Let's solve this carefully step–by–step. --- ### **Given:** * Two long parallel conductors (cylinders) * Each of radius ( a ) * Center-to-center distance ( b ) * ( b \gg a ) * Carry equal and opposite currents ( +I ) and ( -I ) * Relative permeability of the medium: ( \mu_r = \pi ) We are to find **the inductance per unit length, ( L' )**, of this two-wire transmission line. --- ### **1️⃣ Magnetic Field Between Two Long Parallel Wires** For long straight wires carrying opposite currents, the magnetic field at any point is the superposition of the fields due to each wire. But because ( b \gg a ), we can ignore proximity effects and assume the field outside each wire behaves like that of a filament at its center. --- ### **2️⃣ Magnetic Field Around One Wire** Magnetic field at distance ( r ) from a long wire carrying current ( I ): [ B = \frac{\mu I}{2\pi r} ] where ( \mu = \mu_0 \mu_r ). --- ### **3️⃣ Inductance per Unit Length for Two Wires** For two wires carrying equal and opposite currents, the inductance per unit length is given by the standard expression: [ L' = \frac{\mu}{\pi} \cosh^{-1}\left(\frac{b}{2a}\right) ] For large separation (( b \gg a )): [ \cosh^{-1}(x) \approx \ln(2x) ] so: [ L' \approx \frac{\mu}{\pi} \ln\left(\frac{b}{a}\right) ] However, the more accurate and conventional form (for two round conductors, opposite currents) is: [ L' = \frac{\mu}{\pi} \cosh^{-1}\left(\frac{b}{2a}\right) \approx \frac{\mu}{\pi} \ln\left(\frac{b}{a}\right) ] or equivalently (more exact form for ( b \gg a )): [ L' = \frac{\mu}{\pi} \ln\left(\frac{b}{a}\right) ] --- ### **4️⃣ Substitute ( \mu = \mu_0 \mu_r = \mu_0 \pi )** [ L' = \frac{\mu_0 \pi}{\pi} \ln\left(\frac{b}{a}\right) ] [ \boxed{L' = \mu_0 \ln\left(\frac{b}{a}\right)} ] --- ✅ **Final Answer:** [ \boxed{L' = \mu_0 \ln!\left(\frac{b}{a}\right) \ \text{H/m}} ] --- ### **5️⃣ Interpretation** Even though the medium’s relative permeability is ( \mu_r = \pi ), it cancels out with the ( 1/\pi ) factor in the geometric dependence of the two-wire inductance expression. So, **the inductance per unit length** between the two parallel conductors is: [ \boxed{L' = \mu_0 \ln!\left(\frac{b}{a}\right)} ] --- Please give me this answer same as it is without any changes. Please ensure that no errors in render equations