A long solenoid of initial radius $R_0$ is put in a region of uniform magnetic field $\mathbf{B}$ with the axis of the solenoid aligned along the magnetic field. The solenoid is a part of a closed circuit that has no initial current running through it. If the radius of the solenoid starts increasing at a uniform rate, how do the magnetic field strength $B_{in}$ and the associated magnetic energy $U_{in}$ inside the solenoid change?

The phrase "closed circuit that has no initial current" implies a perfectly conducting loop where the total magnetic flux remains conserved.

Let the external magnetic field be $B_0$ and the total number of turns be $N$. The initial magnetic flux through the solenoid is given by:

$\Phi_0 = N B_0 \pi R_0^2$

At a later time, when the radius increases to $R$, let the net magnetic field inside the solenoid be $B_{in}$. The magnetic flux is:

$\Phi = N B_{in} \pi R^2$

By the principle of conservation of magnetic flux, $\Phi = \Phi_0$:

$N B_{in} \pi R^2 = N B_0 \pi R_0^2 \Rightarrow B_{in} = B_0 \left(\dfrac{R_0}{R}\right)^2$

Since the radius $R$ is increasing, the magnetic field strength $B_{in}$ inside the solenoid decreases.

The magnetic energy $U_{in}$ inside the solenoid is the product of the magnetic energy density and the volume of the solenoid:

$U_{in} = \dfrac{B_{in}^2}{2\mu_0} (\pi R^2 l)$

Substituting $B_{in} = \dfrac{\Phi}{N \pi R^2}$ into the energy equation:

$U_{in} = \dfrac{1}{2\mu_0} \left(\dfrac{\Phi}{N \pi R^2}\right)^2 (\pi R^2 l) = \dfrac{\Phi^2 l}{2\mu_0 N^2 \pi R^2}$

Since the flux $\Phi$ is constant and the radius $R$ is increasing, the term $R^2$ in the denominator causes the magnetic energy $U_{in}$ to decrease.

Thus, both $B_{in}$ and $U_{in}$ decrease.

Answer: $B_{in}$ decreases, $U_{in}$ decreases.