A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity $$(g=10m/s^{2})$$ through a uniform horizontal magnetic fie ld of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire when it travells a vertical distance of200 m is ____ mV.

A conductor of length $$L$$ moving with velocity $$v$$ perpendicular to a magnetic field of strength $$B$$ has an induced EMF given by the formula:
$$\mathcal{E} = B\,L\,v\quad-(2)$$

First, find the speed $$v$$ of the wire after falling through a vertical distance $$h=200\text{ m}$$ under gravity $$g=10\text{ m/s}^2$$. Using the kinematic relation with initial speed zero:
$$v^2 = u^2 + 2\,g\,h = 0 + 2\times 10\times 200 = 4000\quad-(1)$$
Hence
$$v = \sqrt{4000} = 20\sqrt{10}\text{ m/s}$$

Convert the magnetic field from gauss to tesla: 1 Gauss = $$10^{-4}$$ T. Thus
$$B = 0.5\text{ Gauss} = 0.5\times 10^{-4}\text{ T} = 5\times 10^{-5}\text{ T}$$

Substitute $$B=5\times 10^{-5}\text{ T}$$, $$L=20\text{ m}$$ and $$v=20\sqrt{10}\text{ m/s}$$ into equation $$(2)$$:
$$\mathcal{E} = (5\times 10^{-5})\times 20\times 20\sqrt{10} = 0.02\sqrt{10}\text{ V}$$

Convert volts to millivolts:
$$0.02\sqrt{10}\text{ V} = 20\sqrt{10}\text{ mV}$$

Therefore, the induced EMF is $$20\sqrt{10}\text{ mV}$$, which is Option C.