Displacement current is the current that is produced by the rate of change of the electric displacement field. It differs from the normal current that is produced by the motion of the electric charge. Displacement current is the quantity explained in Maxwell’s Equation. It is measured in Ampere. Displacement currents are produced by a time-varying electric field rather than moving charges.

In this article we will learn about, displacement current, its characteristics, and others in detail.

## What is Displacement Current?

Electricity and magnetism are related to each other. As the electric current travels through a wire, it creates magnetic field lines around the wire. This type of current is called conduction current, which is created by the movement of electrons through a conductor such as an electrical wire.

Whereas a displacement current is a type of current linked with Maxwell’s Equation and is produced by a time-varying electric field.

### Displacement Current Definition

A physical quantity related to Maxwell’s equation that has the property of the electric current is called the Displacement Current. Displacement current is defined as the rate of change of the electric displacement field (D).

Maxwell’s equation includes displacement current that proves the Ampere Circuit Law. It is measured in Ampere.

### Current in Capacitor

A charging capacitor has no conduction of charge but the charge accumulation in the capacitor changes the electric field link with the capacitor that in turn produces the current called the Displacement Current.

I_{D}= J_{D}S = S(∂D/∂t)where,

- S is the area of the Capacitor Plate
- I
_{D}is the Displacement Current- J
_{D}is the Displacement Current Density.- D is related to Electric Field E as,

D = εE

- ε is the Permittivity of material between plates

## Displacement Current Equation

Maxwell’s Equation defines the displacement current which has the same unit as the electric current, the Maxwell field equation is represented as,

∇×H = J + J_{D}where,

H is related to magnetic field B as B = μH

μ is the permeability of the material between the plates

J is the Conducting Current Density.

J_{D}is the Displacement Current Density.

We know that

**∇.(∇×H) = 0**

**∇.J = -∂ρ/∂t**

**∇.J = -∇.∂D/∂t**

Using Gauss’s Law

**∇.D = ρ**

Here, ρ is the electric charge density.

Thus the displacement current density equation is,

J_{D}= ∂D/∂t

### Characteristics of Displacement Current

In an electric circuit, there are two types of current that are conduction current and the other is Displacement current. Various characteristics of displacement current are mentioned below:

- Displacement current does not appear from the actual movement of the electric charge as in the case of the conduction current but is produced by time changing electric field.
- Displacement current is a vector quantity.
- Electromagnetic waves propagate with the help of displacement current

## Electromagnetic Waves (EM)

Electromagnetic Waves are a combination of electric and magnetic field waves produced by moving charges. EM waves are created by the oscillation of electrically charged particles (accelerating charges). The electric field associated with the accelerating charge vibrates due to which a vibrating magnetic field is generated. These vibrating electric and magnetic fields give rise to EM waves. Both the electric and magnetic fields in an electromagnetic wave will fluctuate in time, one causing the other to change.

### Nature of Electromagnetic Waves

In an electromagnetic wave, the electric field and magnetic field are perpendicular to each other and at the same time are perpendicular to the direction of propagation of the wave, this nature of EM wave is known as Transverse nature. In these transverse waves, the direction of disturbance or displacement in the medium is perpendicular to that of the propagation of the wave. The particles of the medium oscillate in a direction perpendicular to the direction of propagation of the wave. Because of this EM waves are transverse in nature.

The electric field of the EM wave is represented as

E_{Y}= E_{0}sin (kx – ωt)where,

E_{Y}= electric field along the y-axis,= direction of propagation of the wavex

Wave numberk = (2π / λ).

The magnetic field of the EM wave is represented as

B_{Z}= B_{0}sin(kx – ωt)where,

B= magnetic field along the z-axis_{Z}= direction of propagation of the wavex

Wave numberk = (2π / λ)

## Faraday’s Law

This law is stated as when an electromotive force is induced in a coil or a circuit when there is a magnetic field varying with time, or there is a rate of change in magnetic flux through the coil. Using Lenz’s law we can determine the direction of the induced emf. It suggests that the induced emf always opposes the cause of its formation in the first place.

The relation between the emf ε in a wire and the electric field E in the wire is given by,

**ε = ∫ E . dl **

Where, dl is the element of the contour of the surface, combining this with the definition of flux,

**φ**_{B}** = ∫ B . dA**

The integral form of the above equation can be written as,

**∫ E . dl = -d/dt ∫ B . dA**

## Maxwell-Ampere Law

The progress in the theory of displacement current can be traced back to a famous physicist named James Clerk Maxwell. Maxwell is well known for Maxwell’s Equations. The combination of four equations demonstrates the fundamentals of electricity and magnetism. For displacement current, we will be focusing on one of these equations known as the Maxwell-Ampere law.

Before Maxwell, Andre-Marie Ampere had developed the famous equation known as Ampere’s law. This law relates the magnetic field (B) surrounding a closed loop to the conduction current (I) traveling through that loop multiplied by a constant known as the permeability of free space (μ_{0}).

∫B . ds = μ_{0}I

Whenever there is continuous conduction current Ampere’s law holds true, but there are cases when problems arise in the law as it’s written. For example, a circuit with a capacitor in it. When the capacitor is charging and discharging, current flows through the wires creating a magnetic field, but between the plates of the capacitor, there is no presence of current flow. According to Ampere’s law, there can be no magnetic field created by the current here, but we know that a magnetic field does exist. Maxwell realized this discrepancy in Ampere’s law and modified it in order to resolve the issue.

∫B . ds = μ_{0}(I + ε_{0}(dφ_{E }/dt))

This final form of the equation is known as the Maxwell-Ampere law.

The part Maxwell added to it is known as displacement current (I_{d}), and the formula is,

I_{d}= ε_{0}(dφ_{E}/dt)

The above equation consists of two terms multiplied together. The first is known as the permittivity of free space (ε_{0}), and the second is the derivative with respect to time and electric flux (φ_{E}). Electric flux is the rate of flow of an electric field through a given area. By taking its derivative with respect to time, we consider the change in that rate of flow over time.

## Displacement Current using Maxwell’s Equation

Maxwell adjusted the main relation of Ampere’s Circuital Law with an additional term. This made the relationship complete and wholesome with both static and time-varying parts present to play their part. Determination of displacement can be done using,

At first, take the magnetic field intensity without the magnetism,

**B = μH**

Addition of an additional term (J_{d}) in the above equation,

**∇ × H = J + J**_{d}

Performing the divergence action,

**∇ . (∇ × H) = 0 = ∇ . J + ∇ . J**_{d}

To define the J_{d} term,

**∇ . J**_{d}** = -∇ . J**

**∇ . J**_{d}** = ∂p**_{v}** / ∂t**………….(1)

Using Gauss’s law, ∇ . D = p, Where D is the displacement vector and p is the charge density. Equation (1) becomes,

**∂p**_{v}** / ∂t = (∂ / ∂t) (∇ . D) = ∇ . (∂D / ∂t)**

**∇ . J**_{d}** = ∇ . (∂D / ∂t)**

Therefore,

**J**_{d}** = ∇ . (∂D / ∂t)**

This derived term is known as the displacement current density formula. This comes in use when time-varying fields are required. For a constant displacement vector, i.e. a constant charge density the displacement current density vanishes.

So, the integral form of Maxwell’s equation is,

∫E . da = Q / ε_{0}

∫B . da = 0

∫E . dl = -∫δB / δt. (da)

∫B . dl = μ_{0}l + μ_{0}ε_{0}∫(∂E / ∂t)

## Need for Displacement Current

Ampere’s circuital law for conduction of current during charging of a capacitor was found inconsistent. Therefore, Maxwell modified Ampere’s circuital law by introducing the concept of displacement current.

Displacement currents play a central role in the propagation of electromagnetic radiation, such as light and radio waves through empty space. It is required to make the conduction current lead in the circuit. Conduction current in wires can be made to lead the voltage by means of displacement current inside the capacitor and it has vast uses in induction motors, industrial appliances, and in our day-to-day life.

**Read More,**

## Solved Examples on Displacement Current

**Example 1: Instantaneous displacement current of 2.0 A is set up in the space between two parallel plates of a 1 μF capacitor. find the rate of change in potential difference across the capacitor.**

**Solution:**

In a Capacitor

V = q/C

dV/dt = i/C

dV/dt = 2.0A / 1μF

= 2 x 10

^{6}V/sThe rate of change in potential difference across the capacitor is, 2 x 10

^{6}V/s

**Example 2: A parallel plate capacitor with plate area A and separation between the plates d, is charged by a constant current I. Consider a plane surface of area A/4 parallel to the plates and drawn between the plates. What is the displacement current through this area? **

**Solution: **

Electric field between the plates of Capacitor

E= q/Aε

_{0 }= It / Aε_{0}Electric flux through area A/4 is,

φ

_{E}= (A/4)E = It / 4ε_{0}Then displacement current is,

I

_{D}= ε_{0}(dφ_{E}/ dt)= ε

_{0}d/dt (It / 4ε_{0})= I/4

Hence, the displacement current through the required area is I/4.

**Example 3: A parallel plate capacitor with circular plates of radius (R) is being charged. At the instant, the displacement current in the region between the plates enclosed between R/2 and R is given by**

**Solution:**

Displacement Current is given by

I = ε

_{0}(dφ_{E}/ dt)= ε

_{0}A(dE / dt) (Since, φ=A.E)= A (d/dt) (q/A

_{o})= (A/A

_{o}) i (Since, dq/dt = i)Now, A = π(R/2)

^{2}, A_{o}= πR^{2}= π(R/2)

^{2}/ πR^{2}I

^{ }= 1/4 iThus, the required displacement current is 1/4 i, where

is the conduction current.i

**Example 4: A coil that has 700 turns develops an average induced voltage of 50 V. What must be the change in the magnetic flux that occur to produce such a voltage if the time interval for this change is 0.7 seconds? **

**Solution: **

Given:

Number of Turn (N) = 700

Induced Voltage (e) = 50 V

Time Interval (dt) = 0.7 s

By using Faraday’s Law,

e = N (dφ/dt)

Therefore, Change in flux (dφ) is,

dφ = e×dt / N

= 50×0.7 / 700

dφ = 0.05 Wb

**Example 5: The magnetic flux linked with a coil having 250 turns is changed from 1.4 Wb to 2 Wb in 0.45 seconds. Calculate the induced emf in the coil.**

**Solution:**

Given:

Number or Turns (N) = 250

Initial Flux (φ

_{1}) = 1.4 WbFinal flux (φ

_{2}) = 2 WbTherefore, change in flux (dφ) can be given by,

dφ = φ

_{2}– φ_{1}= 2 – 1.4 = 0.6 WbTime Interval (dt) = 0.45 s

According to Faraday’s Law,

Induced Emf (e) = N (dφ/dt)

e = 250 × (0.6 / 0.45)

e = 333.33 V

**Example 6: A 0.20 m wide and 0.60 m long rectangular loop of wire is oriented perpendicular to a uniform magnetic field of 0.30 T. What is the magnetic flux through the loop? **

**Solution: **

Given:

Length of Rectangular Loop (L) = 0.60 m

Breadth of Rectangular Loop (B) = 0.20 m

Area of Rectangular Loop (A) = 0.60 x 0.20 = 0.12 m

^{2}Magnetic Field (B) = 0.30 T

We know that,

Magnetic Flux (φ) = BA

φ = 0.3 × 0.12

φ = 0.036 Wb

## FAQs on Displacement Current

**Q1: What is Displacement Current? **

**Q1: What is Displacement Current?**

**Answer:**

Displacement current is the current in the insulated region of the charged capcitor due to the changing electric flux. It is a property of Maxwell’s Equation.

For a charged capacitor, the electric field between the plates is given by,

E = Q / ε

_{0 }AQ = ε

_{0}φ_{E}Displacement Current is given by,

i_{d}= dQ/dt

i_{d}= ε_{0}(dφ_{E }/dt)

### Q2: What is the SI unit of Displacement Current?

**Answer: **

The SI unit of Displacement Current is the same as the SI unit of Conduction current, i.e. Ampere (A)

### Q3: How does the Displacement Current is produced?

**Answer: **

Displacement current is produced by the varying electromotive force and the time-varying electric field also produces the displacement current.

### Q4: What is the Expression for the Displacement Current?

**Answer:**

The displacement current expression is,

I_{D}= J_{D}S = S(∂D/∂t)where,

is the area of Capacitor PlateSIis the Displacement Current_{D}Jis the Displacement Current Density._{D}is related to Electric Field E as,D

D = εE

is the Permittivity of material between platesε

### Q5: State Ampere-Maxwell law

**Answer: **

Ampere-Maxwell law states that,

“The line integral of magnetic field around a closed loop is equal to μ

_{0}times the sum of the conduction current and displacement current flowing therough the closed loop.”

### Q6: What are the Uses of Displacement Current?

**Answer:**

Various uses of the displacement current are,

- Electromagnetic waves like radio waves, light waves, etc propagate through empty space using displacement current.
- Moving and changing electric field produces the displacement current.
- It helps us to explain Maxwell’s Equation.