Physics INTRODUCTION and DISPLACEMENT CURRENT for CBSE-NCERT

### Topic covered

color{blue}{star} INTRODUCTION
color{blue}{star} DISPLACEMENT CURRENT

### INTRODUCTION

color{blue} ✍️As we know that an electric current produces magnetic field, so two current-carrying wires exert a magnetic force on each other.

color {blue}➢Further, we have seen that a magnetic field changing with time gives rise to an electric field.

color{blue} ✍️While applying the Ampere’s circuital law to find magnetic field at a point outside a capacitor connected to a time-varying current, Maxwell noticed an inconsistency in the Ampere’s circuital law. He suggested the existence of an additional current, called by him, the "displacement current" to remove this inconsistency.

color {blue}➢Maxwell formulated a set of equations involving electric and magnetic fields, and their sources, the charge and current densities.

color {blue}➢These equations are known as Maxwell’s equations. Together with the Lorentz force formula , they mathematically express all the basic laws of electromagnetism.

color {blue}{➢➢}The most important prediction to emerge from Maxwell’s equations is the existence of "electromagnetic waves," which are (coupled) time varying electric and magnetic fields that propagate in space.

color {blue}{➢➢}The speed of the waves, according to these equations, turned out to be very close to the speed of light ( 3 ×10^8 m//s),obtained from optical measurements.

color {blue}➢This led to the remarkable conclusion that light is an electromagnetic wave.

### DISPLACEMENT CURRENT

color{blue} ✍️As we know that an electrical current produces a magnetic field around it.

color {blue}{➢➢}Maxwell showed that for logical consistency, a changing electric field must also produce a magnetic field.

color {blue}{➢➢}This effect is of great importance because it explains the existence of radio waves, gamma rays and visible light, as well as all other forms of electromagnetic waves.

color{blue} ✍️To see how a changing electric field gives rise to a magnetic field, let us consider the process of charging of a capacitor and apply Ampere’s circuital law given by (Chapter 4)

color {blue}{phi B•d1= mu_0i(t)}

...........(8.1)

color {blue}{➢➢}to find magnetic field at a point outside the capacitor.

color {blue}{➢➢}Figure 8.1(a) shows a parallel plate capacitor C which is a part of circuit through which a time-dependent current i(t ) flows .

color{blue} ✍️Let us find the magnetic field at a point such as P, in a region outside the parallel plate capacitor. For this, we consider a plane circular loop of radius r whose plane is perpendicular to the direction of the current-carrying wire, and which is centred symmetrically with respect to the wire [Fig. 8.1(a)].

color {blue}{➢➢}From symmetry, the magnetic field is directed along the circumference of the circular loop and is the same in magnitude at all points on the loop so that if B is the magnitude of the field, the left side of Eq. (8.1) is B (2π r).

So we have

color {blue}{B (2pir)= mu_0i(t)}

...........(8.2)

color {blue}➢Now, consider a different surface, which has the same boundary.

color{blue} ✍️This is a pot like surface [Fig. 8.1(b)] which nowhere touches the current, but has its bottom between the capacitor plates; its mouth is the circular loop mentioned above.

color {blue}{➢➢}Another such surface is shaped like a tiffin box (without the lid) [Fig. 8.1(c)].

color {blue}{➢➢}On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the left hand side of Eq. (8.1) has not changed but the right hand side is zero and not μ_0i, since no current passes through the surface of Fig. 8.1(b) and (c).

color {blue}{➢➢}So we have a contradiction calculated one way, there is a magnetic field at a point P; calculated another way, the magnetic field at P is zero.

color {blue}{➢➢}Since the contradiction arises from our use of Ampere’s circuital law, this law must be missing something. The missing term must be such that one gets the same magnetic field at point P, no matter what surface is used.
We can actually guess the missing term by looking carefully at Fig. 8.1(c).

color {blue}{➢➢}If the plates of the capacitor have an area A, and a total charge Q, the magnitude of the electric field E between the plates is (Q//A)//ε_0 (see Eq. 2.41).

color{blue} ✍️The field is perpendicular to the surface S of Fig. 8.1(c). It has the same magnitude over the area A of the capacitor plates, and vanishes outside it. So what is the electric flux ΦE through the surface S ? Using Gauss’s law, it is

color {blue}{Φ_E=|E|A= 1/ε_0 Q/A A - Q/(ε_0)}

..............(8..3)

color {blue}{➢➢}Now if the charge Q on the capacitor plates changes with time, there is a current i = (dQ//dt), so that using Eq. (8.3), we have

(dΦ_E)/(dt) = d/(dt) (Q/(ε_0))= 1/(ε_0) (dQ)/(dt)

color {blue}{➢➢}This implies that for consistency,

color {blue}{ε_0(dΦ_E)/(dt)=i}

.................(8.4)

color {blue}{➢➢}This is the missing term in Ampere’s circuital law.

color{blue} ✍️If we generalise this law by adding to the total current carried by conductors through the surface, another term which is ε_0 times the rate of change of electric flux through the same surface, the total has the same value of current i for all surfaces.

color {blue}{➢➢}If this is done, there is no contradiction in the value of B obtained anywhere using the generalised Ampere’s law.
B at the point P is non-zero no matter which surface is used for calculating it. B at a point P outside the plates [Fig. 8.1(a)] is the same as at a point M just inside, as it should be.

color {blue}{➢➢}The current carried by conductors due to flow of charges is called "conduction current."

color{blue} ✍️The current, given by Eq. (8.4), is a new term, and is due to changing electric field (or electric displacement, an old term still used sometimes). It is, therefore, called displacement current or Maxwell’s displacement current.

color {blue}{➢➢}Figure 8.2 shows the electric and magnetic fields inside the parallel plate capacitor discussed above. The generalisation made by Maxwell then is the following. The source of a magnetic field is not just the conduction electric current due to flowing charges, but also the time rate of change of electric field.

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color {blue}{➢➢}More precisely, the total current i is the sum of the conduction current denoted by i_c, and the displacement current denoted by id (= ε_0 (dΦ_E// dt)). So we have

color {blue}{ i = i_e + i_d = i_e + epsilon_o ( d Φ_E)/( dt)}

.............(8.5)

color{blue} ✍️In explicit terms, this means that outside the capacitor plates, we have only conduction current i_c = i, and no displacement current, i.e., i_d = 0.

color {blue}{➢➢}On the other hand, inside the capacitor, there is no conduction current, i.e., i_c = 0, and there is only displacement current, so that i_d = i. The generalised (and correct) Ampere’s circuital law has the same form as Eq. (8.1), with one difference: “the total current passing through any surface of which the closed loop is the perimeter” is the sum of the conduction current and the displacement current.

color {blue}{➢➢}The generalised law is

color {blue}{phi B•d1= mu_0 i_e +mu_0 ε_0(dΦ_E)/(dt)}

...............(8.6) and is known as Ampere-Maxwell law.

color {blue}{➢➢}In all respects, the displacement current has the same physical effects as the conduction current.

color {blue}{➢➢}In some cases, for example, steady electric fields in a conducting wire, the displacement current may be zero since the electric field E does not change with time.

color {blue}{➢➢}In other cases, for example, the charging capacitor above, both conduction and displacement currents may be present in different regions of space.

color {blue}{➢➢} In most of the cases, they both may be present in the same region of space, as there exist no perfectly conducting or perfectly insulating medium.

color {blue}{➢➢}Most interestingly, there may be large regions of space where there is no conduction current, but there is only a displacement current due to time-varying electric fields.

color {blue}{➢➢}In such a region, we expect a magnetic field, though there is no (conduction) current source nearby.

color{blue} ✍️The prediction of such a displacement current can be verified experimentally. For example, a magnetic field (say at point M) between the plates of the capacitor in Fig. 8.2(a) can be measured and is seen to be the same as that just outside (at P).

color {blue}{➢➢}The displacement current has (literally) far reaching consequences. One thing we immediately notice is that the laws of electricity and magnetism are now more symmetrical.
Faraday’s law of induction states that there is an induced emf equal to the rate of change of magnetic flux.

color {blue}{➢➢}Now, since the emf between two points 1 and 2 is the work done per unit charge in taking it from 1 to 2, the existence of an emf implies the existence of an electric field.

color {blue}{➢➢}So, we can rephrase Faraday’s law of electromagnetic induction by saying that a magnetic field, changing with time, gives rise to an electric field. Then, the fact that an electric field changing with time gives rise to a magnetic field, is the symmetrical counterpart, and is a consequence of the displacement current being a source of a magnetic field.

color {blue}{➢➢}Thus, time - dependent electric and magnetic fields give rise to each other! Faraday’s law of electromagnetic induction and Ampere-Maxwell law give a quantitative expression of this statement, with the current being the total current, as in Eq. (8.5).

color {blue}{➢➢}One very important consequence of this symmetry is the existence of electromagnetic waves, which we discuss qualitatively in the next section.

### MAXWELL’S EQUATIONS

Q 3138245102

A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series with a resistor R = 1 MΩ across a 2V battery (Fig. 8.3). Calculate the magnetic field at a point P, halfway between the centre and the periphery of the plates, after t = 10–3 s. (The charge on the capacitor at time t is q (t) = CV [1 – exp (–t//τ )], where the time constant τ is equal to CR.)
Class 12 Chapter 8 Example 1
Solution:

The time constant of the CR circuit is τ = CR = 10^(–3) s. Then, we have

q(t) = CV [1 – exp (–t//τ)]
= 2 × 10^(–9) [1– exp (–t//10^(–3))]
The electric field in between the plates at time t is

E = (q(t))/(ε_0A) = q/(piε_0) ; A = pi (1)^2 m^2 = area of the plates.

Consider now a circular loop of radius (1//2) m parallel to the plates passing through P. The magnetic field B at all points on the loop is along the loop and of the same value.
The flux ΦE through this loop is

ΦE = E × area of the loop

=Exx pi xx (1/2)^2 = (piE)/4 = q/(4ε_0)

The displacement current

i_d = ε_0(dΦ_E)/(dt) = 1/4 (dq)/(dt) = 0.5 xx10^(-6) exp (-1)

at t = 10^(–3s). Now, applying Ampere-Maxwell law to the loop,

B xx2pixx(1/2) = mu_0 (i_c+i_d) = mu_0 (0+i_d) = 0.5xx10^(-6) mu_0exp(-1)

or, B = 0.74xx 10^(-13)T