c
(c) \(B = \frac{{{\mu _0}Ni}}{l}\) where \(N =\) Total number of turns, \(l =\) length of the solenoid
\( \Rightarrow 0.2 = \frac{{4\pi \times {{10}^{ - 7}} \times N \times 10}}{{0.8}} \Rightarrow N = \frac{{4 \times {{10}^4}}}{\pi }\)
Since \(N\) turns are made from the winding wire so length of the wire \((L)\) \( = 2\pi r \times N\) [ \(2\pi r = \) length of each turns]
\( \Rightarrow L = 2\pi \times 3 \times {10^{ - 2}} \times \frac{{4 \times {{10}^4}}}{\pi } = 2.4 \times {10^3}\,m\)