= Solution
A <density> deficit in a <flux tube> is not by itself an <instability> calculation. For example, a straight magnetized tube in otherwise field-free gas at the same <temperature> has lateral balance $p_i+B_i^2/(2\mu_0)=p_e$. The <perfect gas> law then gives $\rho_i<\rho_e$, so the tube rises unless some other force balances its weight deficit. This configuration simply lacks <magnetostatic equilibrium>. A <magnetic buoyancy instability>, in contrast, starts from a force-balanced atmosphere and asks whether a small allowed displacement grows. A balanced horizontal magnetic layer can be stable or unstable; the presence of <magnetic pressure> alone does not decide this.
Assume a smooth equilibrium, with downward gravitational <acceleration> $g>0$, and write $\mathcal R$ for the gas constant per unit mass. The ambient total <pressure> and force balance are
$$
P(z)=p(z)+\frac{B_0(z)^2}{2\mu_0},\qquad
P'(z)=-\rho(z)g,\qquad p=\mathcal R\rho T.
$$
We use ideal <magnetic flux freezing> on the displacement time scale and rapid lateral <pressure> equilibration. A bodily displacement perpendicular to a straight horizontal tube does not change its length or bend its field. If its area is $\mathcal A$, conservation of <magnetic flux> and mass per unit length gives $B\mathcal A=\mathrm{constant}$ and $\rho\mathcal A=\mathrm{constant}$. Thus the tube preserves
$$
f=\frac{B}{\rho}.
$$
Efficient heat exchange fixes its <temperature> to the ambient <temperature> at its new height, not to its original <temperature>. This is the thermal assumption that distinguishes the requested criterion from the adiabatic one.
Take a tube initially at $z_0$ and let $f_0=B_0(z_0)/\rho(z_0)$. After a small upward displacement $\xi$, let $\rho_i(z)$ be the <density> it would have in lateral balance at $z=z_0+\xi$. Its preserved $f_0$ and imposed local <temperature> give
$$
\mathcal R T(z)\rho_i(z)+\frac{f_0^2\rho_i(z)^2}{2\mu_0}=P(z).
$$
At the initial position $\rho_i=\rho$. Differentiate this relation at $z_0$:
$$
\left(c_T^2+v_A^2\right)\rho_i'+\mathcal R\rho T'=-\rho g,
\qquad c_T^2=\frac p\rho=\mathcal RT,\quad
v_A^2=\frac{B_0^2}{\mu_0\rho}.
$$
For the ambient atmosphere the same differentiation must also differentiate $f(z)=B_0(z)/\rho(z)$. It gives
$$
\left(c_T^2+v_A^2\right)\rho'+\mathcal R\rho T'
+\frac{\rho B_0}{\mu_0}f'=-\rho g.
$$
Subtracting isolates the actual displaced <density> contrast:
$$
\rho_i(z_0+\xi)-\rho(z_0+\xi)
=\frac{\rho B_0}{\mu_0(c_T^2+v_A^2)}f'\xi+O(\xi^2).
$$
The pressure-gradient force on the tube balances the weight of the displaced ambient gas, so its leading vertical <buoyancy> <acceleration> is
$$
\ddot\xi=-\frac{gB_0}{\mu_0(c_T^2+v_A^2)}
\frac{d}{dz}\left(\frac{B_0}{\rho}\right)\xi.
$$
The coefficient multiplying $\xi$ is positive precisely when the tube becomes lighter on upward displacement. Thus the <isothermal interchange criterion for magnetic buoyancy> is
$$
\boxed{B_0\frac{d}{dz}\left(\frac{B_0}{\rho}\right)<0,
\quad\text{equivalently}\quad
\frac{d}{dz}\left(\frac{B_0^2}{\rho^2}\right)<0.}
$$
Where $B_0>0$, this is simply $d(B_0/\rho)/dz<0$, or $d\log B_0/dz<d\log\rho/dz$. The squared form is independent of the field's choice of direction. Equality is neutral at this order, and the reversed inequality is restoring for these displacements. A layer with an interval satisfying the strict inequality admits local growing interchange disturbances under the assumed <pressure>/thermal equilibration. The calculation does not address field-bending undular modes, finite heat-transfer times or magnetic <diffusion>; those can change the criterion. In particular, the <localized interchange criterion for a magnetized atmosphere> with an adiabatic gas response is not interchangeable with this fast-thermal-exchange result.
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