Invariance of magnetic moment in slowly-changing magnetic fields (plasma physics) I am trying to understand how the magnetic moment is invariant in slowly changing magnetic fields. There is a proof in the textbook I am using, but I'm stuck on how $-e\int\frac{\partial B}{\partial t}dS$ becomes $\pi r_{L}|e|\dot{B}$ and how the final $\delta(W_{\perp}/B) = 0$ is obtained.
I was thinking of doing some sort of Taylor expansion on the rate of change of B and ignore higher-order terms but I'm just not sure where to go. Any ideas to help on this proof?

 A: 
I am trying to understand how the magnetic moment is invariant in slowly changing magnetic fields.

Try starting with a static magnetic field that only varies spatially along, say, the z-direction.  Imagine a cylindrically symmetric geometry where the magnetic fields are gaining in strength as you move further along positive z.  You can show, from the divergence of the magnetic field that:
$$
\frac{ 1 }{ \rho } \frac{ \partial }{ \partial \rho } \left( \rho B_{\rho} \right) + \frac{ \partial B_{z} }{ \partial z } = 0 \tag{0}
$$
where $\rho$ is the radial component/direction.  If we assume $B_{z}$ is independent of $\rho$, then we can show that:
$$
B_{\rho} = - \frac{ 1 }{ 2 } \rho \ \frac{ \partial B_{z} }{ \partial z } \tag{1}
$$
We can rewrite this into two components, x and y, where we just replace $\rho$ with x and y.  We can approximate the spatial positions as:
$$
\begin{align}
  x & = \rho_{cs} \sin{\Omega_{cs} t} \tag{2a}  \\
  y & = \frac{ q_{s} }{ \lvert q_{s} \rvert } \rho_{cs} \cos{\Omega_{cs} t} \tag{2b}
\end{align}
$$
where $q_{s}$ is the charge of species $s$, $\rho_{cs} = \tfrac{ m_{s} v_{\perp} }{ \lvert q_{s} \rvert B }$ is the gyroradius of species $s$, $m_{s}$ is the mass of species $s$, $B$ is the magnetic field magnitude, $v_{\perp}$ is the velocity orthogonal to $\mathbf{B}$, and $\Omega_{cs} = \tfrac{ \lvert q_{s} \rvert B }{ m_{s} }$ is the cyclotron frequency of species $s$.
We know the z-component of the Lorentz force for such a scenario is given by:
$$
F_{z} = q_{s} \left( v_{x} B_{y} - v_{y} B_{x} \right) = - \frac{ q_{s} }{ 2 } \left( \frac{ \partial B_{z} }{ \partial z } \right) \left( v_{x} \ y - v_{y} \ x \right) \tag{3}
$$
Using Equations 2a and 2b and their derivatives and some arithmetic on Equation 3, we can show that:
$$
F_{z} = - \left( \frac{ \lvert q_{s} \rvert }{ 2 } \Omega_{cs} \rho_{cs}^{2} \right) \left( \frac{ \partial B_{z} }{ \partial z } \right) \tag{4}
$$
where the first term in ()'s is the particle magnetic moment, $\mu_{s}$.  Note that he magnetic moment can also be written as:
$$
\mu_{s} = \frac{ m_{s} \ v_{\perp}^{2} }{ 2 \ B } = \frac{ w_{\perp} }{ B } \tag{5}
$$
The above is for the force parallel to the increasing magnetic field magnitude, i.e., along the z-direction.  There is also an azimuthal force on the particle given by:
$$
F_{\phi} = q_{s} \ v_{z} \ B_{\rho} \tag{6}
$$
We can show that the rate of change of the perpendicular kinetic energy, $w_{\perp}$, is the rate at which work is done by the azimuthal force on the particle, given by:
$$
\frac{ d w_{\perp} }{ dt } = v_{\phi} F_{\phi} = - \left( \frac{ q_{s} }{ \lvert q_{s} \rvert } v_{\perp} \right) \left( q_{s} \ v_{z} \ B_{\rho} \right) \tag{7}
$$
Next we replace $B_{\rho}$ using Equation 1 and assume $\rho \rightarrow \rho_{cs}$ to give:
$$
\frac{ d w_{\perp} }{ dt } = \left( \frac{ m_{s} \ v_{\perp}^{2} }{ 2 \ B } \right) v_{z} \ \frac{ \partial B_{z} }{ \partial z } = \mu_{s} \ v_{z} \ \frac{ \partial B_{z} }{ \partial z } \tag{8}
$$
Recall that we assumed a static magnetic field (i.e., $\tfrac{ \partial \mathbf{B} }{ \partial t } = 0$), which let's us take advantage of the following:
$$
\frac{ d \mathbf{B} }{ dt } = \frac{ \partial \mathbf{B} }{ \partial t } + \mathbf{v} \cdot \nabla \mathbf{B} + \mathbf{v} \left( \nabla \cdot \mathbf{B} \right) = \mathbf{v} \cdot \nabla \mathbf{B} \tag{9}
$$
Now let us look directly at the time rate of change of $\mu_{s}$, which is given by:
$$
\frac{ d \mu_{s} }{ dt } = \frac{ d }{ dt } \left( \frac{ w_{\perp} }{ B } \right) = \frac{ 1 }{ B } \frac{ d w_{\perp} }{ dt } - \left( \frac{ w_{\perp} }{ B^{2} } \right) \frac{ d B }{ dt } \tag{10}
$$
Using Equations 8, 9, and 10 we can show that:
$$
\frac{ d \mu_{s} }{ dt } = \frac{ 1 }{ B } \frac{ d w_{\perp} }{ dt } - \frac{ w_{\perp} \ v_{z} }{ B^{2} } \frac{ \partial B_{z} }{ \partial z } = \frac{ w_{\perp} \ v_{z} }{ B^{2} } \frac{ \partial B_{z} }{ \partial z } - \frac{ w_{\perp} \ v_{z} }{ B^{2} } \frac{ \partial B_{z} }{ \partial z } = 0 \tag{11}
$$
