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+### Statement
+
+$12.1.7.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$12.1.7^\ast$. Solve problem 12.1.6 in the case where the wave propagates in a medium with a dielectric permittivity $\varepsilon$. The velocity of the wave in the medium is $c/\sqrt{\varepsilon}$.
+
+### Solution
+
+The method of solution is identical to that in problem 12.1.6. In the rectangular contour, let's denote the sides as $ab = L$ and $aa' = x$. Faraday's law of electromagnetic induction in magnitude for the contour is:
+$$\mathscr{E}_i = \frac{d\Phi}{dt}$$
+where $\mathscr{E}_i$ is the induced electromotive force (EMF) created by the electric field $E$, and $\Phi$ is the magnetic flux through the contour.
+
+From the geometry of the setup:
+$$\mathscr{E}_i = E L$$
+$$d\Phi = B L dx$$
+
+Substituting these two formulas into the induction law, we get after canceling $L$:
+$$E = B \frac{dx}{dt}$$
+
+Here, the rate of change of distance $dx/dt$ is simply the propagation velocity of the wave in the medium, $v = c/\sqrt{\varepsilon}$. Therefore:
+$$E = B \frac{c}{\sqrt{\varepsilon}} \implies B = \frac{E\sqrt{\varepsilon}}{c}$$
+This is the magnetic induction in the SI system.
+
+In the CGS (Gaussian) system, Faraday's law of electromagnetic induction contains an additional factor of $1/c$ on the right side:
+$$EL = \frac{1}{c} \frac{d\Phi}{dt} = \frac{1}{c} B L \frac{dx}{dt}$$
+Substituting the velocity $dx/dt = c/\sqrt{\varepsilon}$:
+$$EL = \frac{1}{c} B L \frac{c}{\sqrt{\varepsilon}} = B L \frac{1}{\sqrt{\varepsilon}}$$
+Canceling $L$, we find the magnetic induction in the CGS system:
+$$E = B \frac{1}{\sqrt{\varepsilon}} \implies B = E\sqrt{\varepsilon}$$
+
+#### Answer
+$B = \frac{E\sqrt{\varepsilon}}{c}$ (in SI); $B = E\sqrt{\varepsilon}$ (in CGS)
+
+#### Answer
+
+[Insert a concise answer or boxed result]