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+### Statement
+
+$11.3.13.$ [Insert the problem statement]
+
+### Solution
+
+Self‑inductances
+
+For a long solenoid, the total self‑inductance is $L = \mu_0 n^2 \pi r^2 l $
+Therefore:
+
+$L_1 = \mu_0 n_1^2 \pi r_1^2 l_1, \qquad
+L_2 = \mu_0 n_2^2 \pi r_2^2 l_2$
+
+Mutual inductance
+
+The mutual inductance M is defined from the flux that one coil sends through the other.
+The magnetic field created by the outer solenoid (of length$ l_2$) in its interior is uniform and is given by$ B_2 = \mu_0 n_2 I_2$ This field passes through the cross‑section of the inner solenoid ($\pi r_1^2$) along the common length, which is $l_2$
+The flux linked by the inner solenoid due to current $I_2 $is:
+
+$\Phi_{12} = (B_2)(\pi r_1^2)(n_1 l_2) = \mu_0 n_1 n_2 \pi r_1^2 l_2 I_2$
+
+By definition, $M = \Phi_{12}/I_2$:
+
+$M = \mu_0 n_1 n_2 \pi r_1^2 l_2$
+
+Total inductance according to the direction of the currents
+
+In a series connection the total inductance is:
+
+$L = L_1 + L_2 \pm 2M$
+
+where the + sign corresponds to currents in the same direction (fields reinforcing each other) and the - sign to currents in opposite directions (fields opposing each other).
+
+Substituting the expressions:
+
+$\boxed{L_{\pm} = \mu_0 \pi \Bigl( n_1^2 r_1^2 l_1 + n_2^2 r_2^2 l_2 \pm 2 n_1 n_2 r_1^2 l_2 \Bigr)}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]