Column I gives certain situations in which a straight metallic wire of resistance is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS.
| Column I | Column II |
|---|---|
| (A) A charged capacitor is connected to the ends of the wire | (p) A constant current flows through the wire |
| (B) The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion | (q) Thermal energy is generated in the wire |
| (C) The wire is placed in a constant electric field that has a direction along the length of the wire | (r) A constant potential difference develops between the ends of the wire |
| (D) A battery of constant emf is connected to the ends of the wire | (s) Charges of constant magnitude appear at the ends of the wire |
Answer: A → II, B → III, B → IV, C → IV, D → I, D → II, D → III
Solution
(A) Charged capacitor connected: The capacitor discharges through the wire, producing a time-varying (not constant) current. Thermal energy is generated due to resistance. Matches (q) only.
(B) Wire moving in magnetic field: A motional emf is induced (), creating a constant potential difference across the wire. Charges accumulate at the ends to maintain this potential. Matches (r) and (s).
(C) Wire in constant electric field: The electric field causes charge redistribution, with charges accumulating at the ends to create an internal field that cancels the external field. A constant potential difference develops. Matches (r) and (s).
(D) Battery connected: A constant emf drives a constant current through the wire (Ohm's law). Thermal energy is generated (). A constant potential difference (equal to the emf) exists across the wire. Matches (p), (q), and (r).
Pairs: A→(q); B→(r),(s); C→(r),(s); D→(p),(q),(r).
(B) Wire moving in magnetic field: A motional emf is induced (), creating a constant potential difference across the wire. Charges accumulate at the ends to maintain this potential. Matches (r) and (s).
(C) Wire in constant electric field: The electric field causes charge redistribution, with charges accumulating at the ends to create an internal field that cancels the external field. A constant potential difference develops. Matches (r) and (s).
(D) Battery connected: A constant emf drives a constant current through the wire (Ohm's law). Thermal energy is generated (). A constant potential difference (equal to the emf) exists across the wire. Matches (p), (q), and (r).
Pairs: A→(q); B→(r),(s); C→(r),(s); D→(p),(q),(r).
JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Match the columns