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Physics
According to modified Ampere's circuital law (iD=. displacement current)
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Q. According to modified Ampere's circuital law
$\left(i_{D}=\right.$ displacement current)
Electromagnetic Waves
A
$\oint \vec{B} \cdot \vec{d l}=\mu_{0}\left(i_{C}+\varepsilon_{0} \frac{d \phi_{E}}{d t}\right)$
B
$\oint \vec{B} \cdot \vec{d l}=\mu_{0} \varepsilon_{0} \frac{d \phi_{E}}{d t}$
C
$\oint \vec{B} \cdot \vec{d l}=\mu_{0} i$
D
$\oint \vec{B} \cdot \vec{d /}=\mu_{0}\left(i_{C} \frac{d \phi_{E}}{d t}+i_{D}\right)$
Solution:
According to modified Ampere's circuital law.
$\oint \vec{B} \cdot \vec{d l}=\mu_{0}\left(i_{C}+\varepsilon_{0} \frac{d \phi_{E}}{d t}\right)$