The problem involves a conducting circular loop placed in a uniform magnetic field with its plane perpendicular to the field. The radius of the loop is expanding at a constant rate, and we are asked to find the magnitude of the induced emf in the loop at an instant when the radius of the loop is $$2 \mathrm{~cm}$$.

The magnetic flux through a circular loop of radius $$r$$ and area $$A = \pi r^2$$ placed in a uniform magnetic field $$B$$ perpendicular to the plane of the loop is given by:

$$\Phi_B = B A = B \pi r^2$$

The induced emf in the loop is given by Faraday's law of electromagnetic induction:

$$\mathcal{E} = -\frac{d\Phi_B}{dt}$$

In this case, the radius of the loop is expanding at a constant rate of $$10^{-3} \mathrm{~m/s}$$, which means that the rate of change of the area of the loop is:

$$\frac{dA}{dt} = \frac{d}{dt} (\pi r^2) = 2 \pi r \frac{dr}{dt} = 2 \pi (0.02 \mathrm{~m}) (10^{-3} \mathrm{~m/s}) = 4 \times 10^{-5} \mathrm{~m^2/s}$$

The magnetic flux through the loop is changing at this rate, and the induced emf in the loop is given by:

$$\mathcal{E} = \left|\frac{d\Phi_B}{dt}\right| = \left|\frac{dB}{dt} \frac{dA}{dt}\right| = \left|B \frac{dA}{dt}\right| = \left|0.4 \mathrm{~T} \times 4 \times 10^{-5} \mathrm{~m^2/s}\right| = 16 \pi \mu \mathrm{V}$$

Therefore, the magnitude of the induced emf in the loop at an instant when the radius of the loop is $$2 \mathrm{~cm}$$ is $$50.24 $$ $$ \simeq $$ 50 $\mu \mathrm{V}$.