= Solution
<Self-inductance> relates current to its own <magnetic flux linkage>: $\Lambda=LI$, so the induced <electromotive force> is $-L\dot I$ for fixed geometry. A long <solenoid> has $B=\mu_0NI/l$ and area $\pi a^2$. Its $N$ turns give
$$
\boxed{L=\frac{\mu_0N^2\pi a^2}{l}.}
$$
The <homopolar generator> uses a disc rotating through this axial field; integrate the motional <Lorentz force> from shaft to rim:
$$
\mathcal E=\int_0^a\omega rB\,dr=\frac12\omega B a^2=C\omega I,\qquad C=\frac{\mu_0Na^2}{2l}.
$$
For a <homopolar disc dynamo>, choose the wiring and rotation sense giving positive magnetic feedback. <Kirchhoff's voltage law> is
$$
L\dot I+RI=C\omega I,\qquad I(t)=I(0)\exp\!\left[\frac{C\omega-R}{L}t\right]
$$
for prescribed constant $\omega$. Thus
$$
\boxed{\omega_c=\frac{R}{C}=\frac{2Rl}{\mu_0Na^2}.}
$$
A nonzero seed grows when $\omega>\omega_c$, decays when $\omega<\omega_c$, and is marginal at equality. This explains the printed $\omega\geq\omega_c$ as the threshold including its boundary; equality does not itself give exponential growth. An exactly zero initial current remains zero in the ideal homogeneous equation, and reversing the feedback sense prevents the instability.
In the <torque-controlled homopolar dynamo equilibrium>, a radial current element exerts a braking <torque> $dG_B=I B r\,dr$, so $G_B=IBa^2/2=C I^2$. For steady rotation and nonzero steady current, $G=CI^2$ and $\omega=\omega_c$. Hence the field intensity magnitude is
$$
\boxed{|H|=\frac Nl\sqrt{\frac GC}=\sqrt{\frac{2NG}{\mu_0a^2l}}.}
$$
The two current signs give opposite magnetic polarities. Finally,
$$
\boxed{P_{\rm mechanical}=G\omega_c=CI^2\frac RC=RI^2=P_{\rm Joule}.}
$$
The steady magnetic and rotational <energies> do not change. This is a steady-state balance, not an assertion that every dynamical trajectory must approach that state.
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