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Question

(A) Using Ampere's law for one of the wires to obtain the magnetic field $B$ at a distance $r$ from the wire:$\oint \vec{B} \cdot d\vec{l} = \mu_0 I$$

Vedclass · Accepted Answer

$L = \frac{\mu_0 l}{\pi} \ln \left( \frac{d-a}{a} \right)$

Vedclass · Answer

(A) Using Ampere's law for one of the wires to obtain the magnetic field $B$ at a distance $r$ from the wire:$\oint \vec{B} \cdot d\vec{l} = \mu_0 I$$B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$The magnetic flux $\phi$ through the rectangular area of length $l$ between the two wires (from $r=a$ to $r=d-a$) due to one wire is:$\phi_1 = \int_a^{d-a} B \cdot l \cdot dr = \int_a^{d-a} \left( \frac{\mu_0 I}{2\pi r} ight) l \cdot dr = \frac{\mu_0 I l}{2\pi} \ln \left( \frac{d-a}{a} ight)$Since the currents are in opposite directions,the magnetic fields produced by both wires in the region between them point in the same direction. Thus,the total flux $\phi_{	ext{total}}$ is twice the flux due to one wire:$\phi_{	ext{total}} = 2 \phi_1 = \frac{\mu_0 I l}{\pi} \ln \left( \frac{d-a}{a} ight)$The inductance $L$ is defined as $L = \frac{\phi_{	ext{total}}}{I}$:$L = \frac{\mu_0 l}{\pi} \ln \left( \frac{d-a}{a} ight)$

Two long parallel wires whose centres are at a distance $d$ apart carry equal currents in opposite directions. If the flux within the wires is neglected,the inductance of such an arrangement of wire of length $l$ and radius $a$ will be

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