Match the following List-I with List-II. \(\begin{array}{cccc} \hline & \text { List-I } & & \text { List-II…
Match the following List-I with List-II.
\(\begin{array}{cccc}
\hline & \text { List-I } & & \text { List-II } \\
\hline \text { A. } & \oint E d A & \text { (i) } & 0 \\
\hline \text { B. } & \oint B d A & \text { (ii) } & -\frac{d \phi_B}{d t} \\
\hline \text { C. } & \oint E d l & \text { (iii) } & \frac{Q}{E_0} \\
\hline \text { D. } & \oint B d l & \text { (iv) } & \mu_0\left(i_c+i_d\right)
\end{array}\)
\(\begin{array}{cccc} \text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iii) } & \text { (ii) } & \text { (i) } & \text { (iv) }\end{array}\)
\(\begin{array}{cccc} \text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iv) } & \text { (i) } & \text { (iii) } & \text { (ii) } \end{array}\)
\(\begin{array}{cccc} \text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iii) } & \text { (i) } & \text { (ii) } & \text { (iv) }\end{array}\)
\(\begin{array}{cccc} \text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iii) } & \text { (i) } & \text { (iv) } & \text { (ii) } \end{array}\)
Solution
\(\mathrm{A} \rightarrow\) III According to Gauss's law, total electric flux passing through a closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the total charge enclosed within the surface. i.e., \(\quad \phi=\oint \mathbf{E}. d \mathbf{A}=\frac{Q}{\varepsilon_0}\)
\(\mathrm{B} \rightarrow \mathrm{I}\), According to gauss' law of magnetism, total magnetic flux passing through a bar magnet around its enclosed surface is zero because the magnetic lines of forces emerging from North-pole is equal to magnetic lines of force entering into South direction.
i.e., \(\oint B. d A=0\)
\(\mathrm{C} \rightarrow\) II, The electromotive force in a wire is the line integral,
\(E=\int \mathbf{E} \cdot d \mathbf{l}\)...(i)
But, by Faraday's law of electromagnetic induction, (EMI)
\(\varepsilon=-\frac{d \phi B}{d t}\)
From Eqs. (i) and (ii), we get
\(\int \mathbf{E} \cdot d \mathbf{l}=-\frac{d \phi B}{d t}\)
\(\mathrm{D} \rightarrow \mathrm{IV}\), According to the Maxwell, when electric field and hence electric flux charges with time, then an additional current comes into existence to the conduction current called displacement current. Therefore, the total current across the loop,
\(I=I_c+I_d\)
Hence, the modified form of the Ampere's law is
\(\oint \mathbf{B} \cdot d \mathbf{l}=\mu_0\left[I_c+I_d\right]\)