A $4 \mu \mathrm{~F}$ capacitor is charged to 10 V . The battery is then disconnected and a pure 10 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is

In an LC circuit, energy conservation requires that the maximum current occurs when all initial capacitor energy transfers to the inductor. The initial energy stored in the capacitor is $U_C = \frac{1}{2}CV^2 = \frac{1}{2}(4 \times 10^{-6}\ \mathrm{F})(10\ \mathrm{V})^2 = 2 \times 10^{-4}\ \mathrm{J}$.

When the current reaches its maximum $I_{\text{max}}$, the inductor energy equals the initial capacitor energy: $\frac{1}{2}LI_{\text{max}}^2 = \frac{1}{2}CV^2$. Solving for $I_{\text{max}}$ yields $I_{\text{max}} = V\sqrt{\frac{C}{L}}$.

Substituting $L = 10\ \mathrm{mH} = 10^{-2}\ \mathrm{H}$ gives $I_{\text{max}} = 10\ \mathrm{V}\sqrt{\frac{4 \times 10^{-6}\ \mathrm{F}}{10^{-2}\ \mathrm{H}}} = 10\sqrt{4 \times 10^{-4}} = 10 \times 0.02 = 0.2\ \mathrm{A}$.

Final answer: $\boxed{0.2\ \mathrm{A}}$