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

And:

`\begin{gathered} B_2=((4\pi\ast10^(-7))/(4\pi))((2\ast25)/(0.05))\Rightarrow B_2=(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=(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.