Remark: This is not a homework question...It is pure out of theoretical interest. I asked this the mathematics-community a couple days ago and got no answer, so I figured I'd try here.

Most standard physics textbooks compute the force two infinite wires exert on each other, but they remain silent about the case where the wires are finite. Let's say we have two parallel wires carrying a current of equal magnitude in the same direction, both of which have a length $d$ and also seperated by a distance $d$. I now want to find out the force one wire exerts on another, using the Biot-Savart Law.

Let the left wire be positioned at the origin of the $xy$-plane, going along the $y$-axis, and let the other wire be a distance $d$ to the right. We assume the currents are flowing in the positive $y$-direction. Then we first choose a source element (on the left wire) of infinitesimal length $dy$ described by the position vector $\mathbf{r_0} = y_0 \hat{j}$. This constitutes a current source of $I d\vec{l} = (Idy) \hat{j}$.

We then pick an arbitrary field point $P$ on the other wire with position vector $\mathbf{r_p} = x\hat{i} + y \hat{j}$. Then the position vector $\mathbf{r}$ pointing from the source point to the field point is given as \begin{align*} \mathbf{r} = \mathbf{r_p} - \mathbf{r_0} = x\hat{i} + (y - y_0) \hat{j}, \end{align*} with $\sqrt{x^2 + (y-y_0)^2}$ being the length of this vector. If we now calculate the crossproduct $d\vec{l} \times \mathbf{r}$, we can write it as \begin{align*} dy \hat{j} \times (x\hat{i} + (y - y_0)\hat{j}) = -dy x \hat{k} \end{align*} Now comes the tricky part. I think I need to setup a double integral, because we are working with infinitesimal force elements $d\mathbf{F}$, each which is given as $d\mathbf{F} = I d\vec{l} \times \mathbf{B}$. But we also have that \begin{align*} d\mathbf{B} = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \hat{r}}{r^2} = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \mathbf{r}}{r^3} = -\frac{\mu_0 I}{4 \pi} \frac{dyx}{\sqrt{x^2 + (y-y_0)^2}} \hat{k} \end{align*} Hence I need to somehow integrate over $d\mathbf{F}$ and $d\mathbf{B}$. Does anyone have an idea how to do this?


1 Answer 1


Let's call the circuit in the origin circuit one and it's line element $\mathrm{d}l_1=(0,\mathrm{d}y_1,0)$ and the one to it's right $r_2=(d,y_2,0)$ then the force between them is $\mathrm{d}F_{12}=i \mathrm{d}l_2 \times B_1$ where

$$B_1=\frac{\mu_0i}{4\pi}\int_{l_1} \frac{\mathrm{d}l_1\times \Delta r}{(\Delta r)^3}$$

and $\Delta r=(d,(y_2-y_1),0)$ so we have that

$$\mathrm{d}l_1\times \Delta r=(0,0,-\mathrm{d}y_1 d)$$

so we get as you wrote:

$$B_1=-\frac{\mu_0 i d}{4 \pi}\int_{0}^{d} \frac{\mathrm{d}y_1}{(d^2+(y_2-y_1)^2)^{\frac{3}{2}}}$$

Ok now let's call $y_2-y_1=t$ so $\mathrm{d}t=-\mathrm{d}y_1$ then we can write

$$B_1=\frac{\mu_0 i d}{4\pi}\int_{y_2}^{y_2-d}\frac{\mathrm{d}t}{(d^2+t^2)^{\frac{3}{2}}}$$

we now make the substitution

$$t=d\cdot \sinh(u)$$

and we obtain

$$\mathrm{d}t=d\cdot \cosh(u)\mathrm{d}u$$

and then

$$B_1=\frac{\mu_0 i d}{4\pi}\int \mathrm{d}u \frac{d \cosh(u)}{d^3 \cosh(u)^3}$$

in which we used

$$\cosh(u)^2-\sinh(u)^2=1$$ $$B_1=\frac{\mu_0 i }{4\pi d}\int \frac{\mathrm{d}u}{\cosh(u)^2}$$

now $\frac{1}{\cosh^2(u)}$ is the derivative of $\tanh(u)$ so

$$\int \frac{\mathrm{d}u}{\cosh(u)^2}=\tanh(u)$$

we get then

$$B_1=\frac{\mu_0 i }{4\pi d}\tanh\left(a\sinh\left(\frac{y_2-y_1}{d}\right)\right)+\text{const}$$

where we have substituted back all parameters

$$u=a\sinh\left(\frac{t}{d}\right) \\ t=y_2-y_1$$

so knowing that (where $a\sinh(x)$ is the inverse function of $\sinh(x)$):



$$B_1=\frac{\mu_0 i }{4\pi d} \frac{\frac{y_2-y_1}{d}}{\sqrt{(\frac{y_2-y_1}{d})^2+1}}+\text{const}=\frac{\mu_0 i }{4\pi} \frac{1}{\sqrt{(y_2-y_1)^2+d^2}}+\text{const}$$

now we calculate it between $y_1=0$ and $y_1=d$ which yields

$$B_1=\frac{\mu_0 i }{4\pi} \left[ \frac{1}{\sqrt{(y_2-d)^2+d^2}}-\frac{1}{\sqrt{y_2^2+d^2}}\right]$$

to calculate the force we take $B_1=(0,0,B_1 \hat{z})$ and we operate the following:

$$\mathrm{d}F_{12}=i\mathrm{d}l_2 \times B_1=i(B_1\mathrm{d}y_2,0,0)$$

now we have to integrate on the circuit two:

$$F_{12}=\frac{\mu_0 i^2 }{4\pi} \int_{0}^{d}\mathrm{d}y_2\left[ \frac{1}{\sqrt{(y_2-d)^2+d^2}}-\frac{1}{\sqrt{y_2^2+d^2}}\right]=\frac{\mu_0 i^2 }{4\pi} \left[I(y_2-d)-I(y_2)\right]$$

and know we do the same trick as before

$$t=y_2-d \ \text{or}\ t=y_2\ \text{for the second piece}$$ $$t=d\cdot \sinh(u)$$ $$\mathrm{d}y_2=\mathrm{d}t=d\cdot \cosh(u)\mathrm{d}u$$


$$I=\int \mathrm{d}u\cdot d \cdot \cosh(u) \frac{1}{\sqrt{d^2\cosh^2(u)}}=u=a\sinh\left(\frac{t}{d}\right)$$

we finally get

$$F_{12}=\frac{\mu_0 i^2 }{4\pi}\left[a\sinh\left(\frac{y_2-d}{d}\right)-a\sinh\left(\frac{y_2}{d}\right)\right]_0^d$$

which curiously enough is zero for this choice of parameters! I hope that helped!

  • $\begingroup$ Why do I have $-dyx \hat{k}$ as a result of my crossproduct? What does the $x$ in there mean? Does that mean I have to integrate from $0$ to $x$? $\endgroup$
    – Kamil
    Commented Mar 29, 2015 at 21:16
  • $\begingroup$ well $x$ in this case is $d$ the separation between the two wires which is not involved in the integration since the wires are vertical, the cross product is done between the line element of the wire $dl_{wire}=(0,dy_1,0)$ and the distance vector between the two wires $\Delta r=(d,(y_2-y_1),0)$ and it's calculated using the usual rules of the cross product which yelds $-dy_1\cdot d$ $\endgroup$
    – Fra
    Commented Mar 29, 2015 at 21:21
  • $\begingroup$ Ah I see, thanks. By the way I think you made a mistake. You have $\int_0^d \frac{dy_1}{(d^2 +(y_2 - y_1)^2)^{3/2}}$. But why do you have a $3/2$ as an exponent there? I had just a square root. Also, can I ask how you deduced the integration bounds? $\endgroup$
    – Kamil
    Commented Mar 29, 2015 at 21:28
  • $\begingroup$ the integration bounds are determined by the wire position (vertical from zero to $d$) and it's lenght which is $d$. For the factor $\frac{3}{2}$ it comes from the biot-savart law which involves the distance (square root) cubed (factor 3) at the denominator! $\endgroup$
    – Fra
    Commented Mar 29, 2015 at 21:30
  • $\begingroup$ Yes, but I mean the next integral. It goes from $y_2$ to $y_2 -d$. Shouldnt the upper bound be $d - y_2$ ? $\endgroup$
    – Kamil
    Commented Mar 29, 2015 at 21:31

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