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What is the electric force between tow rods with different lengths that carry currents $I_1$ and $I_2$? Indeed, by the use of the Biot-Savart law I obtained a relation to calculate it, but there is something unusual!
When the lengths of the rods ($L_1$ and $L_2$) are not the same, It yields $F_{12}$ $\neq$ $F_{21}$ (where $F_{12}$ is the force applied by the first rod to the second and $F_{21}$ is the force applied by the second rod to the first. I think this inequality contradicts Newton's law that says $F_{12}$ must be equal to $F_{21}$. It would be really appreciative if someone guide me.

rudolph1024
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Mahdi Rouholamini
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  • @Mahdi Do you mean the Lorentz force? – Gonenc May 28 '15 at 16:37
  • @SebastianRiese: I think your thought experiment proves that if the Biot-Savart law & the Lorentz force satisfy Newton's Third Law for infinitesimal elements, then they satisfy it for any current configuration. Your logic is impeccable, but as I showed in my answer, Biot-Savart & Lorentz do not in fact satisfy Newton's Third Law on an element-by-element basis. – Michael Seifert May 28 '15 at 16:52
  • @MichaelSeifert That seems reasonable. After all, static, open currents are unphysical (and thus may have unphysical consequences). I should have thought about it a bit longer. You are absolutely right. – Sebastian Riese May 28 '15 at 16:55

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The Biot-Savart Law does not obey Newton's Third Law for two open current segments. It only works if at least one of the loops is closed.

Proof: The force between two wire segments can be written as a double path integral using the Biot-Savart Law and the Lorentz force Law. Labeling the current distributions #1 and #2, the force on #2 due to #1 works out to be: $$ \vec{F}_{21} = \frac{\mu_0 I_1 I_2}{4 \pi} \int_1 \int_2 \frac{d\vec{l}_2 \times (d \vec{l}_1 \times \hat{r}_{21})}{r_{21}^2}, $$ where $\vec{l}_1$ and $\vec{l}_2$ are the parametrized curves and $\vec{r}_{21} = \vec{l}_2 - \vec{l}_1$. Applying a vector identity, we can split this into two terms: $$ \frac{\mu_0 I_1 I_2}{4 \pi} \left[ \int_1 \int_2 \frac{d\vec{l}_1 (d \vec{l}_2 \cdot \hat{r}_{21})}{r_{21}^2} - \int_1 \int_2 \frac{ \hat{r}_{21} (d \vec{l}_1 \cdot d\vec{l}_2)}{r_{21}^2} \right] $$ The second term is manifestly symmetric between 1 and 2 (with the appropriate sign flip since $\hat{r}_{21}$ changes direction when we switch 1 and 2.) The first term, however, is more of a problem: it can be written as $$ \frac{\mu_0 I_1 I_2}{4 \pi} \int_1 d\vec{l}_1 \left[ \int_2 \frac{d \vec{l}_2 \cdot \hat{r}_{21}}{r_{21}^2} \right] = \frac{\mu_0 I_1 I_2}{4 \pi} \int_1 d\vec{l}_1 \left[ \int_2 d\left( - \frac{1}{r_{21}} \right) \right] \\ = \frac{\mu_0 I_1 I_2}{4 \pi} \int_1 d\vec{l}_1 \left[ - \frac{1}{r_{21}} \right]_\text{endpts. of 2} $$ If loop 2 is closed, the quantity in square brackets is zero (since the "endpoints" are in fact the same point.) However, if this is not the case, then the integral will not in general vanish; and it will in general be different from the corresponding term in the force on #1 due to #2, which would be $$ \frac{\mu_0 I_1 I_2}{4 \pi} \int_2 d\vec{l}_2 \left[ - \frac{1}{r_{21}} \right]_\text{endpts. of 1} $$

Newton's Third Law can be rescued by noting that a "finite current element" must have a time-varying point charge at its ends due to charge conservation. This leads to a time-varying electric field, and therefore a time-varying electromagnetic momentum density. And if the total electromagnetic momentum of the system is changing with time, we would not expect the net force on the system to be zero.

Michael Seifert
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