Two long parallel wires whose centres are a distance d apart carr… - Vidyadip

MCQ

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

✓
$L = \frac{{{\mu _0}{\text{l}}}}{\pi }\,{\log _e}\,\frac{{d - a}}{a}$
B
$L = \frac{{{\mu _0}{\text{l}}}}{\pi }\,{\log _e}\,\frac{d}{a}$
C
$L = \frac{{{\mu _0}{\text{l}}}}{\pi }\,{\log _e}\,\frac{a}{d}$
D
none

✓

Answer

Correct option: A.

$L = \frac{{{\mu _0}{\text{l}}}}{\pi }\,{\log _e}\,\frac{{d - a}}{a}$

a
Using Ampere's law for one of the wires to obtain the magnetic field in terms of current $I$ and

the variable radius $r$ due to one of the wire

$\mu_{0} I =\oint \vec{B} \cdot \overrightarrow{d l}$

$=\int_{0}^{2 \pi} B r d \theta$

$\mu_{0} I =B(2 \pi r)$

$\Rightarrow \quad B =\frac{\mu_{0} I}{2 \pi r}$

Note that because of the symmetry of the setup, the total magnetic flux $(\phi)$ will be two times

generated by one of the wires. Then, we continue to use the relation between self inductance

and flux

$\phi_{\text {total}}=2 \int_{A} \vec{B} \cdot \overrightarrow{d A}$

$=2 \int_{a}^{d-a}\left(\frac{\mu_{0} I}{2 \pi r}\right) l d r$

$\phi_{\text {total}}=\frac{\mu_{0} I l}{\pi} \ln \left(\frac{d-a}{a}\right)$

Inductance, $L=\frac{\phi_{\text {total}}}{I}=\frac{\mu_{0} l}{\pi} \ln \left(\frac{d-a}{a}\right)$

Inductance per unit length, $\phi_{\text {total }}=\frac{\mu_{0}}{\pi} \ln \left(\frac{d-a}{a}\right)$

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