Câu hỏi 5

Ta có : \(I = {{\omega \Phi } \over {\sqrt {{r^2} + {\omega ^2}{L^2}} }} \Rightarrow {1 \over {{I^2}}} = {{{r^2}} \over {{\omega ^2}{\Phi ^2}}} + {{{L^2}} \over {{\Phi ^2}}}\)

Có :f = np ; p = 1 => ω =  2πn ; r = 10π (Ω) \( \Rightarrow {1 \over {{I^2}}} = {{{r^2}} \over {4{\pi ^2}{n^2}{\Phi ^2}}} + {{{L^2}} \over {{\Phi ^2}}}\)

+ \({{{{10}^4}} \over {{n^2}}} = 0 \Rightarrow {1 \over {{n^2}}} = 0 \Rightarrow {1 \over {I_0^2}} = {{{L^4}} \over {{\Phi ^2}}} = 1,5625\) (1)

+ \({{{{10}^4}} \over {{n^2}}} = 100 \Rightarrow {1 \over {{n^2}}} = 0,01 \Rightarrow {1 \over {{I^2}}} = {{{r^2}} \over {100.4{\pi ^2}{\Phi ^2}}} + {{{L^2}} \over {{\Phi ^2}}} = 7,8125\) (2)

+\({{(1)} \over {(2)}} = {{{L^2}} \over {{{{r^2}} \over {100.4{\pi ^2}}} + {L^2}}} = {{1,5625} \over {7,8125}} = {1 \over 5} \Rightarrow L = 0,25H\)