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Question

In Maxwell's revision of Ampere's circuital law, the displacement current density, $\vec{J_D}$, was introduced to ensure consistency and is explicitly defined as being directly proportional to:

The correct answer is

the time rate of change of the electric field, $\frac{\partial \vec{E}}{\partial t}$

Understanding Maxwell's Revision of Ampere's Law and Displacement Current

This question delves into Maxwell's significant modification of Ampere's circuital law. Ampere's law, in its original form, relates the magnetic field around a closed loop to the electric current passing through the loop. However, Maxwell realized this law was incomplete, particularly in situations involving changing electric fields, such as in capacitors during charging. He introduced the concept of displacement current to make the law consistent with the conservation of charge and wave phenomena.

The core of Maxwell's addition was the displacement current density, denoted as $\vec{J_D}$. We need to determine what factor this quantity is directly proportional to, according to Maxwell's theory.

The Concept of Displacement Current Density

Maxwell's equations form the foundation of classical electromagnetism. The revised form of Ampere's circuital law, incorporating Maxwell's contribution, is given by:

$ \nabla \times \vec{H} = \vec{J} + \vec{J_D} $

Here, $\nabla \times \vec{H}$ represents the curl of the magnetic field intensity, $\vec{J}$ is the conduction current density (the flow of charge), and $\vec{J_D}$ is the displacement current density.

Maxwell defined the displacement current density $\vec{J_D}$ based on the time-varying electric field. Specifically, it is related to the time rate of change of the electric displacement field, $\vec{D}$. The definition is:

$ \vec{J_D} = \frac{\partial \vec{D}}{\partial t} $

In many materials (specifically, linear, isotropic dielectrics), the electric displacement field $\vec{D}$ is related to the electric field $\vec{E}$ by $\vec{D} = \epsilon \vec{E}$, where $\epsilon$ is the permittivity of the material. Substituting this into the definition of $\vec{J_D}$, we get:

$ \vec{J_D} = \frac{\partial (\epsilon \vec{E})}{\partial t} = \epsilon \frac{\partial \vec{E}}{\partial t} $

This equation clearly shows that the displacement current density ($\vec{J_D}$) is directly proportional to the time rate of change of the electric field ($\frac{\partial \vec{E}}{\partial t}$). The constant of proportionality is the permittivity ($\epsilon$).

Analysis of Options

Let's analyze the given options in light of this understanding:

  • The curl of the magnetic field intensity, $\nabla \times \vec{H}$: This term, in the modified Ampere's law, equals the sum of conduction current density and displacement current density ($\vec{J} + \vec{J_D}$). While related, $\vec{J_D}$ itself is not directly proportional *only* to this term.
  • The divergence of the magnetic field, $\nabla \cdot \vec{B}$: This represents Gauss's law for magnetism and states that there are no magnetic monopoles ($\nabla \cdot \vec{B} = 0$). It is fundamentally different from the definition of displacement current density.
  • The time rate of change of the electric field, $\frac{\partial \vec{E}}{\partial t}$: As derived above, the displacement current density $\vec{J_D}$ is defined as being proportional to this quantity ($\vec{J_D} = \epsilon \frac{\partial \vec{E}}{\partial t}$). This aligns perfectly with Maxwell's theory.
  • The negative gradient of the electric potential, $-\nabla V$: This quantity, $-\nabla V$, represents the electric field $\vec{E}$ in electrostatic situations (where $\vec{E} = -\nabla V$). However, the displacement current depends on the *time-changing* electric field, not just the static electric field or its potential gradient.

Conclusion

Maxwell's crucial insight was recognizing that a changing electric field creates a magnetic field, just like a moving electric charge does. He quantified this by introducing the displacement current density, $\vec{J_D}$, which is fundamentally linked to how quickly the electric field is changing over time. Therefore, the displacement current density is directly proportional to the time rate of change of the electric field.

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Important Questions from Capacitance

  1. A parallel plate capacitor having cross-sectional area 'A' and separated by distance 'd' is filled by copper plate of thickness b. It's capacitance is :

  2. The unit of capacitance is farad. 1 farad is equal to _________.

  3. Which of the following components store energy in the form of electrical charges?

  4. The capacitance of a capacitor is given by C = Q/V. The capacitance depends on ______.

  5. Whose SI unit is Farad?

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