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Chapter 3 • Theory & Derivations

Guiding Center Drifts in Inhomogeneous Fields & Adiabatic Invariants

Rigorous treatment of charged particle dynamics in spatially inhomogeneous magnetic fields: guiding center perturbation theory, derivation of the gradient-B drift velocity, centrifugal curvature drift, combined vacuum field drift, magnetic gradient parallel to B and the longitudinal mirror force, rigorous proof of the first adiabatic invariant (magnetic moment constancy), loss cone physics and magnetic mirror confinement, longitudinal action second invariant, drift flux third invariant, and Van Allen radiation belt trapping.

§3.1 The Inhomogeneous Guiding Center Approximation & Spatial Scale Separation

1. Spatial Inhomogeneity and Scale Separation

In realistic astrophysical and laboratory magnetic geometries (such as tokamaks, stellarators, and planetary dipoles), magnetic fields are non-uniform in space. Exact particle trajectories cannot be integrated analytically. However, when the magnetic field varies slowly across the dimensions of a single Larmor orbit:

$$\epsilon \equiv \frac{r_L}{L_B} = \frac{r_L}{|\nabla B / B|} \ll 1$$

we can employ the guiding center approximation. The particle motion separates into two disparate spatial and temporal scales:

  • A fast, quasi-periodic circular gyration at frequency $\omega_c$ with radius $r_L$.
  • A slow, secular drift of the gyration center (the guiding center $\vec{R}$) across and along magnetic field lines.

§3.2 Transverse Magnetic Gradients: Derivation of the Gradient-B Drift Velocity v_gradB

1. Physical Mechanism of Gradient Drift

Consider a magnetic field directed along $\hat{z}$ whose strength increases along $\hat{y}$: $\vec{B} = B(y)\hat{z}$, with $\nabla B = \frac{dB}{dy}\hat{y}$.

As a charged particle gyrates in the $xy$-plane, its instantaneous Larmor radius $r_L(y) = \frac{m v_\perp}{|q| B(y)}$ is smaller at larger $y$ (stronger $B$) and larger at smaller $y$ (weaker $B$). This asymmetric curvature prevents the orbit from closing into a circle, creating a steady lateral drift perpendicular to both $\vec{B}$ and $\nabla B$.

2. Mathematical Derivation

Taylor expand the magnetic field around the guiding center $\vec{R}_0 = (x_0, y_0, z_0)$:

$$\vec{B}(\vec{r}) \approx \vec{B}_0 + (\vec{r} - \vec{R}_0)\cdot\nabla\vec{B} = \left( B_0 + y \frac{\partial B}{\partial y} \right)\hat{z}$$

The particle's unperturbed circular orbit around the guiding center is:

$$x(t) = r_L \sin(\omega_c t), \quad y(t) = \pm r_L \cos(\omega_c t)$$ $$v_x(t) = v_\perp \cos(\omega_c t), \quad v_y(t) = \mp v_\perp \sin(\omega_c t)$$

The instantaneous Lorentz force along $y$ is $F_y = -q v_x B(y)$. Taking the time average of $F_y$ over one complete gyration period $\tau_c = 2\pi / \omega_c$:

$$\langle F_y \rangle = -q \langle v_x B(y) \rangle = -q \left\langle v_\perp \cos(\omega_c t) \left[ B_0 + \left(\pm r_L \cos(\omega_c t)\right)\frac{\partial B}{\partial y} \right] \right\rangle$$

Noting that $\langle \cos(\omega_c t) \rangle = 0$ and $\langle \cos^2(\omega_c t) \rangle = \frac{1}{2}$:

$$\langle F_y \rangle = \mp q v_\perp r_L \frac{1}{2} \frac{\partial B}{\partial y} = \mp \frac{1}{2} q v_\perp \left( \frac{m v_\perp}{|q| B_0} \right) \frac{\partial B}{\partial y} = -\frac{1}{2} \frac{m v_\perp^2}{B_0} \frac{\partial B}{\partial y}$$

In coordinate-free vector notation, the net time-averaged transverse force is:

$$\langle \vec{F}_{\nabla B} \rangle = -\mu \nabla B = -\frac{m v_\perp^2}{2 B} \nabla B$$

where $\mu \equiv \frac{m v_\perp^2}{2 B}$ is the particle's magnetic dipole moment. Substituting this force into the general guiding center force drift formula $\vec{v}_F = \frac{1}{q}\frac{\vec{F}\times\vec{B}}{B^2}$:

$$\vec{v}_{\nabla B} = \frac{1}{q} \frac{(-\mu \nabla B) \times \vec{B}}{B^2} = \frac{\mu}{q B^2} (\vec{B} \times \nabla B) = \frac{m v_\perp^2}{2 q B^3} (\vec{B} \times \nabla B)$$

Because of the explicit factor of $q$, ions and electrons drift in opposite directions across magnetic gradients!

§3.3 Curved Magnetic Field Lines: Centrifugal Acceleration & Derivation of Curvature Drift v_c

1. Centrifugal Acceleration in Curved Geometry

Consider magnetic field lines with local radius of curvature $\vec{R}_c$, pointing from the field line toward the center of curvature:

$$\frac{\vec{R}_c}{R_c^2} = -(\hat{b}\cdot\nabla)\hat{b}$$

where $\hat{b} = \vec{B} / B$ is the unit vector along the field. A particle moving along the field with parallel velocity $v_\parallel$ experiences a centrifugal force directed outward:

$$\vec{F}_c = \frac{m v_\parallel^2}{R_c^2} \vec{R}_c$$

2. The Curvature Drift Velocity

Substituting this centrifugal force into the general guiding center force drift formula:

$$\vec{v}_c = \frac{1}{q} \frac{\vec{F}_c \times \vec{B}}{B^2} = \frac{m v_\parallel^2}{q B^2} \frac{\vec{R}_c \times \vec{B}}{R_c^2}$$

Key physical properties of curvature drift:

  • $\vec{v}_c$ depends quadratically on parallel velocity $v_\parallel^2$.
  • $\vec{v}_c$ is inversely proportional to charge $q$: ions and electrons drift in opposite directions, driving a net electric current.

§3.4 Combined Vacuum Curvature and Gradient Drift & Charge Separation Currents

1. The Vacuum Field Relation

In a current-free vacuum magnetic field, Ampère's law demands $\nabla \times \vec{B} = 0$. In curvilinear coordinates with cylindrical symmetry ($\vec{B} = B_\phi(r)\hat{\phi}$):

$$(\nabla \times \vec{B})_z = \frac{1}{r}\frac{\partial(r B_\phi)}{\partial r} = 0 \implies B_\phi(r) \propto \frac{1}{r}$$

Consequently:

$$\frac{\nabla B}{B} = -\frac{\vec{R}_c}{R_c^2} \implies \frac{\vec{R}_c \times \vec{B}}{R_c^2} = \frac{\vec{B} \times \nabla B}{B}$$

2. Total Combined Guiding Center Drift

Summing the gradient-B drift and curvature drift in any vacuum magnetic field:

$$\vec{v}_D = \vec{v}_{\nabla B} + \vec{v}_c = \frac{m v_\perp^2}{2 q B^3}(\vec{B}\times\nabla B) + \frac{m v_\parallel^2}{q B^3}(\vec{B}\times\nabla B)$$ $$\vec{v}_D = \frac{m}{q B^3} \left( v_\parallel^2 + \frac{1}{2}v_\perp^2 \right) (\vec{B} \times \nabla B)$$

In a toroidal magnetic field (such as a simple torus without rotational transform), this combined drift forces ions vertically upward and electrons vertically downward. This catastrophic charge separation sets up a vertical electric field $\vec{E}$, which subsequently drives an outward $\vec{E}\times\vec{B}$ drift that dumps the entire plasma into the outer chamber wall—necessitating helical twisted magnetic fields (tokamaks/stellarators) to short-circuit the charge buildup.

§3.5 Longitudinal Magnetic Gradients: The Magnetic Mirror Force F_parallel = -mu grad_parallel B

1. Magnetic Convergence & Gauss's Law

Consider a cylindrically symmetric magnetic field whose strength increases along the axis of symmetry ($z$-axis): $\frac{\partial B_z}{\partial z} > 0$. Because magnetic field lines must converge as $B$ intensifies, Gauss's law for magnetism $\nabla \cdot \vec{B} = 0$ requires a non-zero radial magnetic field component $B_r$:

$$\nabla \cdot \vec{B} = \frac{1}{r}\frac{\partial(r B_r)}{\partial r} + \frac{\partial B_z}{\partial z} = 0$$

Near the axis of symmetry where $\frac{\partial B_z}{\partial z}$ is approximately independent of $r$:

$$\frac{\partial(r B_r)}{\partial r} \approx -r \frac{\partial B_z}{\partial z} \implies r B_r \approx -\frac{r^2}{2}\frac{\partial B_z}{\partial z} \implies B_r(r, z) \approx -\frac{r}{2}\frac{\partial B_z}{\partial z}$$

2. The Longitudinal Mirror Restoring Force

The particle's azimuthal cyclotron gyration velocity $v_\theta = \mp v_\perp$ couples with this radial magnetic field $B_r$ to produce a Lorentz force along the $z$-axis:

$$F_z = q(\vec{v}\times\vec{B})_z = q(v_\theta B_r - v_r B_\theta) = q v_\theta B_r$$

Substituting $B_r = -\frac{r_L}{2}\frac{\partial B_z}{\partial z}$ and $v_\theta = -\frac{q}{|q|}v_\perp$:

$$F_z = q \left( -\frac{q}{|q|}v_\perp \right) \left( -\frac{r_L}{2}\frac{\partial B_z}{\partial z} \right) = -\frac{1}{2} |q| v_\perp \left( \frac{m v_\perp}{|q| B_z} \right) \frac{\partial B_z}{\partial z} = -\frac{m v_\perp^2}{2 B_z} \frac{\partial B_z}{\partial z}$$

Recalling the magnetic moment $\mu = \frac{m v_\perp^2}{2 B}$, the longitudinal force is:

$$F_\parallel = -\mu \frac{\partial B}{\partial s} = -\mu \nabla_\parallel B$$

This is the fundamental magnetic mirror force. Because $\mu > 0$ and the negative sign is universal, the mirror force always repels charged particles away from regions of stronger magnetic field back toward regions of weaker magnetic field.

§3.6 The First Adiabatic Invariant mu = const, Loss Cone Angle & Magnetic Bottles

1. Invariance of the Magnetic Moment

The total kinetic energy of a charged particle in a static magnetic field is strictly conserved:

$$E = E_\parallel + E_\perp = \frac{1}{2}m v_\parallel^2 + \frac{1}{2}m v_\perp^2 = \frac{1}{2}m v_\parallel^2 + \mu B = \text{const}$$

Differentiating total energy with respect to time along the trajectory:

$$\frac{dE}{dt} = m v_\parallel \frac{dv_\parallel}{dt} + \frac{d(\mu B)}{dt} = v_\parallel F_\parallel + \mu \frac{dB}{dt} + B \frac{d\mu}{dt} = 0$$

Using $F_\parallel = -\mu \frac{\partial B}{\partial s}$ and noting that along the orbit $\frac{dB}{dt} = v_\parallel \frac{\partial B}{\partial s}$:

$$v_\parallel \left(-\mu \frac{\partial B}{\partial s}\right) + \mu \left(v_\parallel \frac{\partial B}{\partial s}\right) + B \frac{d\mu}{dt} = 0 \implies B \frac{d\mu}{dt} = 0$$

Therefore, the magnetic moment $\mu$ is an exact adiabatic invariant:

$$\mu \equiv \frac{m v_\perp^2}{2 B} = \text{const}$$

2. The Magnetic Loss Cone

As a particle travels into a converging magnetic mirror with minimum field $B_{\text{min}}$ and maximum field $B_{\text{max}}$ (mirror ratio $R_m \equiv B_{\text{max}} / B_{\text{min}}$):

$$v_\perp^2(s) = v_{\perp,0}^2 \frac{B(s)}{B_{\text{min}}}$$

By conservation of total energy $v^2 = v_\parallel^2(s) + v_\perp^2(s) = v_0^2$:

$$v_\parallel^2(s) = v_0^2 - v_{\perp,0}^2 \frac{B(s)}{B_{\text{min}}}$$

A particle will be reflected at a turning point ($v_\parallel = 0$) if and only if $B(s)$ reaches a value where $v_\perp^2 = v_0^2$ before reaching $B_{\text{max}}$. The threshold pitch angle $\theta$ at the midplane ($B = B_{\text{min}}$) defining the boundary between trapped and lost particles is the loss cone angle $\theta_m$:

$$\sin^2 \theta_m = \frac{v_{\perp,0}^2}{v_0^2} = \frac{B_{\text{min}}}{B_{\text{max}}} = \frac{1}{R_m} \implies \theta_m = \arcsin\left( \frac{1}{\sqrt{R_m}} \right)$$

Particles with initial pitch angles $\theta < \theta_m$ lie inside the loss cone and escape out the ends of the mirror.

§3.7 Higher Adiabatic Invariants: Longitudinal Action J, Flux Invariant Phi & Radiation Belts

1. The Hierarchy of Adiabatic Invariants

Hamiltonian mechanics dictates that whenever a periodic motion has action integral $J = \oint p\,dq$, the action is an adiabatic invariant under slow perturbations:

  1. First Invariant $\mu$ (Magnetic Moment): Associated with fast cyclotron gyration (period $\tau_c \sim 10^{-6}\text{ s}$): $$\mu = \frac{m v_\perp^2}{2 B} = \text{const}$$
  2. Second Invariant $J$ (Longitudinal Action): Associated with periodic bounce motion between mirror points (period $\tau_b \sim 10^{-2}\text{ s}$): $$J = \oint v_\parallel ds = \text{const}$$
  3. Third Invariant $\Phi$ (Magnetic Flux Drift Invariant): Associated with the slow azimuthal drift of guiding centers around a closed magnetic surface (period $\tau_d \sim 10^2\text{ s}$): $$\Phi = \oint \vec{A} \cdot d\vec{l} = \int \vec{B} \cdot d\vec{S} = \text{const}$$

2. Planetary Magnetospheres & The Van Allen Belts

The Earth's dipolar magnetic field naturally traps energetic protons and electrons in the Van Allen radiation belts. Trapped particles simultaneously execute:

  • Fast cyclotron gyration around dipole field lines ($\tau_c \sim 1\;\mu\text{s}$).
  • North-South bouncing between northern and southern auroral mirror points ($\tau_b \sim 0.1\text{ s}$).
  • Slow longitudinal drift around the Earth (electrons east, ions west) creating the geomagnetic ring current ($\tau_d \sim 10\text{ minutes}$).

Honors Examination Worked Problems & Solutions

Rigorous step-by-step mathematical proofs and solutions to university degree examination problems.

Advanced Honors Exam Problem Example 3.1: Complete Derivation of Gradient-B Drift from Taylor Expansion of Gyration Force over Orbit

Consider a particle of mass $m$ and charge $q$ moving in a magnetic field directed along $\hat{z}$ with a constant gradient along $\hat{y}$: $\vec{B}(y) = [B_0 + y (dB/dy)]\hat{z}$. (a) Write down the instantaneous Lorentz force components $F_x(t)$ and $F_y(t)$ along the unperturbed circular orbit $x(t) = r_L \sin(\omega_c t), y(t) = r_L \cos(\omega_c t)$ for a positive ion. (b) Evaluate the orbit-averaged forces $\langle F_x \rangle$ and $\langle F_y \rangle$ over one full gyration period $\tau_c = 2\pi / \omega_c$. (c) Apply the general force drift equation to compute the drift velocity vector $\vec{v}_D$ and show that it identically matches the formula $\vec{v}_{\nabla B} = \frac{m v_\perp^2}{2 q B^3}(\vec{B}\times\nabla B)$.

Full Rigorous Analytical Solution

(a) Instantaneous Force Components: For a positive ion ($q > 0$), the unperturbed gyration orbit around guiding center $(0,0)$ is:

$$x(t) = r_L \sin(\omega_c t), \quad y(t) = r_L \cos(\omega_c t)$$
$$v_x(t) = \dot{x} = r_L \omega_c \cos(\omega_c t) = v_\perp \cos(\omega_c t)$$
$$v_y(t) = \dot{y} = -r_L \omega_c \sin(\omega_c t) = -v_\perp \sin(\omega_c t)$$

The magnetic field at the particle's position is:

$$\vec{B}(y) = \left( B_0 + r_L \cos(\omega_c t) \frac{dB}{dy} \right) \hat{z}$$

The Lorentz force is $\vec{F} = q(\vec{v}\times\vec{B}) = q(v_y B_z \hat{x} - v_x B_z \hat{y})$:

$$F_x(t) = q v_y(t) B_z(y(t)) = -q v_\perp \sin(\omega_c t) \left[ B_0 + r_L \cos(\omega_c t)\frac{dB}{dy} \right]$$
$$F_y(t) = -q v_x(t) B_z(y(t)) = -q v_\perp \cos(\omega_c t) \left[ B_0 + r_L \cos(\omega_c t)\frac{dB}{dy} \right]$$

(b) Orbit-Averaged Forces: Integrate over one period $\tau_c = 2\pi / \omega_c$:

  1. For $\langle F_x \rangle$:
$$\langle F_x \rangle = -q v_\perp B_0 \langle \sin(\omega_c t) \rangle - q v_\perp r_L \frac{dB}{dy} \langle \sin(\omega_c t)\cos(\omega_c t) \rangle$$

Since $\langle \sin(\omega_c t) \rangle = 0$ and $\langle \sin(\omega_c t)\cos(\omega_c t) \rangle = \frac{1}{2}\langle \sin(2\omega_c t) \rangle = 0$:

$$\langle F_x \rangle = 0$$
  1. For $\langle F_y \rangle$:
$$\langle F_y \rangle = -q v_\perp B_0 \langle \cos(\omega_c t) \rangle - q v_\perp r_L \frac{dB}{dy} \langle \cos^2(\omega_c t) \rangle$$

Since $\langle \cos(\omega_c t) \rangle = 0$ and $\langle \cos^2(\omega_c t) \rangle = \frac{1}{2}$:

$$\langle F_y \rangle = -\frac{1}{2} q v_\perp r_L \frac{dB}{dy}$$

Substitute Larmor radius $r_L = \frac{m v_\perp}{q B_0}$:

$$\langle F_y \rangle = -\frac{1}{2} q v_\perp \left( \frac{m v_\perp}{q B_0} \right) \frac{dB}{dy} = -\frac{m v_\perp^2}{2 B_0} \frac{dB}{dy} = -\mu \frac{dB}{dy}$$

where $\mu = \frac{m v_\perp^2}{2 B_0}$.

(c) Drift Velocity Vector Evaluation: The net effective force is $\langle \vec{F} \rangle = -\mu \frac{dB}{dy} \hat{y} = -\mu \nabla B$. Using the general force drift equation:

$$\vec{v}_D = \frac{1}{q} \frac{\langle \vec{F} \rangle \times \vec{B}}{B_0^2} = \frac{1}{q B_0^2} \left( -\mu \frac{dB}{dy}\hat{y} \times B_0 \hat{z} \right)$$

Since $\hat{y} \times \hat{z} = \hat{x}$:

$$\vec{v}_D = -\frac{\mu B_0}{q B_0^2}\frac{dB}{dy}\hat{x} = -\frac{\mu}{q B_0}\frac{dB}{dy}\hat{x} = -\frac{m v_\perp^2}{2 q B_0^2}\frac{dB}{dy}\hat{x}$$

Now compute using the standard vector formula $\vec{v}_{\nabla B} = \frac{m v_\perp^2}{2 q B^3}(\vec{B}\times\nabla B)$:

$$\vec{B} \times \nabla B = (B_0 \hat{z}) \times \left( \frac{dB}{dy}\hat{y} \right) = B_0 \frac{dB}{dy} (\hat{z}\times\hat{y}) = -B_0 \frac{dB}{dy}\hat{x}$$
$$\vec{v}_{\nabla B} = \frac{m v_\perp^2}{2 q B_0^3}\left( -B_0 \frac{dB}{dy}\hat{x} \right) = -\frac{m v_\perp^2}{2 q B_0^2}\frac{dB}{dy}\hat{x}$$

The two derivations match identically, proving the theorem rigorously from first-order perturbation mechanics.

Final Answer & Physical Verification

Complete rigorous derivation and proof detailed above.

Final Answer & Physical Insight

Complete rigorous derivation and proof detailed above.

Advanced Honors Exam Problem Example 3.2: Magnetic Mirror Loss Cone Fraction and Critical Pitch Angle for Fusion Confinement Geometry

A linear magnetic mirror machine has a midplane field $B_{\text{min}} = 0.50\text{ Tesla}$ and throat coils generating $B_{\text{max}} = 2.50\text{ Tesla}$. (a) Determine the mirror ratio $R_m$ and calculate the critical loss cone half-angle $\theta_m$ in degrees. (b) Assuming an isotropic velocity distribution function $f(\vec{v}) = f(v)$, compute the fraction of particles $F_{\text{loss}}$ that lie within the double loss cone and escape out either end. (c) If a population of deuterium ions ($M_i = 3.34\times 10^{-27}\text{ kg}$) is injected at the midplane with total kinetic energy $E = 10.0\text{ keV}$ at pitch angle $\theta = 45^\circ$, determine their turning point magnetic field $B_{\text{turn}}$ and verify that they are magnetically trapped.

Full Rigorous Analytical Solution

(a) Mirror Ratio and Loss Cone Angle: The mirror ratio is:

$$R_m = \frac{B_{\text{max}}}{B_{\text{min}}} = \frac{2.50\text{ T}}{0.50\text{ T}} = 5.0$$

The critical loss cone angle satisfies:

$$\sin^2 \theta_m = \frac{1}{R_m} = \frac{1}{5.0} = 0.20$$
$$\sin\theta_m = \sqrt{0.20} \approx 0.4472$$
$$\theta_m = \arcsin(0.4472) \approx 26.565^\circ \approx 26.57^\circ$$

(b) Loss Fraction for Isotropic Distribution: For an isotropic velocity distribution, the velocity space solid angle element is $d\Omega = 2\pi \sin\theta d\theta$. A particle escapes if its pitch angle falls within the forward loss cone ($0 \le \theta < \theta_m$) or the backward loss cone ($\pi - \theta_m < \theta \le \pi$). The solid angle of one loss cone is:

$$\Omega_{\text{cone}} = \int_0^{2\pi} d\phi \int_0^{\theta_m} \sin\theta d\theta = 2\pi [1 - \cos\theta_m]$$

The total solid angle for both escape cones is $2 \Omega_{\text{cone}} = 4\pi [1 - \cos\theta_m]$. The fraction of lost particles is:

$$F_{\text{loss}} = \frac{2\Omega_{\text{cone}}}{4\pi} = 1 - \cos\theta_m$$

Since $\sin\theta_m = 1/\sqrt{R_m}$:

$$\cos\theta_m = \sqrt{1 - \sin^2\theta_m} = \sqrt{1 - \frac{1}{R_m}} = \sqrt{1 - 0.20} = \sqrt{0.80} \approx 0.8944$$
$$F_{\text{loss}} = 1 - 0.8944 = 0.1056 = 10.56\%$$

Thus, approximately $10.6\%$ of an isotropic plasma escapes immediately on the first pass, with the remaining $89.4\%$ trapped.

(c) Turning Point of Injected Deuterium Ions: Initial pitch angle is $\theta_0 = 45^\circ$. Since $\theta_0 = 45^\circ > \theta_m = 26.57^\circ$, the ions lie well outside the loss cone and must be trapped! At the midplane:

$$v_{\perp,0}^2 = v_0^2 \sin^2(45^\circ) = 0.50 v_0^2$$

The magnetic moment is:

$$\mu = \frac{m v_{\perp,0}^2}{2 B_{\text{min}}} = \frac{E \sin^2\theta_0}{B_{\text{min}}} = \frac{(10.0\text{ keV})(0.50)}{0.50\text{ T}} = 10.0\text{ keV/T}$$

At the turning point, all kinetic energy is converted into perpendicular energy ($v_\parallel = 0, E_\perp = E$):

$$E = \mu B_{\text{turn}} \implies B_{\text{turn}} = \frac{E}{\mu} = \frac{10.0\text{ keV}}{10.0\text{ keV/T}} = 1.00\text{ Tesla}$$

Because $B_{\text{turn}} = 1.00\text{ T} < B_{\text{max}} = 2.50\text{ T}$, the deuterium ions reflect cleanly at the location where $B = 1.00\text{ T}$, executing stable harmonic bounce oscillations between the two mirror throats.

Final Answer & Physical Verification

Complete rigorous derivation and proof detailed above.

Final Answer & Physical Insight

Complete rigorous derivation and proof detailed above.

Advanced Honors Exam Problem Example 3.3: Invariance of Magnetic Moment in Slowly Time-Varying Magnetic Fields

Consider a charged particle of mass $m$ and charge $q$ gyrating in a spatially uniform magnetic field that increases slowly with time: $\vec{B}(t) = B(t)\hat{z}$, with $\frac{1}{B}\frac{dB}{dt} \ll \omega_c$. (a) From Faraday's law of induction, calculate the induced azimuthal electric field $E_\theta$ around the circular gyro-orbit of radius $r_L$. (b) Compute the net work done by $E_\theta$ on the particle per cyclotron orbit and calculate the time rate of change of perpendicular kinetic energy $\frac{dE_\perp}{dt}$. (c) Rigorously prove that $\frac{d}{dt}\left( \frac{E_\perp}{B} \right) = 0$, confirming the adiabatic invariance of $\mu = m v_\perp^2 / (2B)$.

Full Rigorous Analytical Solution

(a) Induced Azimuthal Electric Field: By Faraday's law of electromagnetic induction in integral form:

$$\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}\int \vec{B}\cdot d\vec{S}$$

For a circular contour of radius $r_L$ centered on the guiding center:

$$\oint \vec{E} \cdot d\vec{l} = E_\theta (2\pi r_L) = -\frac{d}{dt}(\pi r_L^2 B) = -\pi r_L^2 \frac{dB}{dt}$$

(since $r_L$ changes slowly on the cyclotron timescale $\tau_c$). Solving for the induced electric field:

$$E_\theta = -\frac{r_L}{2}\frac{dB}{dt}$$

(b) Work Done per Orbit & Rate of Energy Gain: The work done by the electric field on the particle during one complete gyration is:

$$\Delta E_\perp = \oint q \vec{E}\cdot d\vec{l} = q E_\theta (\pm 2\pi r_L)$$

Because the sense of gyration of the particle ($q > 0$ counter-clockwise, $q < 0$ clockwise) aligns with the accelerating torque of the induced electric field:

$$\Delta E_\perp = |q| |E_\theta| (2\pi r_L) = |q| \left( \frac{r_L}{2}\frac{dB}{dt} \right)(2\pi r_L) = \pi |q| r_L^2 \frac{dB}{dt}$$

Substitute the Larmor radius $r_L = \frac{v_\perp}{\omega_c} = \frac{m v_\perp}{|q| B}$:

$$\pi |q| r_L^2 = \pi |q| \left( \frac{m v_\perp}{|q| B} \right)^2 = \frac{\pi m^2 v_\perp^2}{|q| B^2} = \frac{2\pi m}{|q| B} \left( \frac{1}{2}m v_\perp^2 \frac{1}{B} \right) = \tau_c \left( \frac{E_\perp}{B} \right)$$

where $\tau_c = \frac{2\pi}{\omega_c} = \frac{2\pi m}{|q| B}$ is the cyclotron period. The average time rate of change of perpendicular energy is:

$$\frac{dE_\perp}{dt} = \frac{\Delta E_\perp}{\tau_c} = \frac{E_\perp}{B} \frac{dB}{dt}$$

(c) Proof of Invariance of $\mu$: Differentiate $\mu = \frac{E_\perp}{B}$ with respect to time using the quotient rule:

$$\frac{d\mu}{dt} = \frac{d}{dt}\left( \frac{E_\perp}{B} \right) = \frac{1}{B}\frac{dE_\perp}{dt} - \frac{E_\perp}{B^2}\frac{dB}{dt}$$

Substitute the result $\frac{dE_\perp}{dt} = \frac{E_\perp}{B}\frac{dB}{dt}$:

$$\frac{d\mu}{dt} = \frac{1}{B}\left( \frac{E_\perp}{B}\frac{dB}{dt} \right) - \frac{E_\perp}{B^2}\frac{dB}{dt} = \frac{E_\perp}{B^2}\frac{dB}{dt} - \frac{E_\perp}{B^2}\frac{dB}{dt} = 0$$

This rigorously proves that $\mu = \text{const}$ is invariant under slow temporal variations of the magnetic field! As $B$ increases, the perpendicular energy increases proportionally ($E_\perp \propto B$) while the Larmor radius compresses ($r_L \propto B^{-1/2}$), which is the operational basis of betatron acceleration and adiabatic magnetic compression in fusion devices.

Final Answer & Physical Verification

Complete rigorous derivation and proof detailed above.

Final Answer & Physical Insight

Complete rigorous derivation and proof detailed above.