40: Electromagnetic Waves : Cbse-xi-physics

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CBSE-XI-Physics

40: Electromagnetic Waves

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Section : i

Qstn #1
In a microwave oven, the food is kept in a plastic container and the microwaves is directed towards the food. The food is cooked without melting or igniting the plastic container. Explain.
Ans : The natural frequency of water matches the frequency of microwave. This is the reason that food containing water gets cooked. The natural frequency of the plastic container does not match the frequency of microwave. So, the plastic container is not damaged.
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Qstn #2
A metal rod is placed along the axis of a solenoid carrying a high-frequency alternating current. It is found that the rod gets heated. Explain why the rod gets heated.
Ans : The magnetic field along the axis of a solenoid carrying a high-frequency alternating current changes continuously. Due to the change in the magnetic field, e.m.f (or eddy current) is induced in the metal rod. There will be flow of charge due to the induced e.m.f. The direction of the induced e.m.f changes very frequently due to the high-frequency alternating current in the solenoid. Thus, the rod gets heated up due to the flow of charge in it.
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Qstn #3
Can an electromagnetic wave be deflected by an electric field or a magnetic field?
Ans : No, an electromagnetic wave cannot be deflected by an electric field or a magnetic field. This is because according to Maxwell's theory, an electromagnetic wave does not interact with the static electric field and magnetic field. Even if we consider the particle nature of the wave, the photon is electrically neutral. So, it is not affected by the static magnetic and electric fields.
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Qstn #4
A wire carries an alternating current i = i₀ sin ωt. Is there an electric field in the vicinity of the wire?
Ans : When an alternating current passes through a conductor, the changing magnetic field create a changing electric field outside it. An electromagnetic field is radiated from the surface of the conductor. There is a time-varying electric field outside the conductor. Hence, there is a time-varying electric field in the vicinity of the wire.
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Qstn #5
A capacitor is connected to an alternating-current source. Is there a magnetic field between the plates?
Ans : When an alternating-current source is connected to a capacitor, the electric field between the plates of the capacitor keeps on changing with the applied voltage. Due to the changing electric field, a magnetic field exists in between the plates of the capacitor.
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Qstn #6
Can an electromagnetic wave be polarised?
Ans : An electromagnetic wave is a transverse wave; thus, it can be polarised. An unpolarised wave consists of many independent waves, whose planes of vibrations of electric and magnetic fields are randomly oriented. They are polarised by restricting the vibrations of the electric field vector or magnetic field vector in one direction only.
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Qstn #7
A plane electromagnetic wave is passing through a region. Consider (a) electric field (b) magnetic field (c) electrical energy in a small volume and (d) magnetic energy in a small volume. Construct the pairs of the quantities that oscillate with equal frequencies.
Ans : Let the electromagnetic wave be propagating in the z-direction. The vibrations of the electric and magnetic fields are given by:
E_x= E₀ sin (kz - ωt)
B_y= B₀ sin (kz - ωt)
Let the volume of the region be V.
The angular frequency of the vibrations of the electric and magnetic fields are same and are equal to ω. Therefore, their frequency, `` f=\frac{\omega }{2\pi }``, is same.
The electrical energy in the region,
U_E= `` \left(\frac{1}{2}{\in }_{0}{E}^{2}\right)\times V``
It can be written as:
`` {U}_{\,\mathrm{\,E\,}}=\left(\frac{1}{2}{\in }_{0}\left({{E}_{0}}^{2}{\,\mathrm{\,sin\,}}^{2}\right(kz-\omega t)\right)\times V``
`` {U}_{E}=\left(\frac{1}{2}{\in }_{0}{{E}_{0}}^{2}\times \frac{\left(1-\,\mathrm{\,cos\,}2(kz-\omega t)\right)}{2}\right)\times V``
`` {U}_{E}=\left(\frac{1}{4}{\in }_{0}{{E}_{0}}^{2}\times \left(1-\,\mathrm{\,cos\,}2(kz-\omega t)\right)\right)\times V``
`` ``
The magnetic energy in the region,
`` {U}_{\,\mathrm{\,B\,}}=\left(\frac{{B}^{2}}{2{\mu }_{0}}\right)\times V``
`` {U}_{\,\mathrm{\,B\,}}=\left(\frac{{{B}_{0}}^{2}{\,\mathrm{\,sin\,}}^{2}(kz-\omega t)}{2{\mu }_{0}}\right)\times V``
`` \Rightarrow {U}_{\,\mathrm{\,B\,}}=\left(\frac{{{B}_{0}}^{2}\left(1-\,\mathrm{\,cos\,}(2kz-2\omega t)\right)}{4{\mu }_{0}}\right)\times V``
The angular frequency of the electric and magnetic energies is same and is equal to 2ω.
Therefore, their frequency, `` f\text{'}=\frac{2\omega }{2\pi }=2f``
`` ``, will be same.
Thus, the electric and magnetic fields have same frequencies and the electrical and magnetic energies will have same frequencies.
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#
Section : ii

Qstn #1
A magnetic field can be produced by
(a) a moving charge
(b) a changing electric field
(c) Neither of them
(d) Both of them
digAnsr: d
Ans : (d) Both of them
According to Ampere-Maxwell's Law, a magnetic field is produced due to the conduction current in a conductor and the displacement current. The conduction current is actually the motion of the charge. The displacement current is due to the changing electric field. The displacement current is given by `` {i}_{\,\mathrm{\,d\,}}={\epsilon }_{0}\frac{\,\mathrm{\,d\,}{\Phi }_{E}}{\,\mathrm{\,d\,}t}`` (`` \because `` ϕ_E is the electric flux).
Thus, the magnetic field is produced by the moving charge as well as the changing electric field.
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Qstn #2
A compass needle is placed in the gap of a parallel plate capacitor. The capacitor is connected to a battery through a resistance. The compass needle
(a) does not deflect
(b) deflects for a very short time and then comes back to the original position
(c) deflects and remains deflected as long as the battery is connected
(d) deflects and gradually comes to the original position in a time that is large compared to the time constant
digAnsr: d
Ans : (d) deflects and gradually comes to the original position in a time that is large compared to the time constant
The compass needle deflects due to the presence of the magnetic field. Inside the capacitor, a magnetic field is produced when there is a changing electric field inside it. As the capacitor is connected across the battery, the charge on its plates at a certain time t is given by:
Q = CV (1 `` -`` e^{`` -``t/RC}),
where
Q = charge developed on the plates of the capacitor
R = resistance of the resistor connected in series with the capacitor
C = capacitance of the capacitor
V = potential difference of the battery
The time constant of the capacitor is given, τ = RC
The capacitor keeps on charging up to the time τ. The development of charge on the plates will be gradual after t = RC. The change in electric field will be up to the time the charge is developing on the plates of the capacitor. Thus, the compass needle deflects and gradually comes to the original position in a time that is large compared to the time constant.
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Qstn #3
Dimensions of 1/(µ₀Ïµ₀) is
(a) L/T
(b) T/L
(c) L²/T²
(d) T²/L².
digAnsr: c
Ans : (c) L²/T²
The speed of light, `` C=\frac{1}{\sqrt{{\mu }_{0}{\in }_{0}}}``.
The dimensions of `` \frac{1}{\sqrt{{\mu }_{0}{\in }_{0}}}`` are of velocity, i.e. L/T .
Therefore, `` \frac{1}{{\in }_{0}{\mu }_{0}}`` will have dimensions L²/T².
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Qstn #4
Electromagnetic waves are produced by
(a) a static charge
(b) a moving charge
(c) an accelerating charge
(d) chargeless particles
digAnsr: c
Ans : (c) an accelerating charge
A static charge produces an electrostatic field. A moving charge produces a magnetic field. Electromagnetic waves are produced by an accelerating charge.
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Qstn #5
An electromagnetic wave going through vacuum is described by
E = E₀ sin (kx - ωt); B = B₀ sin (kx - ωt).
Then,
(a) E₀ k = B₀ ω
(b) E₀ B₀ = ωk
(c) E₀ ω = B₀ k
(d) None of these
digAnsr: a
Ans : (a) E₀ k = B₀ ω
The relation between E₀ and B₀is given by `` \frac{{E}_{0}}{{B}_{0}}=c`` ...(1)
Here, c = speed of the electromagnetic wave
The relation between ω (the angular frequency) and k (wave number):
`` \frac{\omega }{k}=c`` ...(2)
Therefore, from (1) and (2), we get:
`` \frac{{E}_{0}}{{B}_{0}}=```` \frac{\omega }{k}=c``
`` \Rightarrow {E}_{0}k={B}_{0}\omega ``
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