Maxwell’s modified form of Ampere’s circuital law is
A. \(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} s}=0\)
B. \(\oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I\)
C. \(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d}}=\mu_{0} \mathrm{I}+\frac{1}{\varepsilon_{0}} \frac{\mathrm{d} q}{\mathrm{dt}}\)
D. \(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}=\mu_{0} \mathrm{I}+\mu_{0} \varepsilon_{0} \frac{\mathrm{d} \varphi}{\mathrm{d} t}\)
Answer:D
Maxwell’s modified form of Ampere’s circuital law is
A. \(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} s}=0\) B. \(\oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I\) C. \(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d}}=\mu_{0} \mathrm{I}+\frac{1}{\varepsilon_{0}} \frac{\mathrm{d} q}{\mathrm{dt}}\) D. \(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}=\mu_{0} \mathrm{I}+\mu_{0} \varepsilon_{0} \frac{\mathrm{d} \varphi}{\mathrm{d} t}\)