Đáp án:

 \(10A\)

Giải thích các bước giải:

Ta có: 

\(\begin{array}{l}
\Phi  = BS = B.\pi {r^2}.\dfrac{\alpha }{{2\pi }} = B.\pi {r^2}.\dfrac{{\sigma {t^2}}}{{2\pi }}\\
 \Rightarrow e =  - \Phi ' =  - 2.B.\pi {r^2}.\dfrac{{\sigma t}}{{2\pi }} =  - B.\pi {r^2}.\dfrac{{\sigma t}}{\pi }\\
l = \pi .r = 3,14.0,24 = 0,7536\left( m \right)\\
R = \dfrac{\alpha }{\pi }.\dfrac{{\rho l}}{S} = \dfrac{{\sigma {t^2}}}{{3,14}}.\dfrac{{1,{{7.10}^{ - 8}}.0,7536}}{{1,{{2.10}^{ - 6}}}}\\
 \Rightarrow R = 3,{4.10^{ - 3}}.\sigma {t^2}\\
I = \dfrac{{\left| e \right|}}{R} = B.\pi {r^2}.\dfrac{{\sigma t}}{{\pi .3,{{4.10}^{ - 3}}.\sigma {t^2}}}\\
 \Rightarrow I = \dfrac{{0,15.0,{{24}^2}}}{{3,{{4.10}^{ - 3}}.t}} = \dfrac{{2,54}}{t} = \dfrac{{2,54}}{{0,25}} = 10A
\end{array}\)