of 14 /14
1 / 14 CBSE Class 12 Physics NCERT Exemplar Sol utions Chapter 8 Electromagnetic Waves LONG ANSWER TYPE QUESTIONS 8.28 An infinitely long thin wire carrying a uniform linear static charge density λ is placed along the z-axis (Fig. 8.1). The wire is set into motion along its length with a uniform velocity . Calculate the poynting vector Ans. Consider a cylindrical Gaussian surface in such a way that the axis of cylinder lies on wire. Electric field intensity due to long straight wire at a distance a and charge density c/m www.schoolconnects.in
• Author

others
• Category

## Documents

• view

7

0

Embed Size (px)

### Transcript of CBSE Class 12 Physics NCERT Exemplar Solutions Chapter 8 ...

8.28 An infinitely long thin wire carrying a uniform linear static charge density λ is
placed along the z-axis (Fig. 8.1). The wire is set into motion along its length with a
uniform velocity . Calculate the poynting vector
Ans. Consider a cylindrical Gaussian surface in such a way that the axis of cylinder lies on
wire. Electric field intensity due to long straight wire at a distance a and charge density
c/m
2 / 14
8.29 Sea water at frequency ν = 4 × 108 Hz has permittivity ε = 80 εo,
permeability μ ≈ μo and resistivity ρ = 0.25 Ω–m. Imagine a parallel plate capacitor
immersed in sea water and driven by an alternating voltage source V(t) = Vo sin (2πνt).
What fraction of the conduction current density is the displacement current density?
Ans. Suppose distance between the parallel plates of capacitor is ‘d’ and the applied voltage
V(t)=V0 sin(2πvt)
Let
=4 0×2×80×0.25×108
8.30 A long straight cable of length l is placed symmetrically along z-axis and has
radius a(<<<l). The cable consists of a thin wire and a co-axial conducting tube. An
alternating current I(t) = Io sin (2πνt) flows down the central thin wire and returns
along the co-axial conducting tube. The induced electric field at a distance s from the
www.sc ho
olc on
ne cts
.
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the
total displacement current Id.
(iii) Compare the conduction current I0 with the displacement current .
Ans. (i) Induced electric field E(s, t) at distance s(s< radius of co- axial cable) is given as
E(s, t)= 0I0v cos2 vt . Displacement current density Jd is given by,
[ s and a are constant]
www.sc ho
olc on
ne cts
Integration of
[ loge1=0]
I0 sin(2πvt)
The negative sign shows that the displacement current Id is opposite to the conduction
www.sc ho
olc on
ne cts
6 / 14
current IC.
Id is in –Z direction as IC is in +Z direction.
(iii)
Where
Required ratio
8.31 A plane EM wave travelling in vacuum along z-direction is given by = E0 sin(kz –
ωt ) and =B0 sin(kz –wt) .
(i) Evaluate over the rectangular loop 1234 shown in Fig 8.2.
(ii) Evaluate over the surface bounded by loop 1234.
www.sc ho
olc on
ne cts
(iv) By using similar process and the equation Prove that
Ans. As the electromagnetic wave is propagating along Z-axis then its electric and magnetic
field vectors are along X and Y axis.
(i) and as in figure below.
Line integral of E over loop 1234,
www.sc ho
olc on
ne cts
.in
(ii) Consider a strip of area ds=h.dz as in figure. Angle between and is zero.
8 / 14
www.sc ho
olc on
ne cts
[-sin(kz1-ωt)(-ω)+sin(kz2-ωt)(-ω)]
[ B=B0 B=BY]
(iv) Consider a loop 1.2,3,4 in Y-Z plane as in figure.
www.sc ho
olc on
ne cts
…III
Now to calculate E = Let us consider the rectangular strip of loop 1,2,3,4 of area ds
each ds= hdz.
[IC=conduction current]
B0h [sin(kz1-ωt)-sin(kz2-ωt)]
1212 / 14
Hence proved.
8.32 A plane EM wave travelling along z direction is described by = E sin(kz –ωt )
and = B0 sin(kz –ωt ) . Show that
(i) The average energy density of the wave is given by
(ii) The time averaged intensity of the wave is given by
Ans. (i) The electromagnetic wave carry energy which is due to electric field vector and
magnetic field vector. In electromagnetic wave, E and B varies with time.
E and B varies with time.
The energy density due to electric field is
The energy density due to magnetic field B is
Total average energy density of electromagnetic wave
E=E0 sin (kz-ωt)
B=B0 sin (kz-ωt)
The average value of B2 over a cycle=
(ii) We know that E0= CB0 and
UB = UE
www.sc ho
olc on
ne cts
www.sc ho
olc on
ne cts