The response of a superconductor to an EM field is easiest to derive from the London equations
https://en.wikipedia.org/wiki/London_equations,
see the section on "London penetration depth".
However, the implicit assumption of steady state doesn't hold, of course.
Rather we have to take also the displacement current ##\dot{D}=\epsilon \dot{E}## into account and write
##-\nabla^2 B=\mathrm{rot} (j+\epsilon\dot{E})=-\frac{n_se^2}{m}B+\epsilon \,\mathrm{rot} \dot{ E} ##
now, using the first London equation, the displacement current
can also be linked to the superconducting current:
##\frac{ d^2 j_s}{dt^2}=\frac{n_s e^2}{m} \frac{ dE}{dt}##.
So,
##-\nabla^2 B=\mathrm{rot} (j+\dot{E})=-\frac{n_se^2}{m}B+\mathrm{rot} \dot{E} =\frac{n_se^2}{m}B-\epsilon \frac{d^2B}{dt^2} ##.
This equation is best analyzed assuming a periodic time dependence ## B=B_0 \exp(-i\omega t)##.
Then
##\nabla^2 B_0= (\frac{n_s e^2}{m}-\epsilon \omega^2)B_0=\lambda_L^{-2}(\omega)B_0##.
You can see that the penetration depth ##\lambda_L## increases with frequency.
Some remarks are in order:
a) I assumed T=0 so that there is no normal conducting electronic contribution.
b) This expression is only true for not too large frequencies ##\omega##.