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Q.189. An electromagnetic wave of frequency v = 3.0 MHz passes from vacuum into a non-magnetic medium with permittivity a = 4.0. Find the increment of its wavelength. 

Ans. The velocity of light in a medium of relative permittivity Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRevThus the change in wavelength of light (from its value in vaccum to its value in the medium) is

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

  

Q.190. A plane electromagnetic wave falls at right angles to the surface of a plane-parallel plate of thickness l. The plate is made of non-magnetic substance whose permittivity decreases exponentially from a value al  at the front surface down to a value a, at the rear one. How long does it take a given wave phase to travel across this plate? 

Ans. From the data of the problem the relative permittivity of the medium varies as  

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Hence the local velocity of light

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Thus the required time Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

  

Q.191. A plane electromagnetic wave of frequency v = 10 MHz propagates in a poorly conducting medium with conductivity σ = = 10 mS/m and permittivity ε = 9. Find the ratio of amplitudes of conduction and displacement current densities. 

Ans. Conduction current density Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Displacement current density Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Ratio of magnitudesIrodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev on puting the values.

 

Q.192. A plane electromagnetic wave E = Em  cos (ωt— kr) propagates in vacuum. Assuming the vectors Em  and k to be known, find the vector H as a function of time t at the point with radius vector r = 0. 

Ans. 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

So integrating (ignoring a constant) and using Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

 

Q.193. A plane electromagnetic wave E = Em  cos (ωt— kr), where Em, = Emey, k = kex, ex, e are the unit vectors of the x, y axes, propagates in vacuum. Find the vector H at the point with radius vector r = xex  at the moment (a) t = 0, (b) t = to. Consider the case when Em  =1 60 V/m, k = 0.51 m-1, x = 7.7 m, and to  = = 33 ns. 

Ans. As in the previous problem 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Thus

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

  

Q.194. A plane electromagnetic wave E = Em, cos (ωt— kx) propagating in vacuum induces the emf gin, in a square frame with side l. The orientation of the frame is shown in Fig. 4.37. Find the amplitude value εind, if E= 0.50 mV/m, the frequency v =5.0 MHz and l = 50 cm. 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Ans. 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Putting the values

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

 

Q.195. Proceeding from Maxwell's equations show that in the case of a plane electromagnetic wave (Fig. 4.38) propagating in vacuum the following relations hold: 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Ans. 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

 

Q.196. Find the mean Poynting vector (8) of a plane electromagnetic wave E = E cos (ωt— kr) if the wave propagates in vacuum. 

Ans. 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

So

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

  

Q.197.. A plane harmonic electromagnetic wave with plane polarization propagates in vacuum. The electric component of the wave has a strength amplitude Em = 50 mV/m, the frequency is v 100 MHz. Find:
 (a) the efficient value of the displacement current density;
 (b) the mean energy flow density averaged over an oscillation period. 

Ans.

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Thus

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev 

 

Q.198. A ball of radius R = 50 cm is located in a non-magnetic medium with permittivity ε = 4.0. In that medium a plane electromagnetic wave propagates,the strength amplitude of whose electric component is equal to Em = 200 Vim. What amount of energy reaches the ball during a time interval t = 1.0 min?

Ans. For the Poynting vector we can derive as in (196)

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev along the direction of propagation.

Hence in time t (which is much longer than the time period T of the wave), the eneigy reaching the ball is

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

  

Q.199. A standing electromagnetic wave with electric component E = Em  cos kx•cos ωt is sustained along the x axis in vacuum. Find the magnetic component of the wave B (x, t). Draw the approximate distribution pattern of the wave's electric and magnetic components (E and B) at the moments t = 0 and t = T/4, where T is the oscillation period. 

Ans. 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Also

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

 

Q.200. A standing electromagnetic wave E = Em  cos kx• cos ωt is sustained along the x axis in vacuum. Find the projection of the Poynting vector on the x axis sx (x, t) and the mean value of that projection averaged over an oscillation period.

Ans. 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Thus

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

 

Q.201. A parallel-plate air capacitor whose electrodes are shaped as discs of radius R = 6.0 cm is connected to a source of an alternating sinusoidal voltage with frequency ω = 1000 s-1. Find the ratio of peak values of magnetic and electric energies within the capacitor. 

Ans. Inside the condenser the peak electrical energy Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

(d = separation between the plates, πR= area of each plate.). 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev  is the maximum voltage 

Changing electric field causes a displacement current

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

This gives rise to a magnetic field B (r) (at a radial distance r from the centre of the plate)

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

 Energy associated with this field is

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Thus the maximum magnetic energy

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Hence

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

The approximation are valid only if  Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev 

 

Q.202. An alternating sinusoidal current of frequency ω = = 1000 s-1 f lows in the winding of a straight solenoid whose crosssectional radius is equal to R = 6.0 cm. Find the ratio of peak values of electric and magnetic energies within the solenoid. 

Ans. Here Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev  then the peak magnetic eneigy is

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Changing magnetic field induces an electric field which by Faraday’s law is given by 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

The associated peak electric eneigy is

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Again we expect the results to be valid if and only if

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev 

 

Q.203. A parallel-plate capacity whose electrodes are shaped as round discs is charged slowly. Demonstrate that the flux of the Poynting vector across the capacitor's lateral surface is equal to the increment of the capacitor's energy per unit time. The dissipation of field at the edge is to be neglected in calculations. 

Ans. If the chaige on the capacitor is Q, the rate of increase of the capacitor’s energy

 Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

N ow electric field betw een the plates (inside it) is, Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

So displacement current Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

This will lead to a magnetic field, (circuital) inside the plates. At a radial distance r

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Hence Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev at the edge.

Thus inward Poynting vector  Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Total flow =Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

  

Q.204. A current I flows along a straight conductor with round cross-section. Find the flux of the Poynting vector across the lateral surface of the conductor's segment with resistance R.

Ans. Suppose the radius of the conductor is R0. Then the conduction current density is

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

where  Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRevi.s the resistivity.

Inside the conductor there is a magnetic field given by

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

∴ Energy flowing in per second in a section of length l is

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

But the resistance Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

 Thus the energy flowing into the conductor = I 2 R.

 

Q.205. Non-relativistic protons accelerated by a potential difference U form a round beam with current I. Find the magnitude and direction of the Poynting vector outside the beam at a distance r from its axis. 

Ans. Here Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev where R = radius of cross section of the conductor and n = chaige density (per unit volume)

Also

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Thus, the moving protons have a charge per unit length

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

This gives rise to an electric field at a distance r given by

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

The magnetic field is Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Thus 

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev radially outward from the axis

 This is the Poynting vector.

 

Q.206. A current flowing in the winding of a long straight solenoid is increased at a sufficiently slow rate. Demonstrate that the rate at which the energy of the magnetic field in the solenoid increases is equal to the flux of the Poynting vector across the lateral surface of the solenoid. 

Ans. Within the solenoid Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev and the rate of change of magnetic energy

 Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

where R = radius of cross section of the solenoid l = length.

Also Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev along the axis within the solenoid.
By Faraday’s law, the induced electric field is

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

or Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

  

Q.207. Fig. 4.39 illustrates a segment of a double line carrying direct current whose direction is indicated by the arrows. Taking into account that the potential Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev and making use of the Poynting vector, establish on which side (left or right) the source of the current is located.

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Ans. Given Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

The electric field is as shown by the dashed lines (— →).
The magnetic field is as shown Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev emeiging out of the paper.Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev is parallel to the wires and towards right.

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Hence source must be on the left. 

 

Q.208. The energy is transferred from a source of constant voltage V to a consumer by means of a long straight coaxial cable with negligible active resistance. The consumed current is I. Find the energy flux across the cross-section of the cable. The conductive sheath is supposed to be thin. 

Ans. The electric field (— → ) and the magnetic field (H→) are as shown. The electric field by Gauss’s theorem is like

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Then Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Magnetic field is Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

 

Q.209. A source of ac voltage V = Vo  cos ωt delivers energy to a consumer by means of a long straight coaxial cable with negligible active resistance. The current in the circuit varies as I = = Io  cos ωt — φ). Find the time-averaged energy flux through the cross-section of the cable. The sheath is thin.

Ans. As in the previous problem 
Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Hence time averaged power flux ( along the z axis )  = Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

On using Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev 

 

Q.210. Demonstrate that at the boundary between two media the normal components of the Poynting vector are continuous, i.e. S1n = S2n.

Ans. Let Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev be along tbe z axis. Then

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

Irodov Solutions: Electromagnetic Waves. Radiation- 1 Notes | EduRev

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