How can two conductors of unequal lengths exert a force of same magnitude on each other?

Question

I understand Newton's 3rd Law states that if Object (conductor) A exerts a force on Object B, Object A will experience an equal and opposite force. However, in a current carrying conductor Lorentz force is given as: $F=\frac{\mu_0I_1I_2l}{2\pi r}$ where $\mu_0$ is the permeability of free space, $I_1$ and $I_2$ are the currents of flowing in both conductors, $l$ is the length of the conductor experiencing the force, and $r$ is the distance between both conductors. Therefore, if there were two conductors of same current but unequal lengths (e.g. 10m and 3m), how would you get the same force magnitude on both wires? I would really appreciate your help in understanding this. Thank you in advance.

Answer 1

The formula you have given is for two wires of infinite length. The formula for two wires of finite length is more complicated, and satisfies Newton's third law. In section 7.2 of my EM textbook, I derive the $\bf B$ field for a wire of length L in the x direction: \begin{equation} {\bf B}(x,y)=\frac{{\bf I\times{\hat j}}}{cy}\left[\frac{(L/2-x)}{\sqrt{L/2-x)^2+y^2}}+\frac{(L/2+x)}{\sqrt{L/2+x)^2+y^2}}\right]. \label{bxy} \end{equation} Using this field, you can find the force on the second wire by integrating I$\bf dl\times B$ over that wire.

Answer 2

I agree with Jerrold Franklin’s answer. I’ll just add a general proof that the magnetostatic force obeys Newton’s third law. Say you have two conductors of respective current densities $j_1,j_2$ producing respective fields $B_1,B_2$. The force of $a$ on $b$ is: $$ F_{a\to b}=\int j_b\times B_ad^3x $$ Newton’s third law is equivalent to: $$ F_{1\to2}+F_{2\to1}=0 $$

The key is that the magnetostatic self force is always zero. You can see this from the Maxwell stress tensor: $$ j\times B=\nabla\cdot \sigma\\ \sigma=\frac{1}{2\mu_0}\left(B\otimes B-\frac{1}{2}B^2 1 \right) $$ Thus the total self force $F=\int j\times B d^3x$ is just a boundary integral which vanishes if the field decays fast enough (which is the case when the current is compactly supported).

Applying this result to $j_1,j_2,j_{1+2}:=j_1+j_2$, you get: $$ F_{1\to1}=0\\ F_{2\to2}=0\\ F_{1+2\to1+2}=0 $$ but the force is quadratic, so: $$ F_{1+2\to1+2}=F_{1\to1}+F_{2\to2}+F_{1\to2}+F_{2\to1} $$ which leads to Newton’s third law.

Btw, this only holds in statics, as in dynamics the field stores momentum and can radiate some to infinity, so you don’t have a zero self force and the third law does not apply anymore.

Hope this helps.

Answer 3

The setup with infinite straight wires is only a special case of the most general setup. It's possible to retrieve it using the expression for the magnetic field generated by an elementary part of a circuit, as done below.

Let's consider two wires where currents with intensity $i_1$ and $i_2$. Each segment of the wire 1 induces on a point of the wire 2 a magnetic field

\begin{equation} d \mathbf{B}_{1}(\mathbf{r}_2) = - \dfrac{\mu}{2\pi} i_1 \dfrac{\mathbf{r}_2 - \tilde{\mathbf{r}}_1}{|\mathbf{r}_2 - \tilde{\mathbf{r}}_1|^2}\times d \tilde{\mathbf{r}}_1 \ , \end{equation}

using $\tilde{}$ to identify the "active" element, and the force acting on it reads

\begin{equation} dd \mathbf{F}_2(\mathbf{r}_2) = - i_2 d\mathbf{B}(\mathbf{r}_2) \times d \mathbf{r}_2 \ , \end{equation}

where the magnetic field at point $\mathbf{r}_2$ is the sum of the contribution of the two wires, $d\mathbf{B} = d\mathbf{B}_1 + d\mathbf{B}_2$,

\begin{equation} dd \mathbf{F}_2(\mathbf{r}_2) = - i_2 \left( d\mathbf{B}_1 (\mathbf{r}_2) + d\mathbf{B}_2 (\mathbf{r}_2) \right) \times d \mathbf{r}_2 \ , \end{equation}

Explicitly writing the expression of the contributions of the magnetic field,

\begin{equation} dd\mathbf{F}(\mathbf{r}_2) = \dfrac{\mu}{2\pi} \left( \dfrac{\mathbf{r}_2 - \tilde{\mathbf{r}}_1}{|\mathbf{r}_2 - \tilde{\mathbf{r}}_1|^2}\times d \tilde{\mathbf{r}}_1 \right) \times d \mathbf{r}_2 \, i_1 i_2 + \dfrac{\mu}{2\pi} \left( \dfrac{\mathbf{r}_2 - \tilde{\mathbf{r}}_2}{|\mathbf{r}_2 - \tilde{\mathbf{r}}_2|^2}\times d \tilde{\mathbf{r}}_2 \right) \times d \mathbf{r}_2 \, i_2^2 \ , \end{equation}

it should be readily clear that the contribution of the elementary element of one circuit on the elementary element of the other circuit (the first term) is the same inverting the role of the "active" and the "passive" circuits, i.e. switching the indices 1 and 2 in the first term of the latter expression. Since the elementary contributions are "equal and opposite", you get macroscopic (integral) "equal and opposite" actions between the two circuits.