1. Use Faraday’s law

For a stationary closed loop in a uniformly changing magnetic field, $|\mathcal E| = \left|\frac{d\Phi}{dt}\right| = A\left|\frac{dB}{dt}\right|$ where $A$ is the area enclosed by the loop.

2. Magnetic field change

Initial field: $B_i = 1000\ \text{Gauss}$ Final field: $B_f = 500\ \text{Gauss}$ So, $\Delta B = 500\ \text{Gauss}$ Since $1\ \text{Gauss} = 10^{-4}\ \text{T}$, $\Delta B = 500\times 10^{-4} = 5\times 10^{-2}\ \text{T}$

Time interval: $\Delta t = 5\ \text{s}$ Hence, $\left|\frac{dB}{dt}\right| = \frac{5\times 10^{-2}}{5} = 10^{-2}\ \text{T/s}$

3. Area of the loop

From the given figure, the total enclosed area is $A = 56\ \text{cm}^2$ Convert to SI units: $A = 56\times 10^{-4} = 5.6\times 10^{-3}\ \text{m}^2$

4. Induced emf

Now, $\mathcal E = A\frac{\Delta B}{\Delta t}$ $\mathcal E = (5.6\times 10^{-3})(10^{-2})$ $\mathcal E = 5.6\times 10^{-5}\ \text{V}$ $\mathcal E = 56\ \mu\text{V}$

5. Match with options

Thus the induced emf is $\boxed{56\ \mu\text{V}}$ which corresponds to Option C.