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I was reading ' Change in B at a current sheet ' in Electricity & Magnetism by Edward M Purcell. this is the concerned excerpt:

Consider a square portion of the sheet, $1~\text{m}$ on a side. The current included is equal to $\mathcal J,$ the length of the current path is $1~\text{m},$ and the average field that acts on this current, assuming that the current is uniformly distributed through the thickness of the sheet, is $\frac{1}{2} ({B_z}^+ + {B_z}^-)\;.$ Therefore the force on this portion of the current distribution is $$\text{Force on $1~\text{m}^2$ of sheet}=\frac{1}{2} ({B_z}^+ + {B_z}^-) \mathcal J$$ ...We can substitute $({B_z}^+ - {B_z}^-)/\mu_0$ for $\mathcal J,$ so that that the force per square meter can be expressed in this way: \begin{align}\text{Force per $\text{m}^2$}&= \left(\frac{{B_z}^+ + {B_z}^-}{2}\right)({B_z}^+ - {B_z}^-)\cdot \frac{1}{\mu_0}\\& =\frac{1}{2\mu_0}[({B_z} ^+)^2- ({B_z}^- )^2]\cdot\end{align} The force is perpendicular to the surface & proportional to the area, like the stress caused by hydrostatic pressure.

I am a bit confused; the magnetic fields ${B_z}^+$ and ${B_z}^-$ are created by the the current in the sheet; how the current itself get affected by the same magnetic field it created? How can the field exert force on the current-carrying sheet as it is itself created by that current?

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Consider a sheet of thickness $t\gt 0$ and normal $\hat y$ with a volume current density of $\vec J=J\hat x.$

The magnetic field can be $\vec B_0$ at the surface $y=0$ and it can be $\vec B_t$ at the surface $y=t$ and in between it can be $\vec B(x,y,z)=\vec B_0+\left(\vec B_t-\vec B_0\right)y/t.$

When you verify that it solves Maxwell you find out that $\vec B_t-\vec B_0=\mu_0Jt\hat z.$

Now, you asked about that average force. If you don't expect single charge to exert a force on itself, then you could compute the force on a charge due to the total field, the one that is $\vec B(x,y,z)=\vec B_0+\left(\vec B_t-\vec B_0\right)y/t.$ And each region of charge has some little charge $dq$ and some velocity $\vec v$ and so there is a little bit of force felt that is $dq \vec v\times \vec B.$ And if you find the total force on all the charges in the region you find $$\vec F=\sum_i dq_i \vec v_i\times \vec B.$$

You can multiply and divide by the volume of each region and get $$\vec F=\sum_i \left(\frac{dq_i \vec v_i}{\Delta x_i\Delta y_i\Delta z_i}\right)\times \vec B\Delta x_i\Delta y_i\Delta z_i.$$

Or using the definition of $\vec J$ and the continuum limit you get $$\vec F\frac{1}{m^2}=\int_{y=0}^{y=t}\vec J\times \vec B(x,y,z)dy,\text{ so}$$

$$\vec F\frac{1}{m^2}=\vec J\times\left( \vec B_0t+\left(\vec B_t-\vec B_0\right)t^2/2t\right).$$ Exactly what you need.

It's the same point as your other question (Force on a layer of charge: How can electric field impart force on the charges which created the field?), if you ignore the force of the moving charge on itself as if it were zero, then you can compute the force on a charge in a region as if it reacts to the total magnetic field for the region, in this case $\vec B(x,y,z)=\vec B_0+\left(\vec B_t-\vec B_0\right)y/t$ and then the force on the sheet is the sum of the force on each part of the sheet which ends up depending only on the current and the field, and then since the sheet has a linearly changing field inside it an a uniform current inside it then it comes out to depend on how much current there is across the whole thickness and the average of the magnetic fields on the two sides.

But really it's just the Lorentz Force.

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  • $\begingroup$ Thanks from the deepest that you responded this time also! $\endgroup$ – user36790 Dec 13 '15 at 13:01
  • $\begingroup$ But I think, you've mistakenly forgotten to include $t$ after integration that is, IMO, the force must be $$\vec F\frac{1}{m^2}=\vec J\times\left( \vec B_0 t+\left(\vec B_t-\vec B_0\right)t^2/2t\right).$$ $\endgroup$ – user36790 Dec 13 '15 at 13:08
  • $\begingroup$ if you ignore the force of the moving charge on itself as if it were zero- hmmm.... don't you think it should be zero? $\endgroup$ – user36790 Dec 13 '15 at 13:21
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I think it's because there's two parallel sheets. The caption for figure 6.23 is The magnetic field between plane-parallel current sheets. Here's the pages concerned:

enter image description here

Even then there's arguably some confusion, because a current doesn't actually "create" a magnetic field. You should be able to work this out by imagining you're a charged particle. When I set you down near an electron, you experience a linear force, and you say the electron has an electric field. But if I throw you past the electron, you also experience a rotational force, and say the electron has a magnetic field too. However your motion doesn't create a magnetic field for the electron. Because all along the electron had an electromagnetic field, and so did you. See Jefimenko for more on that. Whilst Purcell is described as the bible of electromagnetism, IMHO it does cause issues.

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    $\begingroup$ I don't think Purcell discusses the para for fig 6.23; see he uses ${B_z}^+$ & ${B_z}^-$. So, he is rather talking about 6.21, IMO. $\endgroup$ – user36790 Dec 9 '15 at 14:29
  • $\begingroup$ There is no force on a single wire, or on a single sheet. $\endgroup$ – John Duffield Dec 10 '15 at 8:05

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