Question

Answer 1

B8)

Tiết diện của dây: \(S=\dfrac{\rho.l}{R}=\dfrac{0,4.10^{-6}.3}{2}=6.10^{-7}\left(m^2\right)=0,6\left(mm^2\right)\)

Bán kính thiết diện dây:\(S=\pi R^2\Rightarrow R=\sqrt{\dfrac{S}{\pi}}\simeq0,4\left(mm\right)\)

Ý A

B9)\(R_1=R_2\Leftrightarrow\rho_1.\dfrac{l}{S_1}=\rho_2.\dfrac{l}{S_2}\)

\(\Leftrightarrow\rho_1.\dfrac{l}{\dfrac{\pi d_1^2}{4}}=\rho_2.\dfrac{l}{\dfrac{\pi d^2_2}{4}}\)

\(\Leftrightarrow d_2^2=\dfrac{\rho_2.d_1^2}{\rho_1}\simeq6.10^{-7}\left(m\right)\Rightarrow d_2\simeq7,861.10^{-4}\left(m\right)=0,786\left(mm\right)\)

Ý D