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.
Another depiction of the Lorentz force law, showing both the force F12 on wire 1 due to magnetic field of wire 2, and the equal and opposite force F21 on wire 2 due to magnetic field of wire 1.

In magnetostatics, the force of attraction or repulsion between two current-carrying wires (see Figure 1) 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.

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

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

where kA is the magnetic force constant, r is the separation of the wires, and I1, I2 are the DC currents carried by the wires. This is a good approximation for finite lengths if the distance between the wires is small compared to their lengths. 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,

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

with μ0 the magnetic constant, defined in SI units as

 \mu_0  \ \overset{\underset{\mathrm{def}}{}}{=}\  4 \pi \times 10^{-7} \  newtons / (ampere)2.

Thus, in vacuum, the force per meter of length between the two parallel conductors carrying a current of 1 A and spaced apart by 1 m, is exactly 2 × 10−7 N/m.

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

 \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

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

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 x1,x2 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 (x1,D,0) and the differential element of wire 2 is at (x2,D,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} \ ((dx_2,0,0) \ \mathbf{ \times } \ (x_1-x_2,D,0) )} {|(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 x2 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, we want to know the attractive force per unit length of wire 1. Therefore, assume wire 1 has a large but finite length L1. 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 Fm shown above.

References and notes

  1. ^ Raymond A Serway & Jewett JW (2006). Serway's principles of physics: a calculus based text (Fourth Edition ed.). Belmont, CA: Thompson Brooks/Cole. p. 746. ISBN 053449143X. 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. 
  2. ^ Paul M. S. Monk (2004). Physical chemistry: understanding our chemical world. New York: Chichester: Wiley. p. 16. ISBN 0471491810. http://books.google.com/books?vid=ISBN0471491802&id=LupAi35QjhoC&pg=PA16&lpg=PA16&ots=IMiGyIL-67&dq=ampere+definition+si&sig=9Y0k0wgvymmLNYFMcXodwJZwvAM. 
  3. ^ BIPM definition
  4. ^ "Magnetic constant". 2006 CODATA recommended values. NIST. http://physics.nist.gov/cgi-bin/cuu/Value?mu0. Retrieved 2007-08-08. 
  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. ^ Tai L. Chow (2006). Introduction to electromagnetic theory: a modern perspective. Boston: Jones and Bartlett. p. 153. ISBN 0763738271. http://books.google.com/books?id=dpnpMhw1zo8C&pg=PA153&lpg=PA153&dq=%22ampere's+law+of+force%22&source=web&ots=uZOFz9dWv7&sig=NJp3UQvbCOvcVm7eJN4IUdlC9bs. 
  7. ^ Ampère's Force Law Scroll to section "Integral Equation" for formula.

Further reading

See also