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www.gneet.com, , PHYSICS NOTES, , ELECTROMAGNETIC WAVES, DISPLACEMENT CURRENT, Current in capacitors, Consider the charging capacitor in the figure., , We have drawn two loops name as L which is outside the loop and Loop R which is in, between the parallel plates of capacitor., The capacitor is in a circuit that transfers charge (on a wire external to the capacitor) from, the left plate to the right plate, charging the capacitor and increasing the electric field, between its plates. The same current enters the right plate (say I ) as leaves the left plate., Although current is flowing through the capacitor, no actual charge is transported through, the vacuum between its plates., Ampere’s circuital is not applicable for loop L and we can find magnetic field at point P, using Ampere’s circuital law, ⃗ ∙ ⃗⃗⃗, ∮𝐵, 𝑑𝑙 = 𝜇0 𝐼, Now if we consider an imaginary cylindrical surface. No conduction current enters cylinder, surface R, while current I leaves through surface L. Thus Ampere’s law is not applicable, and magnetic field at point P must be zero. 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., Nonetheless, a magnetic field exists between the plates as though a current were present, there as well., For consistency of Ampere's Circuital law requires a displacement current ID = I to flow, across surface R., The explanation is that a displacement current ID flows in the vacuum, and this current, produces the magnetic field in the region between the plates according to Ampere’s law, If Q is the charge on capacitor plate and area of plates of capacitor is A, Electric field between plates is, 1, GORE COACHING CLASSES

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www.gneet.com, , PHYSICS NOTES, , 𝑄, 𝜀0 𝐴, When capacitor is getting charged rate of change in electric field is, 𝜕𝐸, 1 𝑑𝑄, =, 𝜕𝑡 𝜀0 𝐴 𝑑𝑡, 𝜕𝐸, 𝜀0 𝐴, = 𝐼𝐷, 𝜕𝑡, Here ID is called displacement current, In integral form, 𝜕𝐸⃗, ⃗⃗⃗⃗ = 𝐼𝐷, 𝜀0 ∫, 𝑑𝑎, 𝜕𝑡, 𝑑𝜙𝐸, 𝜀0 ∫, = 𝐼𝐷, 𝑑𝑡, 𝐸=, , This current does not have significance in the sense of being the motion of charges., The generalization 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. More precisely, the total current I is the sum of the conduction, current denoted by IC and the displacement current denoted by ID, Adding integral form of displacement current in Ampere’s law we get, 𝑑𝜙𝐸, ⃗ ∙ ⃗⃗⃗, ∮𝐵, 𝑑𝑙 = 𝜇0 𝐼𝐶 + 𝜇0 𝜀0 ∫, 𝑑𝑡, and is known as Ampere-Maxwell law., , Electromagnetic waves, According to Maxwell, an accelerated charge is a source of electromagnetic radiation., In an electromagnetic wave, electric and magnetic field vectors are at right angles to each, other and both are at right angles to the direction of propagation., They possess the wave character and propagate through free space without any material, medium. These waves are transverse in nature. Fig shows the variation of electric field, E along Y direction and magnetic field B along Z direction and wave propagation in + X, direction, , According to Maxwell’s theory, these electric and magnetic field do not come into, existence instantaneously. In the region closer to the oscillating change, the phase, 2, GORE COACHING CLASSES

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www.gneet.com, , PHYSICS NOTES, , difference between electric field E and Magnetic field B is π/2 and their magnitude quickly, decreases as 1/r3 ( where r = distance from source) these components are called Inductive, component., At larger distance E and B are in phase and the decrease in their magnitude is, comparatively slower with distance, as per 1/r. These components are called radiated, components, , Characteristics of Electromagnetic waves, (1) Representation in form of equations:, Electromagnetic wave shown in figure at time t, the y component is EY of electric, field given by equation EY = E0 sin(ωt – kx), In vector form E = EYj = [E0 sin(ωt – kx) ] j, Similarly Magnetic component is given as B =[B0 (ωt – kx) ] k, (2) Relation between magnitude of E and B is E = Bc, Here c is velocity of light, (3) The velocity of electromagnetic waves in vacuum, 1, 𝑐=, √𝜀0 𝜇0, The velocity of electromagnetic waves in medium, 1, √𝜀𝜇, Here ε = permittivity of the medium and μ = permeability of the medium, 𝑣=, , From definition of refractive index, 𝑛=, 𝑛=, Since 𝜀𝑟 =, , 𝜀, 𝜀0, , 𝑐, 𝑣, , 𝜀 𝜇, √𝜀𝜇, =√, 𝜀0 𝜇0, √𝜀0 𝜇0, , = 𝑘 dielectric constant of medium and relative permeability 𝜇𝑟 =, , 𝜇, 𝜇0, , 𝑛 = √𝜀𝑟 𝜇𝑟, (4) Electromagnetic waves are transverse in nature, (5) Electromagnetic waves posses energy and they carry energy from one place to the, other ., (6) Electromagnetic waves exerts pressure on a surface when they are incident on it,, called radiation pressure, If ∆U is the energy of electromagnetic waves incident on a surface of area A in time, ∆t , in direction normal to the surface and if all energy is absorbed then change in, momentum, 3, GORE COACHING CLASSES

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www.gneet.com, , PHYSICS NOTES, , ∆𝑝 =, (7) Energy density of electromagnetic wave, 2, 𝜌 = 𝜀0 𝐸𝑟𝑚𝑠, and 𝜌 =, , Δ𝑈, 𝑐, , 2, 𝐵𝑟𝑚𝑠, , 𝜇0, , (8)The intensity of radiation (I) is defined as the radiant energy passing through unit, area normal to the direction of propagation in one second, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐼=, = 𝑃𝑜𝑤𝑒𝑟, (𝑡𝑖𝑚𝑒)(𝑎𝑟𝑒𝑎), If radiation is passing through unit area with velocity c then volume in one second = c, Thus energy volume = ρc from the value of ρ 𝑤𝑒 𝑔𝑒𝑡, 2, 𝐼 = 𝜀0 𝑐𝐸𝑟𝑚𝑠, Similarly, 2, 𝑐𝐵𝑟𝑚𝑠, 𝐼=, 𝜇0, (9) E ×B gives the direction of propagation of the electromagnetic wave, , ELECTROMAGNETIC SPECTRUM, Sr. Name, No., 1, γ – rays, 2, , x − rays, , 3, , Ultra−violet, , 4, , Visible light, , 5, , Infra−red (IR), , 6, , Microwaves, , 7, , Radio, frequency, wave, , Source, Radioactive nuclei, nuclear, reactions, High energy electrons suddenly, stopped by a metal target, Atoms and molecules in, electrical discharge, incandescent solids,, Fluorescent, lamps, molecules of hot bodies, Electronic device, (Vacuum tube), Charges accelerated through, conducting wires, , Wavelength in, (m), 10−14, to, −10, 10, 1 × 10−10 to, 3 × 10−8, 6 x 10−10 to, 4 × 10−7, 4 x 10−7 to, 8 x 10−7, 8 x 10−7 to, 3x 10−5, 10−3, to, 0.3, 10, to, 4, 10, , Frequency, range (Hz), 3 × 1022 to, 3x 1018, 3 × 1018 to, 1 × 1016, 5 x 1017 to, 8 × 1014, 8 x 1014 to, 4 x 1014, 4 x 1014 to, 1 × 1013, 3 x 1011 to, 1 x 109, 3 x 107 – 3, x 104, , Electromagnetic spectrum covers a wide range of wavelengths (or) frequencies. The whole, electromagnetic spectrum has been classified into different parts and sub parts, in order, of increasing wavelength and type of excitation. All electromagnetic waves travel with the, velocity of light. The physical properties of electromagnetic waves are determined by their, wavelength and not by their method of excitation., 4, GORE COACHING CLASSES

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PHYSICS NOTES, , www.gneet.com, , The overlapping in certain parts of the spectrum shows that the particular wave can be, produced by different methods., , Uses of electromagnetic spectrum, The following are some of the uses of electromagnetic waves., , 1. Radio waves : These waves are used in radio and television communication systems., AM band is from 530 kHz to 1710 kHz., Higher frequencies upto 54 MHz are used for short waves bands., Television waves range from 54 MHz to 890 MHz., FM band is from 88 MHz to 108 MHz., Cellular phones use radio waves in ultra high, frequency (UHF) band., , 2. Microwaves : Due to their short wavelengths, they are used in radar communication, system., Microwave ovens are an interesting domestic application of these waves., , 3. Infra red waves :, (i) Infrared lamps are used in physiotherapy., (ii) Infrared photographs are used in weather forecasting., (iii) As infrared radiations are not absorbed by air, thick fog, mist etc, they are used to take, photograph of long distance objects., (iv) Infra red absorption spectrum is used to study the molecular structure., , 4. Visible light : Visible light emitted or reflected from objects around us provides, information about the world. The wavelength range of visible light is 4000 Å to 8000 Å., , 5. Ultra−violet radiations, (i) They are used to destroy the bacteria and for sterilizing surgical instruments., (ii) These radiations are used in detection of forged documents, fingerprints in forensic, laboratories., (iii) They are used to preserve the food items., (iv) They help to find the structure of atoms., , 6. X rays :, (i) X rays are used as a diagnostic tool in medicine., (ii) It is used to study the crystal structure in solids., 5, GORE COACHING CLASSES

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www.gneet.com, , PHYSICS NOTES, , 7. γ−rays : Study of γ rays gives useful information about the nuclear structure and it is, used for treatment of cancer, , Solved Numerical, Q) A 1000 W bulb is kept at the centre of a spherical surface and is at a distance of 10 m, from the surface. Calculate the force acting on the surface of the sphere by the, electromagnetic waves, along with E0, B0 and intensity I. Take the working efficiency of the, bulb to be 2.5% and consider it as a point source, , calculate the energy density on the, surface ., Solution:, The energy consumed every second by a 1000W bulb = 1000J, As the working efficiency of the bulb is equal to 2.5%, the energy radiated by the bulb per, second, 2.5, ∆𝑈 = 1000 ×, 100, ∴ ∆𝑈 = 25 𝐽𝑠 −1, Considering, the bulb at the centre of the sphere, surface area of the sphere, A = 4πR2 = (4)(3.14)(102) = 1256 m2, Intensity I, 𝐸𝑛𝑒𝑟𝑔𝑦, 25, 𝐼=, =, = 0.02𝑊𝑚−2, (𝑡𝑖𝑚𝑒)(𝑎𝑟𝑒𝑎) 1256, 2, 𝐼 = 𝜀0 𝑐𝐸𝑟𝑚𝑠, = 0.02, 1⁄, 2, 0.02, ⟧, ∴ 𝐸𝑟𝑚𝑠 = ⟦, = 2.74 𝑉𝑚−1, 8.85 × 10−12 × 3.0 × 105, NOW, 𝐸𝑟𝑚𝑠, 𝐵𝑟𝑚𝑠 =, 𝑐, 2.74, 𝐵𝑟𝑚𝑠 =, = 9.13 × 10−9 𝑇, 3.0 × 108, 𝐸0 = √2𝐸𝑟𝑚𝑠, 𝐸0 = 1.41 × 2.74 = 3.86 𝑉𝑚−1, 𝐵0 = √2𝐵𝑟𝑚𝑠, 𝐵0 = 1.41 × 9.13 × 10−9 = 1.29 × 10−8 𝑇, The total energy incident on the surface = 25J, ∴ The momentum (∆p) imparted to the surface in one second ( = force), ∆𝑈, 25, ∆𝑝 =, =𝐹=, = 8.33 × 10−8 𝑁, 𝑐, 3 × 108, From I = ρc, energy density, 𝐼, 0.02, 𝜌= =, = 6.67 × 10−11 𝐽𝑚−3, 𝑐 3 × 108, 6, GORE COACHING CLASSES

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www.gneet.com, , PHYSICS NOTES, , Q) The maximum electric field at a distance of 10 m from an isotropic point source of light, is 3.0 V/m. Calculate (a) the maximum value of magnetic field (b) average intensity of the, light at that place and (c) the power of the source, εo = 8.854 × 10-12 C2N-2 m-2, Solution, (a)maximum value of magnetic field, E = Bc, 𝐸, 3.0, 𝐵= =, = 10−8 T, 𝑐 3.0 × 108, (b) average intensity of the light at that place, From formula, 𝐼=, , 2, 𝜀0 𝑐𝐸𝑟𝑚𝑠, , 𝐸02, = 𝜀0 𝑐 ×, 2, , 𝐼 = 8.854 × 10−12 × 3.0 × 108 ×, 𝐼 = 1.195 × 10−2 𝑤𝑚−2, , (3.0)2, 2, , (c) Power, Power = I × Area = I ×4πr2, Power = 1.195×10-2×4×3.14×(10)2 = 15 w, ----------------END-------------To become a member of www.gneet.com give call on 09737240300, , 7, GORE COACHING CLASSES