Electromagnetic Waves physics cbse

PHYSICS CLASS NOTES FOR CBSE Chapter 24. Electromagnetic Waves 01. Introduction Basic Equations of Electricity and Magne...

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PHYSICS CLASS NOTES FOR CBSE Chapter 24. Electromagnetic Waves 01. Introduction Basic Equations of Electricity and Magnetism The whole concept of electricity and magnetism can be explained by the four basic equations we have deal so far. Q E ×ds   (i) (Gauss law for electrostatic) ∈ (ii) (iii) (iv)

  B ×ds    B ×d   i  E ×d   

(Gauss law for magnetism) (Ampere’s law for Magnetism) (Ampere’s law for electrostatic)

The above stated equation are true for non-time varying fields

02. Concept of Displacement Current (Modified Amper’s Law) Maxwell tried to generalis the concept of faradays law that if changing magnetic field can produce changing electric field then the reverse should also be true i.e. changing electric field must produce magnetic field. To understand the concept of displacement current let us try to understand this experiment when the switch was closed at t = 0 both the needles deflected. Magnetic needle (2)

Magnetic needle (1)

Parallel plate capacitor

t 0

=

V

Deflection of needle (1) is under stood as M.F. is produced due to current flowing in the wire. But why did needle 2 deflect? It is lying in between the two plates of capacitor where there is no current. This magnetic field between the plates is due to the changing electric field between the plates (During charging of capacitor).

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CLASS NOTES FOR CBSE – 24. Electromagnetic Waves

Hence maxwell conducted that changing electric field produces a magnetic field For Needle (1) Amper’s law

 B ×d   i

........ (1)

 c

For needle (2) Amper’s law dE B ×d  ∈  dt



........ (2)

Hence there are two methods of producing M.F. (a) Due to flow of electron which is known as conduction current (b) Due to changing electric field combining eq. (1) and eq. (2)







 dE  B ×d   i C      dt 

Modifield ampere’s law NOTE ☞

dE   is known as displacement current dt

03. Final Form of Maxwell’s Equation (a)

q  E ×ds   ∈  B ×ds   d  E ×dl    dt  d   B ×dl   I ∈  dt  

(b) (c) (d)

B

E





The above equation is known as Maxwell’s equation for time varying form. Howover for free space there are no charges and no conduction current the equations that are significant. dB

 E ×d    dt  d  B ×d   ∈  dt E





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CLASS NOTES FOR CBSE – 24. Electromagnetic Waves

04. Transverse Nature of Electro Magnetic Wave and its Properties Electromagnetic waves The idea of electromagnetic waves was given by Maxwell and experimental verification was provided by Hertz and other scientists. A brief history of electromagnetic waves is as follows: On the basis of experimental study of electromagnetic induction, Faraday concluded that a magnetic field changing with time at a point produces a time varying electric field at that point. Maxwell in 1864 pointed out an electric field changing with time at a point also produces a time varying magnetic field. The two fields are mutually perpendicular to each other. This idea led Maxwell to conclude that the mutually perpendicular time varying electric and magnetic fields produce electromagnetic disturbances in space. These disturbances have the properties of wave which are called as electromagnetic waves. According to Maxwell, the electromagnetic waves are those waves in which there are sinusoidal variation of electric and magnetic field vectors at right angle to each other as well as right angles to the direction of wave propagation. An electromagnetic wave is shown in figure.

The velocity of electromagnetic wave in free space is given by  c    ∈

05. Poynting Vector Poynting vector is a vector that describes the magnitude and direction of energy flow rate.    E× B  S     S → Poynting vector The magnitude of poynting vector represents the rate at which energy flows through a unit surface area perpendicular to the direction of wave propogation SI unit J/sm2 or w/m2

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CLASS NOTES FOR CBSE – 24. Electromagnetic Waves

06. Energy Density and Intensity We know that electric and magnetic field have energy and since EM wave have both these components hence it carries energy with it. We know that energy density associated with  EF   ∈E   We know that energy density associated with B MF    Thus total energy of EM wave is given by  B U  UE  UB or  ∈E      putting the values of E.F. and M.F.

B sin t  xc  U   ∈E sint  xc     If we take the average value over a long period of time   x   sin  t        c Av 





  Uav   ∈ E   B  



The above equation can also be written as ∈ E B Uav   or Uav     Intensity : Energy crossing per unit area per unit time perpendicular to the directions of propogation is called intensity of wave. Energy contained in the volume : E.M. wave A

C△T

 U  Uav × vol   ∈E  AC∆T  ∈ E C U intensity     A∆T  ∴

 I   ∈ CE or 

 B   

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CLASS NOTES FOR CBSE – 24. Electromagnetic Waves

07. Electromagnetic Spectrum Orderly arrangement of electromagnetic radiations according to their wavelength or frequency is called as electromagnetic spectrum. Name

Frequency range (Hz)

Wavelength range (m)

γ-ray

 ×    × 

 ×     ×  

X-ray

 ×    × 

 ×     ×  

Ultraviolet

 ×    × 

 ×     ×  

Visible light

 ×    × 

 ×     ×  

Infra-red

 ×    × 

 ×     ×  

Micro-waves

 ×    × 

 ×     ×  

Ultra high radio frequencies

 ×    × 

 ×    

Very high Radio frequencies

 ×    × 

  

Radio frequencies

 ×    × 

  

Visible spectrum Colour

Wavelength (m)

Violet

 ×    × 

Blue

 ×    × 

Green

 ×    × 

Yellow

 ×   × 

Orange

 ×    × 

Red

 ×    × 

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CLASS NOTES FOR CBSE – 24. Electromagnetic Waves

CBSE Exercise (1) (Q 1 to 4) One Mark 1. The charging current for a capacitor is 0.25 A. What is the displacement current across its plates? 2. Arrange the following electromagnetic waves in order of increasing frequency : γ-rays, microwaves, infrared rays and ultraviolet rays. 3. Identify the different types of electromagnetic radiations, which are used (i) to kill germs, (ii) for physical therapy. 4. Which of the following has the shortest wavelength? Microwaves, ultraviolet rays, X-rays. (Q 5 to 7) Two Marks 5. (i) How does oscillating charge produce electromagnetic waves? (ii) Sketch a schematic diagram depicting oscillating electric and magnetic fields of an EM wave propagating along positive Z-direction. 6. How are X-rays produced? Write their two important uses. 7. A capacitor of capacitance C is being charged by connecting it across a DC source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor. (Q 8 to 10) Three Marks 8. Show that



  (a)       (b)    



where the symbols have their usual meanings. 9. Light from a lamp hanging from the ceiling of a room is falling on the billiard table of area 1.5m2. If the light energy falls on the table at a rate 2 × 105W m‒2 for 1 hour, find the radiation pressure on the table due to the falling light.

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CLASS NOTES FOR CBSE – 24. Electromagnetic Waves

10. The oscillating electric filed in a plane electromagnetic wave is given by Ex = 60 sin (ω t ‒ k x) (in V m‒1) The frequency of electric filed is 2 × 107 Hz. (a) Find ω, λ and k and write expression for electric field. (b) Find B0 and write expression for magnetic field. (c) Predict the direction of propagation of electromagnetic wave.

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CLASS NOTES FOR CBSE – 24. Electromagnetic Waves

ANSWER Q1. The displacement current is equal to 0.25 A, as the charging current is 0.25 A. Q2 Microwaves < Infrared rays < Ultraviolet rays < γ-rays. Q3 (i) (ii)

γ-rays Infrared waves

Q4 X-ray has shortest wavelength. Q5 (i)

(ii)

The oscillating charge produces an oscillating electric field and an oscillating electric field produces magnetic field which is then produces an oscillating emf. An oscillating voltage (emf) produces an oscillating magnetic field and so an electromagnetic waves. The propagation of electromagnetic wave is shown in figure below.

Q6 X-rays can be produced by colliding fast moving electron beam on metal target. Uses (i) Medical diagnosis. (ii) Study of crystal structure. Q7 The ammeter will show the momentary deflection. This momentary deflection occurs due to the fact that the conduction current flows through connection wires during the charging of capacitor. This lead to gathering of charge at two plates and hence varying electric field of increasing nature is produced between the plates which in turn produce displacement current in space between two plates. This maintains the continuity with the conduction current. Ic = ID The current inside the capacitor

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CLASS NOTES FOR CBSE – 24. Electromagnetic Waves

 Displacement current, ID and ID =    Q8

    ×  =         

(a) L.H.S



Now,

  



L.H.S. = c × c = c2

Hence,

     

and



     



(b) If c and v are speeds of electromagnetic waves in free space and a medium of refractive index n, then          Now, and              ×      ×      ∴     



 







Q9 Here, rate at which light energy falls on the table, R = 2 × 105 W m‒2 Area of the table, A = 1.5 m2; t = 1 hour = 3,600 s Total light energy falling on the table in 1 hour, U = R A t = 2×105 × 1.5 × 3,600 = 1.08 × 109 J Total momentum transferred to the table in 1 hour, U  ×     kg m s       ×  Force exerted on the table by light radiation,   F =       N   Radiation pressure on the table, F   P =      ×   N m  A 

Q10 (a) Here, Now,

E0 = 50 V m‒1, v = 2 × 107 Hz ω = 2 π v × 2 π × 2 × 107 = 4 π × 107 s‒7

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CLASS NOTES FOR CBSE – 24. Electromagnetic Waves

  ×  λ =      m    ×           m  and   The expression for electric field, Ex = 50 sin (4 π × 107 t ‒ 2.63 x) (b) If c and  are speeds of electromagnetic waves in free space and a medium of refractive index n, then  n =     Now, c =  and              ×      ×      ∴ n =        



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