Let there be two parallel current-carrying infinitely-long wires separated by $d$ carrying currents $i_1$ & $i_2$ respectively in the same direction.

In order to find the force on an infinitesimal part $d\mathbf l$ of the right wire carrying current $i_2$ by the magnetic field of the left wire carrying current $i_1$, Biot-Savart Law & $\mathbf F= id\mathbf l \times \mathbf B$ are used leading to the formula $$ F= \frac{\mu_0\; i_1 i_2}{2\pi d}\;.$$

But recently, I learnt another stuff- magnetic pressure.

A sheet carrying current changes abruptly the magnetic field parallel to the sheet & perpendicular to the current from one side to another side; lesser the thickness of the sheet, more the abrupt discontinuity in the change of magnetic field while moving from one side to another side of the sheet. This is represented in the following formula:

$$B_\text{$\perp$, front}- B_\text{$\perp$, behind} = \mu_0 \mathcal J\;.$$

The force on the current sheet due to these magnetic fields are given by $$\begin{align} F&=\left(\frac{B_\text{$\perp$, front} + B_\text{$\perp$, behind} }{2}\right)(B_\text{$\perp$, front} -B_\text{$\perp$, behind} )\cdot \frac{1}{\mu_0}\\& =\frac{1}{2\mu_0}\left[(B_\text{$\perp$, front})^2- (B_\text{$\perp$, behind} )^2\right]\cdot\end{align}\;.$$

I wanted to use this formula to deduce the force between on the right wire. Is it right to use this formula to find the force on the right wire?

So, $$B_\text{$\perp$, front}=-\frac{\mu_0i_1}{2\pi(d+\delta)}- \frac{\mu_0i_2}{2\pi\delta } \\B_\text{$\perp$, behind}=-\frac{\mu_0i_1}{2\pi(d-\delta)}+ \frac{\mu_0i_2}{2\pi\delta} $$ where $\delta$ is infinitesimal increment or decrement of spatial displacement from the respective wires.

So, putting these values in their respective positions in the equation of the magnetic pressure, I got \begin{align}F&=\frac{1}{2\mu_0}\left[\left(-\frac{\mu_0i_1}{2\pi(d+\delta)}- \frac{\mu_0i_2}{2\pi\delta}\right)^2-\left(-\frac{\mu_0i_1}{2\pi(d-\delta)}+ \frac{\mu_0i_2}{2\pi\delta}\right)^2\right]\\&=\frac{1}{2\mu_0}\frac{\mu_0^2}{4\pi^2}\left[\frac{i_1^2}{(d+\delta)^2}-\frac{i_1^2}{(d-\delta)^2}+\frac{2i_1i_2}{(d+\delta)\delta}+\frac{2i_1i_2}{(d-\delta)\delta}\right]\\&=\frac{1}{2\mu_0}\frac{\mu_0^2}{4\pi^2}\left[\frac{i_1^2 d^2 +i_1^2\delta^2-2i_1^2d\delta-i_1^2d^2-i_1^2\delta^2-2i_1^2d\delta}{(d^2-\delta^2)^2}+ \frac{2i_1i_2d- 2i_1i_2\delta+ 2i_1i_2d+ 2i_1i_2\delta}{(d^2-\delta^2)\delta}\right]\\&=\frac{1}{2\mu_0}\frac{\mu_0^2}{4\pi^2}\left[-\frac{4i_1^2d\delta}{(d^2-\delta^2)^2 }+ \frac{4i_1i_2d}{(d^2-\delta^2)\delta}\right]\\&= \frac{\mu_0}{2\pi^2}\left[\frac{i_1i_2d}{(d^2-\delta^2)\delta}- \frac{i_1^2d\delta}{(d^2-\delta^2)^2}\right]\;.\end{align}

And so, DISASTER ! This is nowhere to the formula I derived using Biot-Savart Law & Lorentz-Force equation. There is also a square of $\pi$ in the denominator whereas in the formula derived earlier by Biot-Savart Law & Lorentz-Force definition, there is only single $\pi$ in the denominator. Adding fuel to the fire, $\delta \to 0$ which means the relation I came to using magnetic pressure is indeterminate :(

So, is it wrong to use the magnetic pressure relation to derive the force of the magnetic field due to the parallel wire carrying current $i_1$ in the same direction?

If not, than what is wrong in my approach?

Note, I'm not telling anyone to please do this for me sort of request; what I really want to know is whether my approach to deduce the force using the relation of magnetic pressure is right or wrong. And if it is right to use that, why it didn't give the same result that I got using Biot-Savart Law & Lorentz-Force relation.


1 Answer 1


The difficulty you're facing is that you're trying to apply a formula for current sheets to line currents. Additionally, your expression for the force on a current sheet is incorrect. It's easy to tell it's incorrect because it's in units of energy, and as a magnetic pressure, the force should be proportional to the area of the current sheet.

The way you're approaching this problem is essentially an energy method. It's possible to derive the Biot-Savart law using the magnetic field energy. Start with the expression for magnetic field energy:

$$ U=\frac{1}{2\mu_0} \int B^2 dV$$

Then displace one of the wires a differential amount, and calculate the change in energy. To make notation clear, I'm denoting the position of a wire as $x$, and the small displacement $dx$

$$dU=\frac{1}{2\mu_0} \int B(x+dx)^2-B(x)^2$$

Then the force is the negative ratio of $dU$ over $dx$

$$F = -\frac{dU}{dx}$$

Technically, this is only the force in the $x$ direction. Realistically, only the distance between wires is relevant, so make $x$ the distance between wires, and perturb that quantity by $dx$.

  • $\begingroup$ Are you saying $$ F=\frac{1}{2\mu_0}\left[(B_\text{$\perp$, front})^2- (B_\text{$\perp$, behind} )^2\right]$$ is incorrect? $\endgroup$
    – user36790
    Commented Dec 14, 2015 at 18:10
  • $\begingroup$ Yes, that's the one. It has a form like the magnetic energy density, so it can't be exactly correct. The correct form is probably something like: $F=\frac{A}{2\mu_0}\left[(B_\text{$\perp$, front})^2- (B_\text{$\perp$, behind} )^2\right]$ $\endgroup$
    – David
    Commented Dec 14, 2015 at 19:24
  • $\begingroup$ I was talking about 1 meter-square area. $\endgroup$
    – user36790
    Commented Dec 14, 2015 at 19:28
  • $\begingroup$ In that case, sure. The more intrinsic difficulty is how that expression assumes magnetic fields that are constant in space (the magnetic field between current sheets is effectively constant), while the Biot-Savart law explicitly use a magnetic field that varies radially away from the wires. $\endgroup$
    – David
    Commented Dec 14, 2015 at 19:38
  • $\begingroup$ That's why I worked with $\delta\;,$ an infinitesimal distance. $\endgroup$
    – user36790
    Commented Dec 14, 2015 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.