Magnetostatics and Newton's third law Here's a simple question that has been driving me to distraction.  Having issues uploading images, so I'm just going to have to describe the system.  Let's say we have a solenoid oriented along the $z$ axis, and it is generating a magnetic field pointing largely in $+\hat{z}$.  We also have a wire sitting a short distance below the solenoid, oriented along the $y$ axis, and carrying a current in the $+\hat{y}$ direction.  The solenoid will then produce a $J\times B$ force on the wire pointing in the $+\hat{x}$ direction.
But shouldn't the wire then produce a force on the solenoid in the $-\hat{x}$ direction?  I just can't see how.  At the solenoid location, the field from the wire will point largely in the $+\hat{x}$ direction.  I don't see how there is a $J\times B$ that can resolve to a direction of $-\hat{x}$.  I think I could be convinced that the wire puts a torque on the solenoid, but that's not the same thing as an equal and opposite force.  Where is the equal and opposite force?
 A: Exact calculation, except with a missing factor of 2, and a simplified geometry
For simplicity, imagine a current $I$ travels in the $+\hat y$ direction from a point $(0,-d,0)$ to $(0,d,0)$ within a solenoid of radius $R \gg d$. The current entires and leaves this wire segment from two other wires wire directed parallel to the $\hat z$ axis direction and extending off to negative infinity in the $z$ direction. The solenoid itself extends infinitely in either $\pm \hat z$ direction. Because $R \gg d$, to a good approximation the field on the three straight wires is a constant $B_{solenoid} \hat z$. Furthermore, at the solenoid's scale, the magnetic fields from the two wires parallel to $\hat z$ cancel out. So, the only non-negligible forces are those between the solenoid and the wire of length $2d$ parallel to the $\hat y$-axis.

*

*The easy one: the force from the solenoid on the wire parallel is $F=2dIB_{solenoid}$ in the $+\hat x$ direction.


*The force of the wire on the solenoid can be calculated directly by integrating over the force from each infinitesimal segment of the wire, but because $d \ll R$ these contributions are all nearly equal. So, the Biot-Savart law reduces to
$$ B_{wire} = \frac{2d\mu_0I}{4 \pi} \frac{\hat y \times r'}{|r'|^3} $$
The force this field induces on the solenoid can be found by integrating over the cylindrical surface of the solenoid. Ignoring the skew of the looping wires, the solenoid consists of a sheet of current density (in amps/meter) $J$, related to $B_{solenoid}$ by $B_{solenoid} = \mu_0 J$. The force on a loop of the solenoid at height $z$ of width $dz$, carrying current $J\, dz$, is
$$
\int_0^{2\pi} (J\, dz) (R\, d \theta) \hat \theta \times B_{wire}
= \frac{RF\, dz}{4 \pi (z^2+R^2)^{3/2}} \int_0^{2\pi} \hat \theta \times (\hat y \times r') \, d \theta
$$
Using some vector triple product identities, and that $r' = (R \cos \theta, R \sin \theta, z)$ and $\hat \theta = (-\sin \theta, \cos \theta, 0)$,
$$
\hat \theta \times (\hat y \times r') = \hat y (\hat \theta \cdot r') - r' (\hat \theta \cdot \hat y) = - r' \cos \theta
$$
making the integral over $\theta$ above
$$
- \int_0^{2\pi} (R \cos^2 \theta, R \sin \theta \cos \theta, z) \, d \theta = -(\pi R,0,2\pi z)
$$
and the total expression for the force on the infinitesimal loop
$$
- \frac{RF\, dz}{4 (z^2+R^2)^{3/2}} (R,0,2z)
$$
The total force is thus the integral of this quantity along the infinite extent of the solenoid. The $\hat y$ and $\hat z$ components components are clearly zero (the latter because the integrand is an odd function). The $\hat x$ component is just
$$
- \frac{R^2 F}{4} \int_{-\infty}^\infty \frac{1}{(z^2+R^2)^{3/2}} \, dz
= -\frac{F}{4} \int_{-\infty}^\infty \frac{1}{(u^2 + 1)^{3/2}} \, du
= -\frac{F}{2}
$$
Well, that shows the force to be in the $-\hat x$ direction, as required, but there's a factor of two missing somewhere.
Assuming that the missing factor of two isn't emblematic of a bigger problem, the calculation in part 2 above can be repeated for a solenoid that extends only from $z=0$ to $z=\infty$. While the magnetic field generated by the solenoid is no longer the same, the quantity $F$ can be reinterpreted simply as $F \equiv 2d \mu_0 d I J$, and the force on the solenoid is
$$
 - \frac{RF}{4} \int_0^\infty \frac{(R,0,2z)}{(z^2+R^2)^{3/2}} \, dz
=  - \frac{F}{4} \int_0^\infty \frac{(1,0,2u)}{(u^2+1)^{3/2}} \, du
= \left( - \frac{F}{4}, 0, - \frac{F}{2} \right)
$$
A: If the wire is located a little below a loop of the solenoid (crossing below its center), then its field will have a (+z) component for negative values of x, and a (-z) component for positive values of x.  The side of the loop on the (-x) side will be pushed in the (-x) direction by the field from the wire, and the one on the (+x) side will also be pushed in the (-x) direction.
