Question

Two long thin parallel wires 13.0 cm apart carry 25 A currents in thesame direction. Determine the magnetic field at point, P, 12.0 cm fromone wire and 5.0 cm from the other

Answer 1

We will have the following:

$\begin{gathered} B_1=(\frac{\mu}{4\pi})(\frac{2l_1}{L_1})\Rightarrow B_1=(\frac{4\pi\ast10^{-7}}{4\pi})(\frac{2\ast25}{0.12}) \\ \\ \Rightarrow B_1=\frac{1}{24000}\Rightarrow B_1\approx1.2\ast10^{-5}T \end{gathered}$

And:

$\begin{gathered} B_2=(\frac{4\pi\ast10^{-7}}{4\pi})(\frac{2\ast25}{0.05})\Rightarrow B_2=\frac{1}{10000} \\ \\ \Rightarrow B_2=1\ast10^{-4}T \end{gathered}$

So, the magnetic field at that point is:

$\begin{gathered} M=B_2-B_1\Rightarrow M=\frac{11}{125000} \\ \\ \Rightarrow M=8.8\ast10^{-5}T \end{gathered}$

So, the magnetic field at the point is approximately 8.8*10^-5 T.