If superconductors have no resistance, what prevents you making the cross-section as small as you can handle, if there will be no power dissipated anyhow?
Is there a limit on the magnetic field that the superconductor can contain, if there is no limit on the current itself?
The super-current is carried by the gradient of the phase of the condensate, and there is a finite energy cost associated with this. If the gradient energy is larger than the BCS condensation energy (the energy gained by forming Cooper pairs), then superconductivity will disappear. This is the critical current.
In order to be more quantitative, consider the Landau-Ginzburg functional $$ {\cal F} = \frac{1}{2m}|(k-qA)\psi|^2+\alpha|\psi|^2+\frac{1}{2}\beta|\psi|^4 $$ where $k$ is the momentum, $q$ is the charge, $A$ is the gauge potential, $\psi$ is the condensate wave function, and $\alpha,\beta$ are parameters ($\alpha<0$ in the BCS phase). We can compute the super-current $$ j= \frac{q}{m}(k-qA)|\psi|^2 $$ and define the superconducting velocity $v_s=(k-qA)/m$. Vary the free energy with respect to $\psi$ to determine the ground state at finite $v_s$. We get $$ j = 2qn_s^0 v_s \left ( 1-\frac{mv_s^2}{2|\alpha|}\right) $$ where $n_s^0$ is the density of paired electrons in the absence of a current. You see that there is a maximum. $\alpha$ and $\beta$ are related to other quantities, so you can express the critical current in terms of properties like the critical field and the penetration depth.
