Q.What is displacement current? Why did Maxwell find it necessary to introduce this concept into Ampère's circuital law?

Showing the 12 most recent of 16 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.Displacement current is produced due to :

   (a) Constant electric field

   (b) Constant magnetic field

   (c) Changing electric field

   (d) Changing magnetic field

   ›Reveal solutionSolution

   (c) Changing electric field

   ✓Final answer

   (c) Changing electric field

   Maxwell introduced the displacement current $i_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}$, which arises from a time-varying (changing) electric field, for example in the region between the plates of a charging capacitor. A constant electric field gives zero displacement current.

2. CBSE 2026Set A1 markMCQ
   Q.The unit of displacement current is (A) Am (B) A (C) ΩA (D) ΩmA

   ›Reveal solutionSolution

   Displacement current is a form of current; its unit is the ampere (A).

   Maxwell introduced the displacement current to complete Ampère's law:

   $I_d=\varepsilon_0\frac{d\Phi_E}{dt}$

   Although it arises from a changing electric flux rather than moving charge, it has the same dimensions and role as an ordinary conduction current, so its SI unit is the ampere (A).

   ✓Final answer

   (B) A.

3. CBSE 2026Set ANNUAL1 mark
   Q.The concept of displacement current was given by a scientist named ______.

   ›Reveal solutionSolution

   Displacement current was proposed by James Clerk Maxwell to fix an inconsistency in Ampere's circuital law for circuits with a changing electric field (as between capacitor plates).

   Maxwell noticed that the original Ampere's law failed for a charging capacitor (no conduction current flows between the plates, yet a magnetic field is still observed there). He resolved this by introducing an additional 'displacement current' term, proportional to the rate of change of electric flux, which restores consistency and also predicts electromagnetic waves.

   ✓Final answer

   James Clerk Maxwell.

4. CBSE 2025Set X11 markMCQ
   Q.'Ampere-Maxwell Law' is written as (symbols have usual meanings) :

   (a) $\oint \vec{B}\cdot \vec{dl} = \mu_0 i + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$

   (b) $\oint \vec{B}\cdot \vec{dl} = \mu_0 i + \varepsilon_0 \dfrac{d\phi_E}{dt}$

   (c) $\oint \vec{B}\cdot \vec{dl} = \mu_0 i$

   (d) $\oint \vec{E}\cdot \vec{dl} = -\dfrac{d\phi_B}{dt}$

   ›Reveal solutionSolution

   (a) $\oint \vec{B}\cdot d\vec{l} = \mu_0 i + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$. Maxwell added the displacement-current term $\mu_0\varepsilon_0\dfrac{d\phi_E}{dt}$ to Ampere's law. Option (

   ✓Final answer

   (a) $\oint \vec{B}\cdot d\vec{l} = \mu_0 i + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$.

   Maxwell added the displacement-current term $\mu_0\varepsilon_0\dfrac{d\phi_E}{dt}$ to Ampere's law. Option (b) is dimensionally wrong (missing $\mu_0$ on the second term), (c) is the original Ampere's law, and (d) is Faraday's law.

5. CBSE 2024Set 55/1/11 markMCQ
   Q.In the four regions, I, II, III and IV, the electric fields are described as : Region I : $E_x = E_0 \sin(kz - \omega t)$ Region II : $E_x = E_0$ Region III : $E_x = E_0 \sin kz$ Region IV : $E_x = E_0 \cos kz$ The displacement current will exist in the region : (A) I (B) IV (C) II (D) III

   ›Reveal solutionSolution

   Displacement current exists wherever the electric field varies with time. Only Region I has an explicit time dependence ($\sin(kz - \omega t)$), so displacement current exists only in Region I. The correct option is (A).

   Concept and Intuition

   Displacement current is not a current of moving charges — it is a term Maxwell added to Ampère's law to account for changing electric fields. The key idea is simple: wherever the electric field changes with time, there is a displacement current density given by $J_d = \varepsilon_0 \frac{\partial E}{\partial t}$.

   So the question reduces to: in which of these four regions does the electric field explicitly depend on time? A field that is constant in time, or one that depends only on position (like a static pattern), produces no displacement current.

   Let’s examine each region carefully.

   ---

   Step-by-step reasoning

   1. Region I: $E_x = E_0 \sin(kz - \omega t)$ This is a travelling wave — the field depends on both position $z$ and time $t$ through the combination $kz - \omega t$. Compute the partial derivative with respect to time:

   $\frac{\partial E_x}{\partial t} = E_0 \cdot (-\omega) \cos(kz - \omega t) = -\omega E_0 \cos(kz - \omega t)$

   This is non-zero (except at isolated instants). Therefore, displacement current exists in Region I.

   2. Region II: $E_x = E_0$ This is a constant, uniform field — no dependence on $t$ at all.

   $\frac{\partial E_x}{\partial t} = 0$

   No displacement current.

   3. Region III: $E_x = E_0 \sin kz$ Here the field depends only on position $z$, not on time.

   $\frac{\partial E_x}{\partial t} = 0$

   No displacement current. (This is a static sinusoidal pattern, like a standing wave at a frozen instant.)

   4. Region IV: $E_x = E_0 \cos kz$ Again, only a spatial dependence — no $t$ appears.

   $\frac{\partial E_x}{\partial t} = 0$

   No displacement current.

   Watch out

   A common mistake is to think that any spatially varying field (like $\sin kz$ or $\cos kz$) implies a time variation. But $\sin kz$ is a snapshot — it does not change with time unless $t$ appears explicitly. Displacement current requires $\partial E/\partial t \neq 0$, not $\partial E/\partial z \neq 0$.

   Tip

   If you ever see an electric field written as $f(kz - \omega t)$ or $f(kz + \omega t)$, it always has a time derivative. If you see $f(kz)$ or $f(\omega t)$ alone, check which variable is missing — only the presence of $t$ matters for displacement current.

   ---

   ✓Final answer

   The displacement current exists only in Region I, so the correct option is (A).

6. CBSE 2024Set A11 markMCQ
   Q.According to the generalised Ampere-Maxwell law, $\oint \vec{B}\cdot\vec{dl}$ is equal to

   (a) $\mu_0 I_c + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$

   (b) $\mu_0 I_c$

   (c) $\varepsilon_0 \dfrac{d\phi_E}{dt}$

   (d) $\mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$

   ›Reveal solutionSolution

   (a) $\mu_0 I_c + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$

   ✓Final answer

   (a) $\mu_0 I_c + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$

   The generalised Ampere–Maxwell law adds the displacement current $I_d=\varepsilon_0\dfrac{d\phi_E}{dt}$ to the conduction current $I_c$:

   $\oint\vec{B}\cdot d\vec{l}=\mu_0\left(I_c+\varepsilon_0\frac{d\phi_E}{dt}\right)=\mu_0 I_c+\mu_0\varepsilon_0\frac{d\phi_E}{dt}.$

7. CBSE 2024Set A1 mark
   Q.Fill in the blank with appropriate word: Variable electric field produce ______ current.

   ›Reveal solutionSolution

   Maxwell showed that a changing electric field acts as a source of magnetic field, just like a conduction current — this is called displacement current.

   Maxwell noted that Ampere's circuital law (in its original form) was inconsistent when applied to a charging capacitor. To fix this, he proposed that a time-varying electric field between the capacitor plates also gives rise to a magnetic field, exactly as a conduction current does. He called this the displacement current, defined as:

   $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$

   where $\Phi_E$ is the electric flux. This modification led to the generalised Ampere-Maxwell law and predicted the existence of electromagnetic waves.

   ✓Final answer

   A variable (changing) electric field produces a displacement current.

8. CBSE 2024Set ANNUAL1 markMCQ
   Q.The formula for displacement current (I_d) is -

   (a) μ0 dφE/dt

   (b) μ0 ε0 dφE/dt

   (c) ε0 dφE/dt

   (d) (1/ε0) dφE/dt

   ›Reveal solutionSolution

   Maxwell introduced displacement current to make Ampere's law consistent for time-varying electric fields (e.g. between capacitor plates); it equals ε₀ times the rate of change of electric flux.

   Maxwell showed that a changing electric flux $\phi_E$ through a surface is equivalent, in its magnetic effects, to a real conduction current. He defined the displacement current as:

   $I_d = \varepsilon_0 \dfrac{d\phi_E}{dt}$

   This term was added to Ampere's circuital law (giving the Ampere-Maxwell law) so that it holds even in regions like the gap between charging capacitor plates, where there is a changing electric field but no conduction current.

   ✓Final answer

   (c) $\varepsilon_0 \dfrac{d\phi_E}{dt}$.

9. CBSE 2023Set 55/1/11 markMCQ
   Q.In the process of charging of a capacitor, the current produced between the plates of the capacitor is :

   (a) $-\varepsilon_0\dfrac{d\Phi_E}{dt}$

   (b) $-\dfrac{1}{\varepsilon_0}\dfrac{d\Phi_E}{dt}$

   (c) $\varepsilon_0\dfrac{d\Phi_E}{dt}$

   (d) $\dfrac{1}{\varepsilon_0}\dfrac{d\Phi_E}{dt}$ where symbols have their usual meanings.

   ›Reveal solutionSolution

   The current between the plates of a charging capacitor is the displacement current, given by $\varepsilon_0 \frac{d\Phi_E}{dt}$, which matches option (c).

   Why This Question Matters

   When a capacitor charges, no actual charge carriers flow across the gap between the plates. Yet a magnetic field is observed around the gap — as if a current were flowing. Maxwell resolved this paradox by introducing the displacement current, a term that accounts for the changing electric field between the plates. This is the key idea behind the question.

   The symbols have their usual meanings: $\Phi_E$ is the electric flux through a surface between the plates, and $\varepsilon_0$ is the permittivity of free space.

   Step-by-Step Reasoning

   1. Recall Maxwell’s correction to Ampere’s law Ampere’s circuital law in its original form ($\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$) fails for a charging capacitor — because the enclosed current changes depending on the surface chosen. Maxwell added a term to fix this:

   $\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I_{\text{conduction}} + I_{\text{displacement}} \right)$

   where $I_{\text{displacement}}$ is the displacement current.

   2. Define the displacement current Maxwell showed that a changing electric field produces a magnetic field just as a conduction current does. The displacement current through a surface is defined as:

   $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$

   Here $\Phi_E = \int \vec{E} \cdot d\vec{A}$ is the electric flux through the surface.

   3. Apply to the region between capacitor plates Between the plates of a charging capacitor, there is no conduction current — no free charges move across the gap. However, the electric field between the plates is changing as charge builds up. Therefore, the current that appears between the plates is purely the displacement current:

   $I_{\text{between plates}} = \varepsilon_0 \frac{d\Phi_E}{dt}$

   4. Check the sign The question asks for the current produced — the magnitude and direction matter. In Maxwell’s equations, the displacement current term appears with a positive sign in Ampere’s law. The negative sign in options (a) and (b) would imply the displacement current opposes the change in flux, which is not the case. So the correct expression has a positive sign.

   Tip

   A quick way to remember: Displacement current = $\varepsilon_0$ times the rate of change of electric flux. The $\varepsilon_0$ is in the numerator, never the denominator.

   5. Eliminate the wrong options
      • (a) $-\varepsilon_0 \frac{d\Phi_E}{dt}$ — wrong sign
      • (b) $-\frac{1}{\varepsilon_0} \frac{d\Phi_E}{dt}$ — wrong sign and wrong placement of $\varepsilon_0$
      • (d) $\frac{1}{\varepsilon_0} \frac{d\Phi_E}{dt}$ — $\varepsilon_0$ should be in the numerator, not denominator

   Watch out

   A common mistake is to confuse displacement current with the displacement current density. The current density is $\vec{J}_d = \varepsilon_0 \frac{\partial \vec{E}}{\partial t}$, and integrating it over area gives $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$. Don’t drop the $\varepsilon_0$ or invert it.

   Final Answer

   ✓Final answer

   The correct option is (c) $\varepsilon_0 \dfrac{d\Phi_E}{dt}$.

10. CBSE 2022Set HE2171 mark
    Q.What is the reason for origin of Displacement current?

    ›Reveal solutionSolution

    Displacement current arises from a changing electric field, and was introduced by Maxwell to make Ampere's law consistent for circuits with capacitors.

    Maxwell noticed that Ampere's circuital law, $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$, gave inconsistent results for a charging capacitor: for an Amperian loop drawn as a surface passing between the capacitor plates (where no conduction current flows), the law would predict zero magnetic field, contradicting the case of a surface cutting the connecting wire. To resolve this, Maxwell proposed that a time-varying electric field between the plates itself acts as a source of magnetic field, just like a current does, and defined the displacement current $I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}$ (proportional to the rate of change of electric flux). So the origin of displacement current is precisely a changing electric field with time.

    ✓Final answer

    A displacement current originates from a time-varying electric field (rate of change of electric flux), $I_d = \varepsilon_0\,d\Phi_E/dt$, e.g. between the plates of a charging capacitor.

11. CBSE 2020Set 55/1/11 mark
    Q.Write the mathematical form of Ampere-Maxwell circuital law.

    ›Reveal solutionSolution

    Ampere-Maxwell law generalizes Ampere's law by adding Maxwell's displacement current term; it states that magnetic circulation equals the sum of conduction and displacement currents: $\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I_{\text{enc}} + \epsilon_0 \frac{d\Phi_E}{dt} \right)$.

    Why Maxwell had to fix Ampere's law

    Ampere's original circuital law worked beautifully for steady currents: the line integral of the magnetic field around a closed loop equals $\mu_0$ times the current threading through. But it failed spectacularly when currents changed with time, particularly in situations like a charging capacitor where current flows in the wires but no charge physically crosses the gap between the plates.

    Maxwell realized the flaw: a changing electric field between the capacitor plates should contribute to the magnetic field just as a real current does. He introduced the displacement current — not a flow of charge, but a changing electric flux that has the same magnetic effect. This insight unified electromagnetism and predicted electromagnetic waves.

    The mathematical statement

    The Ampere-Maxwell circuital law can be written in two equivalent forms:

    Integral form:

    $\oint_C \vec{B} \cdot d\vec{l} = \mu_0 \left( I_{\text{enc}} + \epsilon_0 \frac{d\Phi_E}{dt} \right)$

    where:

    • $\oint_C \vec{B} \cdot d\vec{l}$ is the circulation of magnetic field around closed path $C$
    • $I_{\text{enc}}$ is the conduction current passing through any surface bounded by $C$
    • $\epsilon_0 \frac{d\Phi_E}{dt}$ is the displacement current, with $\Phi_E = \int \vec{E} \cdot d\vec{A}$ being the electric flux

    Differential form:

    $\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$

    where:

    • $\nabla \times \vec{B}$ is the curl of the magnetic field
    • $\vec{J}$ is the conduction current density
    • $\frac{\partial \vec{E}}{\partial t}$ is the rate of change of electric field

    The two forms are related by Stokes' theorem. The differential form is one of Maxwell's four equations that govern all classical electromagnetic phenomena.

    Tip

    Remember the displacement current term as "$\epsilon_0$ times the rate of change of electric flux" — it has units of current and creates magnetic fields exactly like real current does, which is why electromagnetic waves can propagate through vacuum.

    ✓Final answer

    The Ampere-Maxwell circuital law is $\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I_{\text{enc}} + \epsilon_0 \frac{d\Phi_E}{dt} \right)$ in integral form, or $\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$ in differential form.

12. CBSE 2020Set 55/2/11 markMCQ
    Q.Displacement current exists only when (A) electric field is changing. (B) magnetic field is changing. (C) electric field is not changing. (D) magnetic field is not changing.

    ›Reveal solutionSolution

    Displacement current arises from a changing electric field, not a changing magnetic field. The correct option is (A).

    Concept and Intuition

    The idea of displacement current was introduced by James Clerk Maxwell to fix a logical gap in Ampère’s law. Ampère’s law originally said that a magnetic field is produced by a conduction current (moving charges). But consider a charging capacitor: between the plates, no charges flow, yet a magnetic field is still observed there. Something must be “acting” like a current in that gap.

    Maxwell realised that what changes between the plates is the electric field — it builds up as charge accumulates. He proposed that a changing electric field itself generates a magnetic field, exactly as a current would. He called this effect displacement current. So displacement current exists only when the electric field is changing with time.

    Watch out

    A common mistake is to think displacement current relates to a changing magnetic field — that’s electromagnetic induction (Faraday’s law), not displacement current. The two are symmetric but distinct.

    Step-by-Step Reasoning

    1. Recall the definition of displacement current

       Displacement current density is given by $\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$.

       The total displacement current through a surface is $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$, where $\Phi_E$ is the electric flux.

       Both expressions contain a time derivative of the electric field — so if $\vec{E}$ is constant, $\frac{\partial \vec{E}}{\partial t} = 0$ and displacement current is zero.

    2. Examine each option

       • (A) electric field is changing → $\frac{\partial \vec{E}}{\partial t} \neq 0$ → displacement current exists.
       • (B) magnetic field is changing → This is Faraday’s law, not displacement current. Changing magnetic field induces an electric field, but displacement current depends on changing electric field.
       • (C) electric field is not changing → $\frac{\partial \vec{E}}{\partial t} = 0$ → no displacement current.
       • (D) magnetic field is not changing → Irrelevant; displacement current doesn’t depend on magnetic field changes.

    3. Confirm with Maxwell’s correction to Ampère’s law

       Maxwell’s equation: $\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + \epsilon_0 \frac{d\Phi_E}{dt})$.

       The term $\epsilon_0 \frac{d\Phi_E}{dt}$ is the displacement current. It is zero unless $\Phi_E$ changes — i.e., unless the electric field changes.

    Tip

    A neat way to remember: Displacement current ↔ Displacement of electric field lines — if the field lines are stationary, there’s no displacement current.

    ✓Final answer

    The correct option is (A) — displacement current exists only when the electric field is changing.