The formula for the Force on a current carrying wire is,
[tex]\begin{gathered} F=IBlsin(\theta) \\ where, \\ F=Force \\ I=Current \\ B=Magnetic\text{ }Field \\ l=Length\text{ }Of\text{ }The\text{ }Wire \\ \theta=Angle\text{ }Between\text{ }Length\text{ }And\text{ }Magnetic\text{ }Field \end{gathered}[/tex][tex]\begin{gathered} Here, \\ B=5\times10^{-5}Tesla \\ l=115m \\ I=400A \\ \theta=90\mathring{} \\ So, \\ F=IBlsin(\theta)=400\times5\times10^{-5}\times115\times sin(90\mathring{\text{ }}) \\ So,\text{ }F=230000\times10^{-5}=2.3N \end{gathered}[/tex]So, Force on the wire is 2.3N
