= Solution
The original diagram places $R_2$ in parallel with the <inductor>; $R_1$ connects their common node to the source only when the switch is in position 2. Let $v(t)$ be their shared <voltage> and $i_L(t)$ the downward <inductor> <electric current>. Since $i_L(0^-)=0$ and a finite <voltage> cannot change it instantaneously, the <inductor> initially behaves as an open branch. Thus
$$
I_2(0^+)=\frac{V_0}{R_1+R_2}=\frac{100}{110}\,\mathrm A=0.909\,\mathrm A.
$$
At long times, an ideal <inductor> carrying constant <electric current> has $v=L\dot i_L=0$, shorting $R_2$. Therefore $I_2(\infty)=0$ and $i_L(\infty)=V_0/R_1=10\,\mathrm A$.
For the full transient, <Kirchhoff's current law> gives $(V_0-v)/R_1=i_L+v/R_2$ with $v=L\dot i_L$. Equivalently, the source seen by the <inductor> has the <Thévenin equivalent>
$$
V_{\rm th}=\frac{V_0R_2}{R_1+R_2},\qquad
R_{\rm th}=R_1\parallel R_2=\frac{R_1R_2}{R_1+R_2}.
$$
Solving $L\dot i_L+R_{\rm th}i_L=V_{\rm th}$ gives
$$
\boxed{i_L(t)=\frac{V_0}{R_1}(1-e^{-t/\tau}),\quad
I_2(t)=\frac{V_0}{R_1+R_2}e^{-t/\tau},\quad
\tau=\frac{L}{R_{\rm th}}=86.9\,\mathrm{\mu s}.}
$$
The relevant resistance is the parallel combination with the ideal <voltage source> suppressed, not $R_1+R_2$ and not $R_1$ alone.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-solenoid-switching-transients.png]
{title=Current redistribution when the solenoid is connected and disconnected}
{description=The two panels use their own time origins. On connection the inductor current grows while the parallel-resistor current decays. On disconnection both branches carry equal and opposite currents as stored magnetic energy dissipates.}
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