40: Electromagnetic Waves : Cbse-xi-physics

Home Learn Fast select Course

Change Unit Change Subject Last Menu

CBSE-XI-Physics

40: Electromagnetic Waves

with Solutions - page 2

Note: Please signup/signin free to get personalized experience.

10 minutes can boost your percentage by 10%

Note: Please signup/signin free to get personalized experience.

Qstn #6
An electric field
E →and a magnetic field
B →exist in a region. The fields are not perpendicular to each other.
(a) This is not possible.
(b) No electromagnetic wave is passing through the region.
(c) An electromagnetic wave may be passing through the region.
(d) An electromagnetic wave is certainly passing through the region.
digAnsr: c
Ans : (c) An electromagnetic wave may be passing through the region.
For an electromagnetic wave,electric field ,magnetic field and direction of propagation are mutually perpendicular to each other.We can have a region in which electric and magnetic fields are applied at an angle with each other.In transmission lines Different modes exist. In transverse electric (TE) mode-no electric field exist in the direction of propagation. These are sometimes called H modes because there is only a magnetic field along the direction of propagation (H is the conventional symbol for magnetic field).
Page No 339:

Qstn #7
Consider the following two statements regarding a linearly polarised, plane electromagnetic wave:
(A) The electric field and the magnetic field have equal average values.
(B) The electric energy and the magnetic energy have equal average values.
(a) Both A and B are true.
(b) A is false but B is true.
(c) B is false but A is true
(d) Both A and B are false.
digAnsr: a
Ans : (a) Both A and B are true.
For a linearly polarised, plane electromagnetic wave
`` E={E}_{0}\,\mathrm{\,sin\,}\omega (t-\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$c$}\right.)``
`` B={B}_{0}\,\mathrm{\,sin\,}\omega (t-\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$c$}\right.)``
The average value of either E or B over a cycle is zero ( average of sin(`` \theta ``) over a cycle is zero).
Also the electric energy density (u_E) and magnetic energy density (u_B)are equal.
`` {u}_{\,\mathrm{\,E\,}}=\frac{1}{2}{\in }_{0}{E}^{2}=\frac{{B}^{2}}{2{\mu }_{0}}={u}_{\,\mathrm{\,B\,}}``
Energy can be found out by integrating energy density over the entire volume of full space.
As the energy of the electromagnetic wave is equally shared between electric and magnetic field so their average values will also be equal.
Page No 339:

Qstn #8
A free electron is placed in the path of a plane electromagnetic wave. The electron will start moving
(a) along the electric field
(b) along the magnetic field
(c) along the direction of propagation of the wave
(d) in a plane containing the magnetic field and the direction of propagation
digAnsr: a
Ans : (a) along the electric field
As the electron is at rest initially, only the electric field will exert force on it. There will be no magnetic force on the electron in the stating. Hence, the electron will start moving along the electric field.
Page No 339:

Qstn #9
A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E.
(a) p = 0, E ≠ 0
(b) p ≠ 0, E = 0
(c) p ≠ 0, E ≠ 0
(d) p = 0, E = 0
digAnsr: c
Ans : (c) p ≠ 0, E ≠ 0.
When an electromagnetic wave strikes a material surface, it transports the momentum, as well as the energy, to the surface. The striking electromagnetic wave exerts pressure on the surface. The total energy transferred to the surface by the electromagnetic wave is given by E = pc. Therefore, p ≠ 0, E ≠ 0.
Page No 339:

#
Section : iii

Qstn #1
An electromagnetic wave going through vacuum is described by
E = E₀ sin (kx - ωt).
Which of the following is/are independent of the wavelength?
(a) k
(b) ω
(c) k/ω
(d) kω
digAnsr: c
Ans : (c) k/ω
The given quantities can be expressed as:
k is given by
`` k=\frac{2\pi }{\lambda }``
ω is given by
`` \omega =2\pi \nu ``
`` c=\nu \lambda ``
`` \Rightarrow \omega =2\pi \frac{c}{\lambda }``
k/ω is given by
`` \frac{k}{\omega }=\frac{2\pi /\lambda }{2\pi c/\lambda }=\frac{1}{c}``
kω is given by
`` k\times \omega =\frac{2\pi }{\lambda }\times \frac{2\pi c}{\lambda }=\frac{4{\pi }^{2}c}{{\lambda }^{2}}``
Thus, k/ω is independent of the wavelength.
Page No 339:

Qstn #2
Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor
(a) increases
(b) decreases
(c) does not change
(d) is zero
digAnsr: a,b
Ans : (a) increases
(b) decreases
Displacement current inside a capacitor,
`` {i}_{\,\mathrm{\,d\,}}={\epsilon }_{0}\frac{d{\varphi }_{\,\mathrm{\,E\,}}}{dt}``, where
ϕ_Eis the electric flux inside the capacitor.
Up to the time the electric flux changes, there will be a displacement current. This is possible when the charge on a capacitor changes. Therefore, the displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor or electric field increases or decreases inside the capacitor.
Page No 339:

Qstn #3
Speed of electromagnetic waves is the same
(a) for all wavelengths
(b) in all media
(c) for all intensities
(d) for all frequencies
digAnsr: c
Ans : (c) for all intensities
For any given medium, the speed (c) of an electromagnetic wave is given by
c = νλ,
where
ν = frequency of the electromagnetic wave
λ = wavelength of the electromagnetic wave
As the frequency and wavelength are changed, the speed of the electromagnetic wave changes. So, the speed of an electromagnetic wave is not same for all wavelengths and all frequencies in any medium. The velocity of an electromagnetic wave changes with change in medium. Also, the speed of an electromagnetic wave is same for all the intensities in any medium.
Page No 339:

Qstn #4
Which of the following have zero average value in a plane electromagnetic wave?
(a) Electric field
(b) Magnetic field
(c) Electric energy
(d) Magnetic energy
digAnsr: a,b
Ans : (a) Electric field
(b) Magnetic field
In a plane electromagnetic wave, the electric and the magnetic fields oscillate sinusoidally. For an electromagnetic wave propagating in the z-direction, the electric and magnetic fields are given by:
E_x = E₀ sin (kz - ωt)
B_y = B₀ sin (kz - ωt)
These are sinusoidal functions. Therefore, for a fixed value of z, the average value of the electric and magnetic fields are zero.
Page No 339:

Qstn #5
The energy contained in a small volume through which an electromagnetic wave is passing oscillates with
(a) zero frequency
(b) the frequency of the wave
(c) half the frequency of the wave
(d) double the frequency of the wave
digAnsr: d
Ans : (d) double the frequency of the wave
The energy per unit volume of an electromagnetic wave,
`` u=\frac{1}{2}{\in }_{0}{E}^{2}+\frac{{B}^{2}}{2{\mu }_{0}}``
The energy of the given volume can be calculated by multiplying the volume with the above expression.
`` U=u\times V=\left(\frac{1}{2}{\in }_{0}{E}^{2}+\frac{{B}^{2}}{2{\mu }_{0}}\right)\times V`` ...(1)
Let the direction of propagation of the electromagnetic wave be along the z-axis. Then, the electric and magnetic fields at a particular point are given by
E_x= E₀ sin (kz - ωt)
B_y= B₀ sin (kz - ωt)
Substituting the values of electric and magnetic fields in (1), we get:
`` U=\left(\frac{1}{2}{\in }_{0}\left({{E}_{0}}^{2}{\,\mathrm{\,sin\,}}^{2}\right(kz-\omega t)+\frac{{{B}_{0}}^{2}{\,\mathrm{\,sin\,}}^{2}(kz-\omega t)}{2{\mu }_{0}}\right)\times V``
`` \Rightarrow U=\left({\in }_{0}{{E}_{0}}^{2}\frac{(1-\,\mathrm{\,cos\,}(2kz-2\omega t\left)\right)}{4}+\frac{{{B}_{0}}^{2}(1-\,\mathrm{\,cos\,}(2kz-2\omega t\left)\right)}{4{\mu }_{0}}\right)\times V``
From the above expression, it can be easily understood that the energy of the electric and magnetic fields have angular frequency 2ω. Thus, the frequency of the energy of the electromagnetic wave will also be double.
Page No 339:

#
Section : iv

Qstn #1
Show that the dimensions of the displacement current
ε0dφEdtare that of an electric current.
Ans : Displacement current,
`` {I}_{\,\mathrm{\,D\,}}=\frac{{\in }_{0}d{\phi }_{\,\mathrm{\,e\,}}}{dt}``
Electric flux,
`` {\varphi }_{\,\mathrm{\,e\,}}=EA``
`` \left[{\varphi }_{e}\right]\mathit{=}\left[E\right]\left[A\right]``
`` =\left[\frac{\mathit{1}}{\mathit{4}\pi {\mathit{\in }}_{\mathit{0}}}\frac{q}{{r}^{\mathit{2}}}\right]\left[A\right]``
`` \,\mathrm{\,Also\,},\left[{\mathit{\in }}_{0}\right]=\left[{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{-3}{\,\mathrm{\,T\,}}^{4}{\,\mathrm{\,A\,}}^{2}\right]``
`` \Rightarrow \left[{\varphi }_{\,\mathrm{\,e\,}}\right]=\left[{\,\mathrm{\,M\,}}^{1}{\,\mathrm{\,L\,}}^{3}{\,\mathrm{\,T\,}}^{-4}{\,\mathrm{\,A\,}}^{-2}\right]\left[\,\mathrm{\,AT\,}\right]\left[{\,\mathrm{\,L\,}}^{-2}\right]\left[{\,\mathrm{\,L\,}}^{2}\right]``
`` =\left[{\,\mathrm{\,ML\,}}^{3}{\,\mathrm{\,T\,}}^{-3}{\,\mathrm{\,A\,}}^{-1}\right]``
`` ``
Displacement current,
`` \left[{I}_{\,\mathrm{\,D\,}}\right]=\left[{\in }_{0}\right]\left[{\varphi }_{\,\mathrm{\,e\,}}\right]\left[{\,\mathrm{\,T\,}}^{-1}\right]``
`` \left[{I}_{\,\mathrm{\,D\,}}\right]=\left[{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{-3}{\,\mathrm{\,T\,}}^{4}{\,\mathrm{\,A\,}}^{2}\right]\left[{\,\mathrm{\,ML\,}}^{3}{\,\mathrm{\,T\,}}^{-3}{\,\mathrm{\,A\,}}^{-1}\right]\left[{\,\mathrm{\,T\,}}^{-1}\right]``
`` \left[{I}_{\,\mathrm{\,D\,}}\right]=\left[\,\mathrm{\,A\,}\right]``
[I_D]=[current]
Page No 339:

Qstn #2
A point charge is moving along a straight line with a constant velocity v. Consider a small area A perpendicular to the motion of the charge. Calculate the displacement current through the area when its distance from the charge is x. The value of x is not large, so that the electric field at any instant is essentially given by Coulomb’s law.
Figure
Ans : From Coulomb's law:
Electric field strength,
`` E\mathit{=}\frac{kq}{{x}^{\mathit{2}}}``
Electric flux,
`` {\varphi }_{\,\mathrm{\,E\,}}=EA``
`` {\varphi }_{\,\mathrm{\,E\,}}=\frac{kqA}{{x}^{2}}``
Displacement current = I_d
`` {I}_{\,\mathrm{\,d\,}}=\left|{\in }_{0}\frac{d{\varphi }_{\,\mathrm{\,E\,}}}{dt}\right|``
`` {I}_{\,\mathrm{\,d\,}}=\left|{\in }_{0}\frac{d}{dt}\left(\frac{kqA}{{x}^{2}}\right)\right|``
`` {I}_{\,\mathrm{\,d\,}}=\left|{\in }_{0}kqA\frac{d}{dt}{x}^{-2}\right|``
`` {I}_{\,\mathrm{\,d\,}}=\left|{\mathit{\in }}_{\mathit{0}}\frac{\mathit{1}}{\mathit{4}\mathit{\pi }{\mathit{\in }}_{\mathit{0}}}\mathit{\times }\mathit{q}\mathit{\times }\mathit{A}\mathit{\times }\mathit{(}\mathit{-}\mathit{2}\mathit{)}{\mathit{x}}^{\mathit{-}\mathit{3}}\mathit{\times }\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}\right|``
`` {I}_{\,\mathrm{\,d\,}}=\left|\frac{qAv}{2\,\mathrm{\,\pi \,}{x}^{3}}\right|``
Page No 339:

Qstn #3
A parallel-plate capacitor of plate-area A and plate separation d is joined to a battery of emf ε and internal resistance R at t = 0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.
Ans : Electric field strength for a parallel plate capacitor,
`` E=\frac{Q}{{\in }_{0}A}``
Electric flux linked with the area,
`` {\varphi }_{E}\mathit{=}EA\mathit{=}\frac{Q}{{\mathit{\in }}_{\mathit{0}}A}\mathit{\times }\frac{A}{\mathit{2}}=\frac{Q}{2{\in }_{0}}``
Displacement current,
`` {I}_{\,\mathrm{\,d\,}}={\in }_{0}\frac{d{\varphi }_{\,\mathrm{\,E\,}}}{dt}={\in }_{0}\frac{d}{dt}\left(\frac{Q}{{\mathit{\in }}_{\mathit{0}}\mathit{2}}\right)``
`` {I}_{\,\mathrm{\,d\,}}=\frac{1}{2}\left(\frac{d\,\mathrm{\,Q\,}}{dt}\right)...\left(\,\mathrm{\,i\,}\right)``
Charge on the capacitor as a function of time during charging,
`` Q=\epsilon C\left[1-{e}^{\mathit{-}t\mathit{/}RC}\right]``
Putting this in equation (i), we get:
`` {I}_{\,\mathrm{\,d\,}}=\frac{1}{2}\epsilon C\frac{d}{dt}\left(1\mathit{-}{e}^{\mathit{-}t\mathit{/}RC}\right)``
`` {I}_{\,\mathrm{\,d\,}}=\frac{1}{2}\epsilon C\left(\mathit{-}{e}^{\mathit{-}t\mathit{/}RC}\right)\times \left(\mathit{-}\frac{1}{RC}\right)``
`` \,\mathrm{\,C\,}=\frac{A{\mathit{\in }}_{\mathit{0}}}{d}``
`` \Rightarrow {I}_{\,\mathrm{\,d\,}}=\frac{\epsilon }{2R}\times {e}^{\mathit{-}\frac{td}{{\epsilon }_{\mathit{0}}AR}}``
Page No 339:

Qstn #4
Consider the situation of the previous problem. Define displacement resistance R_d = V/i_d of the space between the plates, where V is the potential difference between the plates and i_d is the displacement current. Show that R_d varies with time as
Rd=R(et/τ-1).
Ans : Electric field strength for a parallel plate capacitor = `` E=\frac{Q}{{\in }_{0}A}``
`` \,\mathrm{\,Electric\,}\,\mathrm{\,flux\,},\varphi \mathit{=}E\mathit{.}A\mathit{=}\frac{Q}{{\mathit{\in }}_{0}A}\mathit{.}A=\frac{Q}{{\in }_{0}}``
`` \,\mathrm{\,Displacement\,}\,\mathrm{\,current\,},{i}_{\,\mathrm{\,d\,}}={\in }_{0}\frac{d{\varphi }_{\,\mathrm{\,E\,}}}{dt}={\in }_{0}\frac{d}{dt}\left(\frac{\,\mathrm{\,Q\,}}{{\in }_{0}}\right)``
`` {i}_{\,\mathrm{\,d\,}}=\left(\frac{dQ}{dt}\right)``
`` \,\mathrm{\,Also\,},Q\mathit{=}CV``
`` {i}_{\,\mathrm{\,d\,}}=\frac{d}{dt}\left({E}_{\mathit{0}}{{C}_{e}}^{\mathit{-}t\mathit{/}RC}\right)``
`` {i}_{\,\mathrm{\,d\,}}={E}_{0}C\left(\mathit{-}\frac{\mathit{1}}{RC}\right){e}^{\mathit{-}t\mathit{/}RC}``
`` \,\mathrm{\,Displacement\,}\,\mathrm{\,resistance\,},{R}_{d}=\frac{{E}_{\mathit{0}}}{{i}_{\,\mathrm{\,d\,}}}-R``
`` \Rightarrow {R}_{d}=\frac{{E}_{\mathit{0}}}{\frac{{E}_{\mathit{0}}}{R}{\,\mathrm{\,e\,}}^{\mathit{-}t\mathit{/}RC}}-R``
`` \Rightarrow {R}_{d}=R{\,\mathrm{\,e\,}}^{t\mathit{/}RC}-R``
`` \mathit{\Rightarrow }{R}_{\mathit{d}}\mathit{=}R({\,\mathrm{\,e\,}}^{\mathit{t}\mathit{/}\mathit{R}\mathit{C}}-1)``
Page No 339: