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Displacement current: NEET MCQs with answers

50 questions on this topic of Electromagnetic Waves (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The dimensions of the quantity ε₀ (dΦ_E/dt) are identical to the dimensions of which of the following physical quantities?

  1. Magnetic flux
  2. Electric current
  3. Electric potential
  4. Magnetic field
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Answer: (B) Electric current

The expression ε₀ (dΦ_E/dt) defines the displacement current, I_d. Since it represents a current, its dimensional formula is [A], which is identical to that of electric current.

medium

2. The Ampere-Maxwell law states that ∮B · dl equals:

  1. Zero
  2. μ₀(iC + iD)
  3. μ₀iC only
  4. ε₀ dΦB/dt
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Answer: (B) μ₀(iC + iD)

Maxwell generalised Ampere's circuital law to ∮B · dl = μ₀(iC + iD), including both conduction and displacement currents.

NCERT Class 12

hard

3. A parallel plate capacitor with circular plates of area A is being charged. At an instant, the displacement current density is J_d. Consider a point P between the plates at distance r from the central axis (r < plate radius). The ratio of the induced magnetic energy density (u_B = B² / 2μ₀) to the displacement current density squared (J_d²) at point P is:

  1. μ₀ r² / 8
  2. μ₀ r² / 4
  3. μ₀² r² / 4
  4. μ₀ r / 2
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Answer: (A) μ₀ r² / 8

Inside the circular plates, the induced magnetic field is B = μ₀ J_d r / 2. The magnetic energy density is u_B = B² / (2μ₀) = (μ₀² J_d² r² / 4) / (2μ₀) = μ₀ J_d² r² / 8. Therefore, the ratio u_B / J_d² = μ₀ r² / 8.

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4. A circular parallel plate capacitor of radius R is being charged. What fraction of the total displacement current passes through a coaxial circular area of radius R/2 between the plates?

  1. 1/4
  2. 1/2
  3. 1/8
  4. 1/16
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Answer: (A) 1/4

Assuming a uniform rate of change of electric field across the plate area, the displacement current is directly proportional to the enclosed cross-sectional area. The area of radius R/2 is π(R/2)² = (1/4)πR², which contains exactly 1/4 of the total displacement current.

easy

5. Consider the following two statements: Statement I: Displacement current produces a magnetic field just like conduction current. Statement II: Inside a non-conducting dielectric placed between the plates of a discharging capacitor, the conduction current is non-zero.

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is true but Statement II is false.
  4. Statement I is false but Statement II is true.
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Answer: (C) Statement I is true but Statement II is false.

Statement I is true because Maxwell modified Ampere's law precisely because a changing electric flux creates an effective current that produces a magnetic field. Statement II is false because in an ideal non-conducting dielectric, there are no free charge carriers, so the conduction current is zero.

hard

6. A parallel plate capacitor consists of circular plates of radius R. During charging, at what radial distance r from the central axis (r > R) is the induced magnetic field magnitude equal to half of its value at r = R/2?

  1. 4R
  2. 2R
  3. R / 4
  4. √2 R
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Answer: (A) 4R

Inside the plates (r ≤ R), B_in(r) = (μ₀ i_d r) / (2πR²). At r = R/2, B_in = (μ₀ i_d) / (4πR). Outside the plates (r > R), all displacement current is enclosed, so B_out(r) = (μ₀ i_d) / (2πr). We are given B_out(r) = (1/2) B_in(R/2) = (1/2) × [(μ₀ i_d) / (4πR)] = (μ₀ i_d) / (8πR). Setting (μ₀ i_d) / (2πr) = (μ₀ i_d) / (8πR) yields 2πr = 8πR, which gives r = 4R.

medium

7. A parallel plate capacitor having square plates of side 5 cm is being charged. The rate of change of the electric field between the plates is 4 × 10¹² V/(m·s). What is the displacement current between the plates? (Take ε₀ = 8.85 × 10⁻¹² C²/N·m²)

  1. 2.21 × 10⁻¹ A
  2. 8.85 × 10⁻² A
  3. 8.85 × 10⁻⁴ A
  4. 4.42 × 10⁻² A
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Answer: (B) 8.85 × 10⁻² A

Displacement current is given by i_d = ε₀ A (dE/dt). The area A = (0.05 m)² = 2.5 × 10⁻³ m². Therefore, i_d = 8.85 × 10⁻¹² × 2.5 × 10⁻³ × 4 × 10¹² = 8.85 × 10⁻² A.

medium

8. A parallel plate capacitor is charged by a constant current I. If the separation between the plates is doubled while keeping the charging current I constant, the displacement current between the plates:

  1. remains I
  2. becomes I / 2
  3. becomes 2I
  4. becomes I / 4
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Answer: (A) remains I

By the principle of continuity of current established by Maxwell, the displacement current i_d between the plates is always equal to the conduction current I charging the capacitor, regardless of the plate separation.

medium

9. A voltage V(t) = V₀ cos(ωt) is applied across a parallel plate capacitor of capacitance C. What is the magnitude of the displacement current between the plates at time t = π / (2ω)?

  1. 0
  2. C ω V₀
  3. (1/2) C ω V₀
  4. 2 C ω V₀
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Answer: (B) C ω V₀

The conduction current entering the capacitor equals the displacement current: i_d = dq/dt = C (dV/dt) = -C ω V₀ sin(ωt). At t = π / (2ω), sin(ωt) = sin(π/2) = 1. Therefore, the magnitude of the displacement current is C ω V₀.

medium

10. An ideal capacitor of capacitance C is being discharged through a resistor R. At time t = 0, the charge on the capacitor is Q₀. At time t = RC, the magnitude of the displacement current inside the capacitor is:

  1. Q₀ / (e² R C)
  2. Q₀ / (R C)
  3. Q₀ (1 - 1/e) / (R C)
  4. Q₀ / (e R C)
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Answer: (D) Q₀ / (e R C)

During discharging, Q(t) = Q₀ e^(-t/RC). The discharging conduction current is i(t) = |dQ/dt| = (Q₀ / RC) e^(-t/RC). Because the magnitude of displacement current equals the conduction current at any instant, at t = RC, I_d = (Q₀ / RC) e^(-1) = Q₀ / (e RC).

medium

11. The electric flux through a certain planar loop varies with time as Φ_E = (4t³ + 5t² - 2t + 1) V·m. The magnitude of the displacement current through the loop at t = 2 s is:

  1. 34 ε₀
  2. 42 ε₀
  3. 50 ε₀
  4. 66 ε₀
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Answer: (D) 66 ε₀

Displacement current is given by I_d = ε₀ (dΦ_E/dt). Differentiating Φ_E with respect to t gives dΦ_E/dt = 12t² + 10t - 2. At t = 2 s, dΦ_E/dt = 12(4) + 10(2) - 2 = 48 + 20 - 2 = 66 V·m/s. Thus, I_d = 66 ε₀.

medium

12. A parallel plate capacitor is formed of two circular plates of area 0.05 m². The electric field between the plates is given by E = (4 × 10⁵ - 2 × 10⁴ t) V m⁻¹, where t is in seconds. The magnitude of the displacement current between the plates is:

  1. 4.42 × 10⁻⁸ A
  2. 8.85 × 10⁻⁹ A
  3. 1.77 × 10⁻⁷ A
  4. 8.85 × 10⁻⁷ A
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Answer: (B) 8.85 × 10⁻⁹ A

The displacement current is given by I_d = ε₀ A |dE/dt|. Here, |dE/dt| = 2 × 10⁴ V m⁻¹ s⁻¹ and A = 0.05 m². Therefore, I_d = (8.85 × 10⁻¹²)(0.05)(2 × 10⁴) = 8.85 × 10⁻⁹ A.

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