To determine the RMS (Root Mean Square) value of the current in the circuit, we start by analyzing the given emf and the capacitive reactance.

The given alternating emf is:

$$\mathrm{E} = 110 \sqrt{2} \sin 100 \mathrm{t} \, \text{volts}$$

Here, the peak voltage (or maximum voltage) $$\mathrm{E_{max}}$$ is:

$$\mathrm{E_{max}} = 110 \sqrt{2} \, \text{volts}$$

Next, the RMS value of the voltage, $$\mathrm{E_{rms}}$$, is obtained by dividing the peak voltage by $$\sqrt{2}$$:

$$\mathrm{E_{rms}} = \frac{\mathrm{E_{max}}}{\sqrt{2}} = \frac{110 \sqrt{2}}{\sqrt{2}} = 110 \, \text{volts}$$

We are given a capacitor with a capacitance $$C = 2 \mu \mathrm{F} = 2 \times 10^{-6} \, \text{F}$$ and we need to determine the RMS current. The capacitive reactance $$\mathrm{X_C}$$ is given by:

$$\mathrm{X_C} = \frac{1}{\omega C}$$

where $$\omega$$ is the angular frequency. From the given formula for emf, we see that:

$$\omega = 100 \, \text{rad/s}$$

Therefore, the capacitive reactance is:

$$\mathrm{X_C} = \frac{1}{100 \times (2 \times 10^{-6})} = \frac{1}{200 \times 10^{-6}} = 5000 \, \Omega$$

Now, we can calculate the RMS value of the current $$\mathrm{I_{rms}}$$ using Ohm's law for AC circuits, which states:

$$\mathrm{I_{rms}} = \frac{\mathrm{E_{rms}}}{\mathrm{X_C}}$$

Substituting the known values:

$$\mathrm{I_{rms}} = \frac{110}{5000} = 0.022 \, \text{A} = 22 \, \text{mA}$$

Hence, the RMS value of the current in the circuit is $$22 \, \text{mA}$$.