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Capacitor question

2007 · Shift 1 · Q63
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  5. /2007 · Shift 1 · Q63

Capacitor question

2007 · Shift 1 · Q63

JEE AdvancedPhysicsCapacitorMCQ+3 / −1

Column I gives certain situations in which a straight metallic wire of resistance R 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 ×\times× 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

  1. A
    A →\to→(Q); B →\to→(R); C →\to→(R, S); D →\to→ (P, Q)
  2. B
    A →\to→(Q); B →\to→(R, S); C →\to→(R, S); D →\to→ (P, Q, R)
  3. C
    A →\to→(Q); B →\to→(R, S); C →\to→(R); D →\to→ (Q, R)
  4. D
    A →\to→(Q, R); B →\to→(R, S); C →\to→(R); D →\to→ (P, Q, R)
View written solutionFree

Correct answer: B

  1. Analyze each case in Column I

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

A charged capacitor discharges through the wire.

  • Current flows only temporarily, so not constant current.
  • Since current flows through resistance, thermal energy is generated.
  • The potential difference across the wire is not constant; it decreases with time.
  • Charges at the ends are also not of constant magnitude.

So, A→QA \to QA→Q

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

This is motional emf.

Magnitude of induced emf: E=Bℓv\mathcal{E} = B\ell vE=Bℓv Since B,ℓ,vB, \ell, vB,ℓ,v are constant, the emf is constant.

  • Hence a constant potential difference develops between the ends.
  • Charges accumulate at the ends until electrostatic force balances magnetic force, so charges of constant magnitude appear at the ends.
  • In the absence of a closed circuit, no steady current flows through the wire.
  • Therefore no Joule heating in the wire due to internal current flow.

So, B→(R,S)B \to (R,S)B→(R,S)

(C) The wire is placed in a constant electric field along its length

A conductor in an external constant electric field cannot sustain electric field inside in electrostatic equilibrium.

Free charges redistribute until internal electric field cancels the applied field.

  • Thus a constant potential difference develops between the ends.
  • Also charges of constant magnitude appear at the ends due to polarization.
  • No continuous current flows in electrostatic equilibrium.
  • No thermal energy generation.

So, C→(R,S)C \to (R,S)C→(R,S)

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

A constant emf across resistance RRR gives I=ERI = \frac{\mathcal{E}}{R}I=RE​ which is constant.

  • So constant current flows.
  • Joule heating occurs: P=I2RP = I^2RP=I2R so thermal energy is generated.
  • Since emf is constant, constant potential difference exists across the wire.
  • Charges at ends are not the relevant static accumulation effect here.

So, D→(P,Q,R)D \to (P,Q,R)D→(P,Q,R)

  1. Collect the matches
  • A→QA \to QA→Q
  • B→(R,S)B \to (R,S)B→(R,S)
  • C→(R,S)C \to (R,S)C→(R,S)
  • D→(P,Q,R)D \to (P,Q,R)D→(P,Q,R)
  1. Compare with options

This exactly matches Option B.

  1. Comparison with stored answer

Stored correct answer = B

Derived answer = B

So they agree.

Previous

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