C

To find the average e.m.f. induced in the coil, we can use Faraday's law of electromagnetic induction, which states:
$$ ext{e.m.f.} = -n \frac{\Delta \Phi}{\Delta t} $$
where:
- $n$ is the number of turns in the coil (1,000 turns),
- $\Delta \Phi$ is the change in magnetic flux,
- $\Delta t$ is the change in time over which the flux changes.
Step 1: Calculate the area in square meters:
The area given is $500 cm^2$. Convert this to square meters:
$$ 500 cm^2 = 500 \times 10^{-4} m^2 = 0.05 m^2 $$
Step 2: Calculate the change in magnetic flux:
The initial magnetic flux ($\Phi_i$) when the coil is perpendicular to the magnetic field (at 0°) is:
$$ \Phi_i = B \cdot A = (2 \times 10^{-5} \frac{Wb}{m^2}) \cdot (0.05 m^2) = 1 \times 10^{-6} Wb $$
The final magnetic flux ($\Phi_f$) when the coil is rotated 180° is:
$$ \Phi_f = -B \cdot A = -1 \times 10^{-6} Wb $$
The change in magnetic flux ($\Delta \Phi$) is:
$$ \Delta \Phi = \Phi_f - \Phi_i = -1 \times 10^{-6} Wb - 1 \times 10^{-6} Wb = -2 \times 10^{-6} Wb $$
Step 3: Calculate the e.m.f.
Substitute into the formula:
$$ \text{e.m.f.} = -1000 \times \frac{-2 \times 10^{-6} Wb}{0.2 s} $$
$$ \text{e.m.f.} = 1000 \times \frac{2 \times 10^{-6}}{0.2} = 10 mV $$
Therefore, the average e.m.f. induced in the coil is 10 millivolts.
Therefore, the correct answer is B.