The equation of motion we used for electromagnetic gyration is

$$ m\frac{d\mathbf{v}}{dt} = q(\mathbf{E}+\mathbf{v}\times\mathbf{B}) $$

However, here the magnetic field was assumed to be homogeneous. In reality, this is not the case. The magnetic field often has a gradient, and sometimes these field lines are curved. Due to this, alongside electric drift, gyrating particles also experience a type of magnetic drift. This means that the guiding center of the gyrating particle shifts, or the radius of its gyration orbit changes. We will discuss three types of magnetic drift: gradient drift, general force drift, and curvature drift.

As shown in the figure below, if the gradient of the B-field is upward, the gyro-radius of a particle will decrease upward. Consequently, the gyro-radius will be larger at the bottom compared to the top. Electrons and ions will drift in opposite directions. The gradient of the B-field is perpendicular to the B-field, and the drift is perpendicular to both.

This change in the magnetic field in the upward direction can be described by the following Taylor expansion:

$$ \mathbf{B} = \mathbf{B}_0 + (\mathbf{r}\cdot\nabla)\mathbf{B}_0 $$

where $\mathbf{B}_0$ is the magnetic field at $r=0$, and $\mathbf{r}$ is the distance from the guiding center. The second derivative of the B-field would appear in the expansion’s second term, but all terms from the second derivative onward are considered negligible. The first derivative indicates a directional change in the B-field. Substituting this equation into the first equation above (ignoring the E-field), we get:

$$ m\frac{d\mathbf{v}}{dt} = q(\mathbf{v}\times\mathbf{B}_0) + q[\mathbf{v}\times (\mathbf{r}\cdot\nabla)\mathbf{B}_0] $$

where two types of velocity can be added: gyration velocity and drift velocity: $\mathbf{v}=\mathbf{v}_g+\mathbf{v}_\nabla$. Assuming the drift velocity is much lower than the gyration velocity, we can write

$$ m\frac{d\mathbf{v}}{dt} = q(\mathbf{v}_\nabla\times\mathbf{B}_0) + q[\mathbf{v}_g\times (\mathbf{r}\cdot\nabla)\mathbf{B}_0] $$

In the first term, the gyration velocity can be ignored (since gyration in a homogeneous field is not relevant here), and in the second term, the drift velocity can be ignored because it is very small. Since we are interested in a timescale much longer than a gyration period, we can average over one gyration. In this case, the left side of the above equation will be zero, as over one gyration, the particle accelerates in one direction and decelerates in the opposite direction, canceling each other out. Setting the left side to zero and multiplying both sides by $\mathbf{B}_0/B_0^2$ (similar to the calculation for electric drift) proves that

$$ \mathbf{v}_\nabla = \frac{1}{B_0^2} \langle (\mathbf{v}_g \times (\mathbf{r}\cdot\nabla)\mathbf{B}_0) \times \mathbf{B}_0 \rangle $$

where the angle brackets indicate averaging over one gyration period. Assuming the B-field is only along the x-axis, we get

$$ \mathbf{v}_\nabla = -\frac{1}{B_0} \left\langle \mathbf{v}_g x \frac{dB_0}{dx} \right\rangle $$

where $x=x_0+r_g \sin\omega_g t$, and $\mathbf{v}_g$ can be obtained from its first derivative. Substituting these two relations into the above equation gives the two components of gradient drift velocity as follows:

$$ v_{\nabla x} = -\frac{v_\perp r_g}{B_0} \left\langle \sin\omega_g t \cos\omega_g t \frac{dB_0}{dx} \right\rangle $$

$$ v_{\nabla y} = \frac{v_\perp r_g}{B_0} \left\langle \sin^2\omega_g t \frac{dB_0}{dx} \right\rangle $$

In the x-component, the time average will be zero because there is a sine-cosine product term in its relation. And we know the time average of the sine-squared term is 1/2. Therefore,

$$ \mathbf{v}_\nabla = \pm \hat{j} \frac{v_\perp r_g}{2B_0} \frac{\partial B_0}{\partial x} $$

where $\hat{j}$ is the unit vector along the y-axis. The plus-minus sign is used for opposite charges. For simplicity, the B-field is assumed along the x-axis here. If the B-field could be in any direction, the actual equation would be

$$ \mathbf{v}_\nabla = \frac{m v_\perp^2}{2qB^3} (\mathbf{B}\times \nabla B) $$

From this relation, it is clear that when there is a gradient in the magnetic field, a gradient drift occurs that is perpendicular to both the magnetic field and its gradient. Since there is a charge $q$, the drift will be in opposite directions for opposite charges. The drift velocity is proportional to the kinetic energy of the particle. If the energy is higher, the drift is greater, as particles with more energy have a larger gyro-radius and thus experience more inhomogeneity in the B-field. The further a particle moves, the more varied the B-field it will encounter.

Like polarization drift, since the drifts for the two charges are in opposite directions, a transverse current is generated. The gradient drift current is

$$ \mathbf{j}_\nabla = n_e e (\mathbf{v}_{\nabla i} - \mathbf{v}_{\nabla e}) = \frac{n_e(\mu_i+\mu_e)}{B^2} (\mathbf{B}\times \nabla B) $$

where the magnetic moment

$$ \mu = \frac{mv_\perp^2}{2B} = \frac{W_\perp}{B} $$

is the ratio of the particle’s perpendicular kinetic energy and the magnetic field.

The equation for E-cross-B drift is $\mathbf{v}_E=(\mathbf{E}\times\mathbf{B})/B^2$, and if we substitute the relation $\mathbf{F}=q\mathbf{E}$ between the electric field and Coulomb force, we obtain the general force drift velocity

$$ \mathbf{v}_F = \frac{1}{\omega_g} \left(\frac{\mathbf{F}}{m}\times\frac{\mathbf{B}}{B}\right) $$

This equation can describe many other drift velocities. When the drift velocity is much lower than the gyro-velocity, various force terms can be substituted in the above equation to derive drift velocity. The force terms for gradient drift, polarization drift, and gravitational drift are shown sequentially below.

$$ \mathbf{F}_\nabla = -\mu \nabla B \\ \mathbf{F}_P = -m \frac{d\mathbf{E}}{dt} \\ \mathbf{F}_G = -m \mathbf{g} $$

Since gravitational force is much weaker than all other forces, gravitational drift becomes significant only on the Sun’s surface in the solar system and can otherwise be ignored.

From the general force drift equation, it is clear that for all forces except the Coulomb force, the drift velocity depends on the sign of the charge. In the case of Coulomb force, the charge term in $\omega_g$ cancels with the charge in the Coulomb force relation, and no charge term remains in the final equation. Thus, only drifts produced by forces other than Coulomb force result in transverse current. In each case, the drift velocity depends on the mass of the carrier particle, so the current varies for different carriers.

When the magnetic field is not homogeneous, gradient drift occurs, and if the magnetic field line is curved, curvature drift also arises. Due to the particle’s velocity along the B-field, it experiences a type of centrifugal force:

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

where $\mathbf{R}_c$ is the radius of local curvature. The centrifugal force experienced by a gyrating particle around the curved B-field line is illustrated below.

Substituting the above equation into the general force drift relation gives

$$ \mathbf{v}_R = \frac{mv_\parallel^2}{q} \frac{\mathbf{R}_c\times\mathbf{B}}{R_c^2 B^2} $$

That is, curvature drift depends on the particle’s parallel kinetic energy $W_\parallel=mv_\parallel^2/2$. This drift velocity is perpendicular to both the magnetic field line and its curvature radius (cross product). Here, too, a transverse current is generated. The curvature drift current is

$$ \mathbf{j}_R = n_e e (\mathbf{v}_{Ri} - \mathbf{v}_{Re} = \frac{2n_e(W_{i\parallel}+W_{e\parallel})}{R_c^2B^2} (\mathbf{R_c\times\mathbf{B}}) $$

which aligns with the direction of the curvature drift velocity, i.e., perpendicular to both the B-field and its curvature.

If the magnetic field is symmetric, then $-\nabla B = (B/R_c^2)\mathbf{R}_c$, and in this case, gradient and curvature drift can be combined to give an equation for total magnetic drift:

$$ \mathbf{v}_B = \mathbf{v}_R + \mathbf{v}_\nabla = (v_\parallel^2+\frac{1}{2}v_\perp^2) \frac{\mathbf{B}\times\nabla B}{\omega_g B^2} $$

which is very important for Earth’s magnetosphere. This total magnetic drift is responsible for the ring current found in our magnetosphere.