The Butterworth filter of order n is described by the magnitude squared of its frequency response given by $${\left| {{H_n}\left( {j\Omega } \right)} \right|^2} = \frac{1}{{\left[ {1 + {{\left( {\frac{\Omega }{{{\Omega _{\,C}}}}} \right)}^{2n}}} \right]}}.$$      The value of $$20\log \left| {{H_n}\left( {j\Omega } \right)} \right|$$   at $$\Omega = {\Omega _{\,C}}$$  is

A. -2 dB

B. -3.01 dB

C. -3 dB

D. -3.5 dB

Answer: Option B