Q: The electric field of a plane electromagnetic wave is given by \overrightarrow { y }=E_0 \hat{i} cos(kz)(\omega t) The corresponding magnetic field \overrightarrow { B } is then given by:

\because\overrightarrow { E }\times \overrightarrow { B }||\overrightarrow { V } Given that wave is propagating along positive z-axis and \overrightarrow { E } along positive x-axis. Hence \overrightarrow { B } along y-axis.From Maxwel equation \overrightarrow { V }\times \overrightarrow { E }=-\dfrac{\partial B}{\partial t} i.e. \dfrac{\partial E}{\partial Z}=-\dfrac{\partial B}{\partial t} and B_0=\dfrac{E_0}{C} so, \overrightarrow {B} = \dfrac{\overrightarrow{E_{0}}}{c}\hat{j}sin(kz)cos(\omega t)

Question 1

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of  2.0\, 	imes\, 10^{10} Hz  and amplitude  48\, V\, m^{-1} .(a) What is the wavelength of the wave?(b) What is the amplitude of the oscillating magnetic field?(c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3  	imes 10^{8} m s^{-1} ]

Accepted Answer

(a)  The wavelength is given by  \lambda =\dfrac{c}{v}=\dfrac{3 	imes 10^{8}}{2 	imes 10^{10}}=1.5 	imes 10^{-2}m (b)   B_o=E_o/c=1.6	imes 10^{-7}T (c)  Energy density due to the electric field,  E_E=\dfrac{1}{2}\epsilon_oE^2 Energy density due to the magnetic field,  E_B=\dfrac{1}{2}B^2/\mu_o also,  c=\dfrac{1}{\sqrt{\epsilon_0\mu_o}} on solving above equations,  E_E=E_B

Question 2

List the properties of electromagnetic waves.

Accepted Answer

Electromagnetic waves are composed oscillating electric and magnetic field at right angles to each other.PropertiesElectromagnetic waves are propagated by oscillating electric fields and magnetic field oscillation at right angles to each other.These waves travel with speed  3	imes { 10 }^{ 8 }m{ s }^{ -1 }  in vacuum.They are not deflected by electric or magnetic field.They can show interference or diffraction.They are transverse waves.May be polarized.Need no medium of propagation.wavelength and frequency related as  c=v\lambda   .

Question 3

How electromagnetic waves are produced ?

Accepted Answer

Electromagnetic waves are the combination of electric and magnetic field waves produced by moving charges.The creation of all electromagnetic waves begins with a charged particle. This charged particle creates an electric field (which can exert a force on other nearby charged particles). When it accelerates as part of an oscillatory motion, the charged particle creates ripples, or oscillations, in its electric field, and also produces a magnetic field (as predicted by Maxwell’s equations).Once in motion, the electric and magnetic fields created by a charged particle are self-perpetuating—time-dependent changes in one field (electric or magnetic) produce the other. Both electric and magnetic fields in an electromagnetic wave will fluctuate in time, one causing the other to change.Electromagnetic waves (photons) are produced anytime a charged particle experiences a change in velocity.. The electromagnetic wave (radiation) that is produced when the charge particle changes it’s velocity is how energy is conserved. This energy is light.A simple example of this is in the production of microwaves.Inside of a microwave oven is a component called a magnetron. Inside the magnetron is a cavity which is like a circular race track for electrons. The electrons go around this little circular track at immense speeds. While they are running in the little circles at these very high speeds, a fluctuating magnetic field causes the electrons to wiggle back and forth towards the inside and outside of the race track. This wiggling happens at a high frequency and is what causes the generation of the microwaves.

Question 4

Which of the following statement is false for the properties of electromagnetic waves?

Accepted Answer

In electromagnetic waves,1. The electric field and magnetic field varies continuously with time and have maxima and minima at same place and at same time.2. Both electric and magnetic field have same energy.3. both electric and magnetic field are perpendicular to each other and perpendicular to direction of propagation.4. These waves don't require any material medium to propagate, they can propagate in vacuum as well.So, the false statement will be the statement given in the option  (C) Hence, the correct option is  (C)

Question 5

What is displacement current? Obtain an expression of displacement current for a charged capacitor. Write Ampere-Maxwell's law.

Accepted Answer

When a capacitor is charging (or discharging), current flows in the circuit. However, there is no actual charge transfer in the insulated region between capacitor which is contradictory to the flow of current. Hence, displacement current is the current in the insulated region due to the changing electric flux.For a charged capacitor, electric field between the plates is given by: E = \cfrac{Q}{\varepsilon_o A} Q = \varepsilon_o \phi_E Displacement current is given by: i_d = \cfrac{dQ}{dt} = \varepsilon_o  \cfrac{d\phi_E}{dt} According to Maxwell's Ampere circuital Law, the line integral of magnetic field along a closed path is equal to  \mu_o  times the total current. \oint \vec B . \vec {dl} = \mu_o (i_c + i_d) where  i_c:  Conduction current            i_d:  Displacement current \oint \vec B . \vec {dl} = \mu_o (i_c + \varepsilon_o\cfrac{d\phi}{dt})

Question 6

If P represents radiation pressure, C represents the speed of light, and Q represents radiation energy striking a unit area per second, the non-zero integers x, y and z such that  P^x.Q^y.C^z  is dimensionless, find the values of  x, y  and z.

Accepted Answer

Radiation pressure   P = ML^{-1}T^{-2}   Q = \dfrac{Energy/Area}{time} = (\dfrac{w}{m^{2}}) = MT^{-3}   C =   Speed of Light   = (LT-1)   M^{0}L^{0}T^{0} = (ML^{-1}T^{-2})^{x} (MT^{-3})^{y}(LT^{-1})^{z}  Comparing powers   M ightarrow 0 = x+y \Rightarrow x = -y ...(1)   L ightarrow 0 = -x+z \Rightarrow x = z ...(2)   T ightarrow 0 = -2x-3y-z ...(3)  Solving (1)(2) and (3)   x = 1   y = -1   z = 1   	herefore \dfrac{P^{1}c^{1}}{Q}

Question 7

Which of the following is not true for electromagnetic waves?

Accepted Answer

Electromagnetic wave have constant velocity in a particular medium and it is independant of frequency and it is equal to  V=\sqrt{\dfrac{1}{\mu_0 \epsilon_0}}=3 	imes 10^8 \ m/s  Option C

Question 8

Which of the following is not an electromagnetic wave?

Accepted Answer

Radio waves, X-rays, gamma rays all lie in the electromagnetic spectra, at different frequencies and wavelength regions. However Earthquake P-waves are not electromagnetic waves.

Question 9

The electric field of a plane electromagnetic wave is given by \overrightarrow { y }=E_0 \hat{i} cos(kz)(\omega t)  The corresponding magnetic field  \overrightarrow { B }  is then given by:

Accepted Answer

\because\overrightarrow { E }	imes \overrightarrow { B }||\overrightarrow { V } Given that wave is propagating along positive z-axis and  \overrightarrow { E }  along positive x-axis. Hence  \overrightarrow { B }  along y-axis.From Maxwel equation \overrightarrow { V }	imes \overrightarrow { E }=-\dfrac{\partial B}{\partial t} i.e.  \dfrac{\partial E}{\partial Z}=-\dfrac{\partial B}{\partial t} and B_0=\dfrac{E_0}{C} so,  \overrightarrow {B} = \dfrac{\overrightarrow{E_{0}}}{c}\hat{j}sin(kz)cos(\omega t)

Question 10

In electromagnetic wave, the average energy density is associated

Accepted Answer

Answer:-  CThe energy in an electromagnetic wave is tied up in the electric and magnetic fields. In general, the energy per unit volume in an electric field is given by:energy density  in electric field= \dfrac{1}{2}\epsilon_0 E^2 In a magnetic field, the energy per unit volume is:energy density  in magnetic  field= \dfrac{1B^2}{2\mu_0} An electromagnetic wave has both electric and magnetic fields, so the total energy density associated with an electromagnetic wave is: u= \dfrac{1}{2}\epsilon_0 E^2 + \dfrac{1B^2}{2\mu_0} It turns out that for an electromagnetic wave, the energy associated with the electric field is equal to the energy associated with the magnetic field, so the energy density can be written in terms of just one or the other: u= \dfrac{1}{2}\epsilon_0 E^2 = \dfrac{1B^2}{2\mu_0} This also implies that in an electromagnetic wave, E = cB.