Question:
Find the inductance of a unit length of two parallel wires, each of radius \(a\), whose centres are at distance \(d\) apart and carry equal currents in opposite directions. Neglect the flux within the wire.
Find the inductance of a unit length of two parallel wires, each of radius \(a\), whose centres are at distance \(d\) apart and carry equal currents in opposite directions. Neglect the flux within the wire.
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For two-wire transmission lines, the inductance per unit length is \(\mu_0/\pi \cdot \ln(d/a)\) when \(d \gg a\) (neglecting wire radii).
Updated On: Apr 20, 2026
- \(\dfrac{\mu_0}{2\pi}\ln\!\dfrac{d-a}{a}\)
- \(\dfrac{\mu_0}{\pi}\ln\!\dfrac{d-a}{a}\)
- \(\dfrac{3\mu_0}{\pi}\ln\!\dfrac{d-a}{a}\)
- \(\dfrac{\mu_0}{3\pi}\ln\!\dfrac{d-a}{a}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Magnetic field due to a long straight wire: \(B = \frac{\mu_0 I}{2\pi r}\). Each wire contributes equally to the total flux.
Step 2: Detailed Explanation:
Flux per unit length from one wire (integrating from \(r = a\) to \(r = d - a\)): \(\phi_1 = \frac{\mu_0 I}{2\pi}\ln\!\frac{d-a}{a}\). Total flux from both wires: \(\phi = 2\phi_1 = \frac{\mu_0 I}{\pi}\ln\!\frac{d-a}{a}\). Inductance per unit length: \(L = \frac{\phi}{I} = \frac{\mu_0}{\pi}\ln\!\frac{d-a}{a}\).
Step 3: Final Answer:
\[ \boxed{\frac{\mu_0}{\pi}\ln\!\frac{d-a}{a}} \]
Magnetic field due to a long straight wire: \(B = \frac{\mu_0 I}{2\pi r}\). Each wire contributes equally to the total flux.
Step 2: Detailed Explanation:
Flux per unit length from one wire (integrating from \(r = a\) to \(r = d - a\)): \(\phi_1 = \frac{\mu_0 I}{2\pi}\ln\!\frac{d-a}{a}\). Total flux from both wires: \(\phi = 2\phi_1 = \frac{\mu_0 I}{\pi}\ln\!\frac{d-a}{a}\). Inductance per unit length: \(L = \frac{\phi}{I} = \frac{\mu_0}{\pi}\ln\!\frac{d-a}{a}\).
Step 3: Final Answer:
\[ \boxed{\frac{\mu_0}{\pi}\ln\!\frac{d-a}{a}} \]
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