3 If we consider a constant phase of a nonaxisymmetric pat- tern, Eq. (8) yields R in and R out are the radii of the inner and outer cylinders, Ω in and Ω out are their rotation rates, and B in and B out the az- imuthal magnetic fields at the inner and outer cylinders. The possible magnetic field solutions are plotted in Fig. 1. Note that – unlike Ω , where Ω in and Ω out are the physically rele- vant quantities – for B$_{φ}$ the fundamental quantities are not so much B in and B out , but rather a$_{B}$ and b$_{B}$ themselves. In par- ticular, a field of the form b$_{B}$/R is generated by running an axial current only through the inner region R < R in , whereas a field of the form a$_{B}$R is generated by running a uniform ax- ial current through the entire region R < R out , including the fluid. Given the z -component of the electric current, curl$_{z}$ B = 2 a$_{B}$ , one finds for the current helicity of the background field The first relation describes the phase velocity of the modes i n the axial direction, the second in the azimuthal direction (and only exists for nonaxisymmetric modes). Obviously the wave is traveling upwards if the real part of the eigenfrequency is negative. At a fixed time the phase relations (23) can also be written as Now, k and m are both real numbers, and without loss of gen- erality one of them can be taken to be positive, say k . The other one, m in this case, must be allowed to have both signs though. Negative m describe right-hand spirals (marked here by R), and positive m describe left-hand spirals (marked here by L). If the axisymmetric background field possesses positi ve B$_{z}$ and B$_{φ}$ (as used for the calculations here) then its current helicity is positive, or equivalently, it forms a right spiral. which may be either positive or negative (and of course van- ishes for the current-free case a$_{B}$ = 0 ). However, both signs yield the same instability curves, merely with the previously mentioned left and right spirals interchanged. The inner value B in is normalized with the uniform vertical field, i.e. ∂z $_{∂t}$∣ ∣ φ = − ℜ ( ω ) k , ∂φ $_{∂t}$∣ ∣ z = − ℜ ( ω ) m . (23) ∂z/∂φ = − m/k. (24) curl B · B = 2 a$_{B}$B 0 , (16) β = B in B 0 . (17) curl B · B = 2 β 3 B 2 0 R in . (18) Ha = B in R 0 √ µ 0 ρνη . (19) For µ$_{B}$ = 1 we have The sign of β thus determines the sign of the helicity of the background field. And again, interchanging ± β simply inter- changes left and right spirals ± m . As usual, the toroidal field amplitude is measured by the Hartmann number We begin with a purely azimuthal field, and no electric cur- rents within the fluid, that is, B$_{φ}$ ∝ 1 /R . Figure 2 presents re- sults for µ Ω = 0 . 5 , showing that for Ha > 100 and Re > 200 there exists an m = 1 nonaxisymmetric instability. Note also how both the upper and lower branches of the instability curve tilt to the right, that is, have a positive slope d Re /d Ha . For a given Hartmann number, the instability therefore only ex- ists within a finite range of Reynolds numbers. If Re is too large, the instability disappears again as a consequence of the suppressing action that differential rotation often has on non- axisymmetric modes. III. FROM AMRI TO HMRI R 0 = √ R in ( R out − R in ) is used as the unit of length, η/R 0 as the unit of velocity and B in as the unit of the azimuthal fields. Frequencies, including the rotation Ω , are normalized with the inner rotation rate Ω in . The magnetic Reynolds num- ber Rm is defined as We consider next a purely axial field, the so-called stan- dard MRI. In this case both axisymmetric and nonaxisymmet- ric modes may be excited, with the axisymmetric mode being the one with the overall lowest Reynolds number (Fig. 3). For Pm = 1 this overall minimum occurs for Ha ≃ 10 and Re ≃ 80 . However, for sufficiently large Ha the m = 1 non- axisymmetric mode is actually preferred over the axisymmet- ric mode. The standard MRI for purely axial fields is there- fore not necessarily an axisymmetric mode. The axisymmet- ric mode only dominates for sufficiently weak fields, includ- ing also the global minimum Re value. It also dominates the entire weak-field branch of the instability curves (Fig. 3).The axisymmetric mode here tilts to the left, whereas the nonax- isymmetric mode tilts to the right, as before in Fig. 2. The Lundquist number S is defined by $^{S = Ha}$· √$_{Pm}$$_{.}$ The boundary conditions associated with the perturbation equations are no-slip for $^{u}$, Figure 4 finally shows results combining azimuthal and ax- ial fields, focusing in this case on β = 2 (so a right-handed helicity). We see the same general pattern as before: only the weak-field branch of the m = 0 mode tilts to the left; all non- axisymmetric modes tilt to the right. Up to Ha ≈ 50 the ax- isymmetric mode is preferred, just as before for the standard MRI. For Ha > 50 the m = 1 right spiral is preferred. and perfectly conducting for $^{b}$, These boundary conditions hold for both R = R in and R = R out . Rm = Ω in R 2 0 η . (20) u$_{R}$ = u$_{φ}$ = u$_{z}$ = 0 , (21) db$_{φ}$/dR + b$_{φ}$/R = b$_{R}$ = 0 . (22)