GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 49 of 65

Marks: +1, -0

Two identical copper wires W1 and W2, placed in parallel as shown in the figure, carry currents I and 2I, respectively, in opposite directions. If the two wires are separated by a distance of

4 r

, then the magnitude of the magnetic field

{ B}^{→}

between the wires at a distance

r

from

{W} 1

is W1

$\;{µ_{0} I}/{6 π r}$
$\;{µ_{0}^{2} I^{2}}/{2 π r^{2}}$
$\;{5 µ_{0} I}/{6 π r}$
$\;{6 µ_{0} I}/{5 π r}$

Validate

Solution:

{V}_{ {g}}=\;{ {d} ω}/{ {d} β}

Now,

\;{d β}/{d ω}=\;{d k(ω)}/{d ω}=\;{d}/{d ω} ⋅ \;{1}/{c} √{ω^{2}-ω_{0}^{2}}=\;{1}/{2 c √{ω^{2}-ω_{0}^{2}}} × 2 ω

\;{ {d} β}/{ {d} ω}=\;{ω}/{ {c} √{ω^{2}-ω_{0}^{2}}}

V_{g}=\;{\;{1}/{ω}}/{c √{ω^{2}-ω_{0}^{2}}}=2 × 10^{8} ⇒ \;{c √{ω^{2}-ω_{0}^{2}}}/{ω}=2 × 10^{8}

⇒ √{ω^{2}-ω_{0}^{2}}=\;{2 ω}/{3}

Now,

{V}_{ {p}}=\;{ω}/{β}=\;{ω}/{ {k}}=\;{ω}/{{1}/{ {c}} √{ω^{2}-ω_{0}^{2}}}=\;{ω {c}}/{2 \;{ω}/{3}}=\;{3 {c}}/{2}

{V}_{ {p}}=\;{3}/{2} × 3 × 10^{8}=4.5 × 10^{8} {m} {/} {s}

{V}_{ {p}}=4.5 × 10^{8} {m} {/} {s}

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