= Solution
Take an <electric current> $I_1$ in the long <solenoid>. Away from its ends, <Ampère's circuital law> gives the approximately uniform <magnetic field>
$$
B_1=\mu_0\frac{N_1}{L}I_1.
$$
Here $\mu_0$ is the <vacuum permeability>; an air core is understood. Each turn of the short surrounding <coil> encloses the <magnetic flux> $B_1\pi R^2$. Thus its <magnetic flux linkage> is $\Lambda_2=N_2B_1\pi R^2$, and the <mutual inductance> is
$$
\boxed{M=\frac{\Lambda_2}{I_1}=\mu_0\frac{N_1N_2\pi R^2}{L}.}
$$
The short <coil> lies in the nearly uniform central <magnetic field>; its own axial length therefore does not enter. With consistent winding orientations, <Faraday's law of induction> gives $\mathcal E_2=-M\,dI_1/dt$. Reversing a winding reverses this signed coupling, while the stated magnitude is unchanged.
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