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Question

Physics Question on Electromagnetism

Match List I with List IIList IList II
A. Bdl=μ0ic+μ0ϵ0dϕEdt\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}I. Gauss' law for electricity
B. Edl=dϕBdt\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}II. Gauss' law for magnetism
C. EdA=Qϵ0\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}III. Faraday law
D. BdA=0\oint \vec{B} \cdot d\vec{A} = 0IV. Ampere – Maxwell law

Choose the correct answer from the options given below

A

A-IV, B-I, C-III, D-II

B

A-II, B-III, C-I, D-IV

C

A-IV, B-III, C-I, D-II

D

A-I, B-II, C-III, D-IV

Answer

A-IV, B-III, C-I, D-II

Explanation

Solution

Ampere–Maxwell Law: - Bdl=μ0ic+μ0ϵ0dΦEdt\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d \Phi_E}{dt}. - This matches with A-IV.

Faraday’s Law of Electromagnetic Induction: - Edl=dΦBdt\oint \vec{E} \cdot d\vec{l} = -\frac{d \Phi_B}{dt}. - This matches with B-III.

Gauss’ Law for Electricity: - EdA=Qϵ0\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}. - This matches with C-I.

Gauss’ Law for Magnetism: - BdA=0\oint \vec{B} \cdot d\vec{A} = 0. - This matches with D-II.

So, the correct option is : (3) A-IV, B-III, C-I, D-II