Question:

Match List I with List II
LIST ILIST II
AGauss's Law in ElectrostaticsI\(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\)
BFaraday's LawII\(\oint \vec{B} \cdot d \vec{A}=0\)
CGauss's Law in MagnetismIII\(\oint \vec{B} \cdot d \vec{l}=\mu_0 i_c+\mu_0 \in_0 \frac{d \phi_E}{d t}\)
DAmpere-Maxwell LawIV\(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\)
Choose the correct answer from the options given below:

Updated On: Jul 31, 2024
  • A-I, B-II, C-III, D-IV
  • A-IV, B-I, C-II, D-III
  • A-III, B-IV, C-I, D-II
  • A-II, B-III, C-IV, D-I
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The Correct Option is B

Solution and Explanation

The correct answer is (B) : A-IV, B-I, C-II, D-III
Gauss's Law of electrostatic

Faraday's law
Gauss's law of magnetism
Ampere's Maxwell law

Where : Conduction current
: Displacement current
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Questions Asked in JEE Main exam

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Concepts Used:

Gauss Law

Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.

Gauss Law:

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.

Gauss Law Formula:

As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;

Q = ϕ ϵ0

The Gauss law formula is expressed by;

ϕ = Q/ϵ0

Where,

Q = total charge within the given surface,

ε0 = the electric constant.