Two parallel wires have currents in opposite directions, $I_1$ and $I_2$. They are situated a distance $R$ from one another. What is the force per length $L$ of one wire on the other?

• $I_1$ and $I_2$ exist in opposite directions.
• The two wires are separated by a distance $R$.

• $\frac{\vec{F}}{L}$

• The currents are steady.
• The wires are infinitely long.
• There are no outside forces to consider.

• We represent the magnetic field from a very long wire as

$$\vec{B}=\frac{\mu_0 I}{2 \pi r} \hat{z}$$

• We represent the magnetic force on a little piece of current as

$$\text{d}\vec{F}= I \text{d}\vec{l} \times \vec{B}$$

• We represent the situation with diagram below.

We know that the magnetic field at the location of Wire 2 from Wire 1 is given by the magnetic field of a long wire: $$\vec{B}=\frac{\mu_0 I_1}{2 \pi r} \hat{z}$$

We can reason that the direction of the field is $+\hat{z}$ because of the Right Hand Rule. We also don't care about the magnetic field from Wire 2 at the location of Wire 2, since Wire 2 cannot exert a force on itself. Now, it remains to calculate the magnetic force.

Since we know the magnetic field, the next thing we wish to define is $\text{d}\vec{l}$. Wire 2 has current directed with $\hat{y}$ in our representation, so we can say $$\text{d}\vec{l} = \text{d}y \hat{y}$$

This gives $$\text{d}\vec{l} \times \vec{B} = \frac{\mu_0 I_1}{2 \pi r}\text{d}y \hat{x}$$

Lastly, we need to choose the limits on our integral. We can select our origin conveniently so that the segment of interest extends from $y=0$ to $y=L$. Now, we write:

$$\frac{\vec{F}_{1 \rightarrow 2}}{L} = \int_0^L I_2 \text{d}\vec{l} \times \vec{B} = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi r}\text{d}y \hat{x} = \frac{\mu_0 I_1 I_2 L}{2 \pi r} \hat{x}$$

For now, we write $$\text{d}\vec{l} = \langle \text{d}x, -\text{d}y, 0 \rangle$$

We write the $y$-component with a negative sign so that $\text{d}y$ can be positive. For the separation vector, we write $$\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = \langle 0,0,0 \rangle - \langle x, y, 0 \rangle = \langle -x, -y, 0 \rangle$$

Notice that we can rewrite $y$ as $y=-x-L$. This equation comes from the equation for a straight line, $y=mx+b$, where the slope of the line (or wire in this case) is $m=-1$ and the y-intercept of the wire is at $b=-L$. An alternate solution to this example could also be to rotate the coordinate system so that the x or y axis lines up with wire. If finding $y$ is troublesome, it may be helpful to rotate your coordinate axes.

We can use geometric arguments to say that $\text{d}y=\text{d}x$. See the diagram above for an insight into this geometric argument. We can now plug in to express $\text{d}\vec{l}$ and $\vec{r}$ in terms of $x$ and $\text{d}x$: $$\text{d}\vec{l} = \langle \text{d}x, -\text{d}x, 0 \rangle$$ $$\vec{r} = \langle -x, L+x, 0 \rangle$$ Now, we can take the cross product and find the magnitude of the $\vec{r}$: $$\text{d}\vec{l} \times \vec{r} = \langle 0, 0, \text{d}x(L+x) - (-\text{d}x)(-x) \rangle = \langle 0, 0, L\text{d}x \rangle = L\text{d}x \hat{z}$$ $$r^3 = (x^2 + (L+x)^2)^{3/2}$$ The last thing we need is the bounds on our integral. Our variable of integration is $x$, since we chose to express everything in terms of $x$ and $\text{d}x$. (Earlier we could have equally have chosen to write everything in terms of y and dy though.) We know that our segment begins at $x=-L$, and ends at $x=0$, so these will be the limits on our integral. Below, we write the integral all set up, and then we evaluate using some assistance some Wolfram Alpha. $$\begin{align*} \vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3} \\ &= \int_{-L}^0 \frac{\mu_0}{4 \pi}\frac{IL\text{d}x}{(x^2 + (L+x)^2)^{3/2}}\hat{z} \\ &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z} \end{align*}$$