Electrostatics Question 6

Question 6 - 25 January - Shift 2

Match List I with List II :

A. Gauss’s
Law in
Electrostat
ics
I. $\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}$
B. Faraday’s
Law
II. $\oint \overrightarrow{{}B} \cdot d \overrightarrow{{}A}=0$
C. Gauss’s
Law in
Magnetism
III. $\oint \vec{B} \cdot \vec{l}=\mu_0 i_C+\mu_0 \in_0 \frac{d \phi_E}{d t}$
D. Ampere-
Maxwell
Law
IV. $\oint \overrightarrow{{}E} \cdot d \overrightarrow{{}s}=\frac{q}{\in_0}$

Choose the correct answer from the options given below :

(1) A-IV, B-I, C-II, D-III

(2) A-I, B-II, C-III, D-IV

(3) A-III, B-IV, C-I, D-II

(4) A-II, B-III, C-IV, D-I

Show Answer

Answer: (1)

Solution:

Formula: Electric Flux

Gauss’s Law of electrostatic

$ \phi=\oint \overrightarrow{{}E} \cdot d \overrightarrow{{}s}=\frac{q}{\epsilon_0} $

Faraday’s law $\oint \vec{E} \cdot d \vec{l}=\frac{-d \phi_B}{d t}$

Gauss’s law of magnetism $\oint \vec{B} \cdot d \vec{A}=0$

Ampere’s Maxwell law

$ \oint \vec{B} \cdot d \vec{l}=\mu_0 i_C+\mu_0 \in_0 \frac{d \phi_E}{d t} $

Where $i_C$ : Conduction current

$\epsilon_0 \frac{d \phi_E}{dt}$ : Displacement current