JEE MAIN - Physics (2024 - 30th January Evening Shift - No. 9)
Match List I with List II
List I | List II | ||
---|---|---|---|
(A) | Gauss's law of magnetostatics | (I) | $$\oint \vec{E} \cdot \vec{d} a=\frac{1}{\varepsilon_0} \int \rho d V$$ |
(B) | Faraday's law of electro magnetic induction | (II) | $$\oint \vec{B} \cdot \vec{d} a=0$$ |
(C) | Ampere's law | (III) | $$\int \vec{E} \cdot \vec{d} l=\frac{-d}{d t} \int \vec{B} \cdot \vec{d} a$$ |
(D) | Gauss's law of electrostatics | (IV) | $$\oint \vec{B} \cdot \vec{d} l=\mu_0 I$$ |
Choose the correct answer from the options given below:
Explanation
Let's identify each law listed in List I and match it with the corresponding mathematical expression listed in List II.
Gauss's law of magnetostatics states that the total magnetic flux through a closed surface is zero, as magnetic monopoles do not exist. This is given by the formula:$$ \oint \vec{B} \cdot \vec{da} = 0 $$
So, (A) matches with (II).
Faraday's law of electromagnetic induction states that the induced electromotive force (emf) in any closed loop is equal to the negative of the time rate of change of the magnetic flux through the loop. It is given by:$$ \oint \vec{E} \cdot \vec{dl} = -\frac{d}{dt} \int \vec{B} \cdot \vec{da} $$
Hence, (B) matches with (III).
Ampere's law relates the integrated magnetic field around a closed loop to the electric current passing through the loop. The mathematical expression given in the context of magnetostatics (without the displacement current) is:$$ \oint \vec{B} \cdot \vec{dl} = \mu_0 I $$
Consequently, (C) matches with (IV).
Lastly, Gauss's law of electrostatics states that the total electric flux out of a closed surface is proportional to the charge enclosed within the surface:
$$ \oint \vec{E} \cdot \vec{da} = \frac{1}{\varepsilon_0} \int \rho dV $$
Therefore, (D) matches with (I).
Based on these matches, the correct answer must link A-II, B-III, C-IV, and D-I:
The correct option is:
Option C
A-II, B-III, C-IV, D-I
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