Question

Match List-I with List-II
 

List-I	List-II
(A) Faraday’s Law	(ii) \( \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \)
(B) Conservation Law	(iii) \( \vec{\nabla} \times \vec{A} = 0 \)
(C) Ohm’s Law	(iv) \( \vec{J} = \sigma \vec{E} \)
(D) Gauss’s Law	(i) \( \frac{\rho}{\epsilon_0} \)

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

Hint:

When matching laws to their expressions, remember that Faraday's Law involves the time derivative of the magnetic field, while Ohm's law relates current density to electric field.

Answer 1

The Correct Option is B

Step 1: Understanding the match.
- Faraday's Law \( \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \), so it matches with (II).
- The Conservation Law involves the continuity equation, which matches with \( \vec{J} = \sigma \vec{E} \), so it matches with (IV).
- Ohm's Law \( \vec{J} = \sigma \vec{E} \), so it matches with (IV).
- Gauss's Law is represented by \( \vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0} \), so it matches with (I).
Step 2: Conclusion.
Thus, the correct match is (2).