Match Metallic Wire Situations with Resulting Electrical Effects

To match the physical situations in Column I with the resulting effects in Column II, we analyze each case individually:

1. (A) A charged capacitor is connected to the ends of the wire:

   • The capacitor discharges through the wire of resistance $R$ with a time-dependent transient current given by:
     $i(t) = \frac{V_0}{R} e^{-t/RC}$
   • Because the current is decaying exponentially, it is not constant, nor is the potential difference constant.
   • The passage of current through the resistance produces Joule heating:
     $H = \int_{0}^{\infty} i^2(t) R \, dt = \frac{1}{2} C V_0^2 > 0$
     Thus, thermal energy is generated in the wire.
   • Therefore, (A) matches with (q).

2. (B) The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion:

   • As the wire moves with velocity $\vec{v}$ in a magnetic field $\vec{B}$, free conduction electrons experience a magnetic Lorentz force $\vec{F}_m = -e(\vec{v} \times \vec{B})$ directed along the length of the wire.
   • This causes electrons to accumulate at one end of the wire, leaving an excess of positive charge at the other end.
   • In steady state, the electric field $\vec{E}$ created by these accumulated charges balances the magnetic force:
     $eE = evB \implies E = vB$
   • Consequently, fixed charges of constant magnitude appear at the ends of the wire, giving effect (s).
   • This electric field establishes a constant motional electromotive force (potential difference) between the ends of the wire:
     $\Delta V = E l = B v l$
     giving effect (r).
   • Since the circuit is open, no continuous current flows ($i = 0$), so no thermal energy is generated in the steady state.
   • Therefore, (B) matches with (r) and (s).

3. (C) The wire is placed in a constant electric field directed along its length:

   • The metallic wire acts as an isolated conductor placed in an external electrostatic field $\vec{E}_{\text{ext}}$.
   • Free electrons redistribute along the wire until the induced internal electric field completely cancels the external electric field inside the conductor ($\vec{E}_{\text{net}} = 0$).
   • In this electrostatic equilibrium, charges of constant magnitude accumulate at the opposite ends of the wire, giving effect (s).
   • Since the net electric field inside the wire is zero, the entire wire becomes an equipotential body (i.e., the potential difference between the ends is zero).
   • No continuous current flows in electrostatic equilibrium, and hence no thermal energy is generated.
   • Therefore, (C) matches with (s).

4. (D) A battery of constant emf is connected to the ends of the wire:

   • The battery maintains a constant potential difference $V$ across the ends of the wire, giving effect (r).
   • By Ohm's law, this constant potential difference drives a constant electric current through the wire:
     $I = \frac{V}{R} = \text{constant}$
     giving effect (p).
   • The flow of current through the wire of resistance $R$ continuously dissipates electrical energy into heat at a constant rate $P = I^2 R$, resulting in the generation of thermal energy, giving effect (q).
   • Therefore, (D) matches with (p), (q), and (r).

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Final Match:

• (A) $\to$ (q)
• (B) $\to$ (r), (s)
• (C) $\to$ (s)
• (D) $\to$ (p), (q), (r)