\(S = 5cm \times 5cm\)
B = 4.10-4 (T)
\(\phi = {10^{ - 6}}\left( {Wb} \right)\)
Từ thông qua khung dây: \(\phi = BS\cos \alpha \Rightarrow \cos \alpha = {\phi \over {BS}}\)
Ta có: \({\rm{cos}}\alpha {\rm{ = }}{{{{10}^{ - 6}}} \over {{{4.10}^{ - 4}}.{{\left( {0,05} \right)}^2}}} = 1 \Rightarrow \alpha = 0 \Rightarrow \overrightarrow n \equiv \overrightarrow B \)