Bài 2: 

Ta có: \(B=2\pi.10^{-7}.N.\dfrac{I}{R}\)

=> I= \(\dfrac{B.R}{2\pi.10^{-7^{ }}.N}=\dfrac{6,28.10^{-6^{ }}0,05}{2\pi.10^{-7^{ }}.100}=5.10^{-3}A\)

Bài 2: 

Ta có: B= 2.10^-7.N.\(\dfrac{I}{R}\)

=> I= \(\dfrac{B.R}{2\pi.10^{-7^{ }}.N}\)= \(\dfrac{6,28.10^{-6^{ }}.0,05}{2\pi.10^{-7^{ }}.100}\)=5.10^-3 (A)

Bài 1: 

a) Ta có: B_M= 2.10^-7.\(\dfrac{I}{r}\) = 2.10^-7.\(\dfrac{0,5}{0,05}\) = 2.10^-6 (T)

b) Ta có: B_N= 2.10^-7.\(\dfrac{I}{r}\)

=> r_N= \(\dfrac{2.10^{-7^{ }}.I}{B_N}\)=\(\dfrac{2,10^{-7^{ }}.0,5}{0,05}\)= 0,2 (m)