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Consider an infinitely long, cylindrical wire which goes through the $z$-axis with current, $I$, and radius $a$. Hence, in cylindrical coordinates, the magnetic field is given by $$\begin{aligned} \mathbf{B}(\rho) &= B_\varphi(\rho)\mathbf{\hat{\varphi}} \\ &= \frac{\mu_0 I}{2\pi\rho}\mathbf{\hat{\varphi}},\ \mathrm{for}\ \rho\ge a, \end{aligned}$$ where $\mu_0$ is the permeability of free space and this can be derived using Ampere's law.

Assume the following:

  • The wire is rigid and perfectly insulated from the surroundings.
  • The wire is surrounded by plasma which satisfies the conditions for Ideal MHD to be valid.
  • The plasma's velocity is zero, however, the temperature ($T$) is non-zero and so the particles will have a non-zero velocity.
  • The plasma is completely uniform and so no pressure forces are present.

According to the Ideal MHD equations, the plasma should remain completely static since there is no current for $\rho>a$ and therefore $\mathbf{J}\times\mathbf{B}=\mathbf{0}$.

However, if we model the particles individually, we find that the particles should undergo a Nonuniform B particle drift with velocity given by $$\mathbf{v}_\rho+\mathbf{v}_{\nabla B} = \frac{2k_B T}{q B}\frac{\mathbf{R}_c\times\mathbf{B}}{R_c^2B},$$ where $k_B$ is the Boltzmann constant, $q$ is the charge of the particle and $\mathbf{R_c}$ is the radius of curvature, which is given by $$\mathbf{R}_c=\rho\mathbf{\hat{\rho}}.$$ Therefore, the velocity can be simplified to $$\begin{aligned} \mathbf{v}_\rho+\mathbf{v}_{\nabla B} &= \frac{2k_BT}{q\rho B_\varphi}\mathbf{\hat{z}}\ \mathrm{for}\ \rho\ge a \\ &=4\pi\frac{k_BT}{q\mu_0I}\mathbf{\hat{z}}\ \mathrm{for}\ \rho\ge a. \end{aligned}$$ Hence, there should be a current generated by the plasma in the $z$-direction? Therefore, the Lorentz force ($\mathbf{J}\times \mathbf{B}$) will be non-zero. Hence, the velocity will be non-zero? Therefore, modelling the individual particles gives a completely different result to modelling the plasma using MHD? Have I made a mistake somewhere? Is the particle drift usually sufficiently slow for plasma where MHD is valid that it's okay to ignore it?

Also, isn't it strange that the particle drift is independent of $\rho$. So the particles infinitely far away from the wire have the same drift as those next to the wire? I guess it makes sense because the field is also very weak infinitely far away so there is nothing to prevent them from drifting.

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Okay so a plasma is an ionized gas comprised of roughly equal numbers of positive and negative charges, called quasi-neutrality. This, combined with long-range forces mediating dynamics, results in the plasma exhibiting a collective behavior. Magnetohydrodynamics (MHD) is an approximation for modeling plasmas and there are several types. The simplest version is called ideal MHD and this assumes there is only one fluid with one current and no electric fields in the bulk fluid rest frame. This latter part is related to a phenomena known as the frozen-in condition (e.g., see https://physics.stackexchange.com/a/452325/59023).

Hence, there should be a current generated by the plasma in the z-direction?

Yes, gradient drifts can result in net currents (e.g., see https://physics.stackexchange.com/a/556682/59023).

Therefore, the Lorentz force $\left( \mathbf{J} \times \mathbf{B} \right)$ will be non-zero. Hence, the velocity will be non-zero?

Well, technically the $\left( \mathbf{J} \times \mathbf{B} \right)$ term isn't really the Lorentz force, but yes, it will be non-zero in this scenario. That term is called the Hall term after the Hall effect.

Therefore, modelling the individual particles gives a completely different result to modelling the plasma using MHD?

Absolutely, yes. There are no particles in MHD. MHD is a fluid approximation.

Have I made a mistake somewhere?

If an infinite wire is causing an azimuthal magnetic field and you surround it with a plasma made of positive and negative charges, then yes, there will be drifts. However, the gradient curvature drift is given by: $$ \mathbf{V}_{gc} = \frac{ m_{s} }{ 2 \ q_{s} \ B } \left( v_{\perp}^{2} + 2 \ v_{\parallel}^{2} \right) \left[ \frac{ \mathbf{B} \times \nabla B }{ B^{2} } \right] \tag{0} $$ where $\mathbf{B}$ is the magnetic field vector, $B$ is its magnitude, $q_{s}$ is the charge (including sign) of the particle species $s$, $m_{s}$ is the mass of species $s$, and $v_{\perp}$($v_{\parallel}$) is the perpendicular(parallel) velocity of the particle with respect to $\mathbf{B}$.

So if $\mathbf{B}$ is entirely along $\hat{\phi}$ and the gradient of its magnitude is only in the $\hat{\rho}$ direction, then the direction of $\mathbf{V}_{gc}$ should be in the negative $\hat{z}$ direction, should it not?

Further, the last term, $\tfrac{ \mathbf{B} \times \nabla B }{ B^{2} }$, will have a dependence on the radial distance, proportional to $\rho^{-1}$. So yes, I think there are a few arithmetic mistakes present.

Note that the particles must have a velocity component along the magnetic field for there to be a curvature drift. In a warm plasma, this would automatically be the case from thermal oscillations. In ideal MHD, there are no particles.

Is the particle drift usually sufficiently slow for plasma where MHD is valid that it's okay to ignore it?

Usual? I am not sure that's the correct phrase. In general, MHD applies very well to things that are intrinsically boring. That is, scenarios where it's perfectly valid to ignore electric fields and treat the plasma as a bulk fluid. MHD can also be used for interesting things, but those are more specific/unique.

Also, isn't it strange that the particle drift is independent of $\rho$. So the particles infinitely far away from the wire have the same drift as those next to the wire?

But the drift is not independent of the particle's radial distance from the wire. The gradient is stronger closer to the wire ($\nabla B \propto \rho^{-2}$), so the gradient drift speed will be higher closer to the wire.

I guess it makes sense because the field is also very weak infinitely far away so there is nothing to prevent them from drifting.

No, this is not correct. You should be thinking that at infinity, there is nothing to MAKE the particles drift. They drift because of the gradient in the magnetic field. When the gradient disappears or becomes sufficiently weak, they have no reason to drift.

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