In attempting to solve a community problem posted recently, I derived a formula for the magnetic force between two current-carrying wire segments of arbitrary length (but same for both) and arbitrary spacing. These may well be parts of infinite line segments, but the formula only accounts for the interactions between the two finite segments. The derivation is rather tedious, so I won't post it right now:

The formula ends up being:

\[F = \frac{\mu_0 I_1 I_2}{2 \pi} \Big ( \frac{\sqrt{D^2 + L^2}}{D} -1 \Big )\]

For conductor length much larger than the spacing, the result reduces to:

\[F \approx \frac{\mu_0 I_1 I_2}{2 \pi} \frac{L}{D}\]

The second formula is the well-known expression for the force between two long wires, so that's a good sign. I have also confirmed the result with numerical simulations.

Does anybody else know off the top of their head that this is correct?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest@Laszlo Mihaly @Ashish Siva What do you think?

Log in to reply

I would calculate the vector potential ( easier than Biot Savart), take the gradient and integrate the force over the length of the other wire. Too busy with real work to do it.

Log in to reply

yes, I attempted the derivation and obtained the same result. Musn't be a coincidence. ;)

Log in to reply