Tiêu đề ELECTROMAGNETICS WAVES.pdf ✅

 Digital www.allendigital.in [ 241 ] We have seen that a magnetic field changing with time gives rise to an electric field. Is the converse also true? Does an electric field changing with time give rise to a magnetic field? James Clerk Maxwell (1831-1879), argued that this was indeed the case – not only an electric current but also a time-varying electric field generates magnetic field. 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. To see how a changing electric field gives rise to a magnetic field, let us consider the process of charging of a capacitor. Case-1 Question: Calculate Magnetic Field at P Solution: 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 Using Ampere’s Law  B.dl I = 0 enc BP (2r) = μ0I 0 P I B 2 R  =  Case-2 Question: Calculate Magnetic Field at P Solution: Here we consider a different surface, which has the same boundary. This is a pot like surface which nowhere touches the current, but has its bottom between the capacitor plates. Using Ampere’s Law  B.dl I = 0 enc BP (2r) = μ0(0) ( No current passes through the surface) BP = 0 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. This shows the inconsistency in ampere circuital law. But what is the reason for this inconsistency? We can easily solve this problem by considering the following situation. Electric field between the plates 0 q E A =  Electric flux through shaded area E 0 q EA A A  = =  04 Electromagnetic Waves P I + + + + + + – – – – – – r P I + + + + + + – – – – – – r +q –q EሬԦ + + + + + + – – – – – – –
NEET : Physics [ 242 ] www.allendigital.in  Digital E 0 q  =  Since charge q is varying w.r.t. time, E 0 d 1 dq dt dt  =  E 0 dq d dt dt  =  E d d i dt  = Maxwell defined E d 0 d i dt  = as Displacement Current. So, this is the solution of inconsistency that we observed. So, in Case2 Using Ampere’s Law  B.dl I = 0 enc BP (2r) = μ0(id) ( id passes through the surface) 0 d P i B 2 R  =  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. Conduction current (c) Displacement current (d)  Due to flow of charge in the conducting wire.  Denoted by IC. c dq I dt =  Due to variable electric field between plates of charging capacitor.  Denoted by Id. E d 0 d i dt  = Hence, total current is defined as I = IC + Id So, Ampere’s Circuital Law can be written as – B.dl I I =  + 0 c d ( )  [Also known as Ampere Maxwell's Law] Displacement current also solved the current continuity problem EሬԦ + + + + + + + – – – – – – – Ic Ic Id I = dq dt Eo +q –q Ic Id Ic = Id = I P I + + + + + + – – – – – – r id
Electromagnetic Waves  Digital www.allendigital.in [ 243 ] Illustration 1: Figure shows a capacitor made of two circular plates each of radius 12 cm and separated by 4.0 mm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and is equal to 0.10 A. Find (a) Capacitance (b) Rate of change of potential difference between the plates? (c) Displacement current across the plates? Solution: (a) Area of one of the plates. A = (12 × 10–2 m)2 Distance between the plates, d = 4.0 mm = 4 × 10–3 m Capacitance of the capacitor, C = 0A/d 2 –2 2 12 2 –3 C (12 10 m) C 8.85 10 N m 4 10 m −     =         −  C = 100 × 10–12 F = 100 pF Charging current, dQ d I (CV) dt dt = = or dV I C dt = or dV I dt C = (b) Rate of change of potential difference = dV I dt C = –12 dV 0.10A dt 100 10 F =  = 1 × 109 V/s (c) Displacement current d 0 dE I A dt   =      For a parallel-plate capacitor, 0 0 0 Q / A Q E A  = = =    Where  is surface density of charge. d 0 0 d Q I A dt A     =             d 0 c 0 1 dQ dQ I A I A dt dt =   = =   So d = c = 0.10 A Maxwell's Equations and Properties of Electromagnetic Waves Maxwell’s Equations 1. Gauss’ Law in Electrostatics en 0 Q E dA  =   +q –q EሬԦ 4mm 12cm I = 0.10A
NEET : Physics [ 244 ] www.allendigital.in  Digital 2. Gauss’ Law in Magnetism B dA 0  =  3. Maxwell-Faraday’s Law d B E dl dt   = −  4. Ampere – Maxwell Law E 0 c 0 d B dl = I dt      +      Charge Generates at rest E E = Electric Field Uniform Motion E ,B B = Magnetic Field Accelerated Motion E ,B , EMW EMW = Electromagnetic Waves Electromagnetic Waves Electromagnetic waves consist of sinusoidally time varying Electric and Magnetic field. Electric and magnetic fields oscillate sinusoidally in perpendicular planes as well as the direction the propagation of wave so, EMW are transverse in nature. N S × × × × × × × × × × × × × × × × × × × × × EሬԦ ind changing with time