Electromagnetic Waves Test

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Electromagnetic Waves Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

The charge on a parallel plate capacitor varies as $$q = {q}_{0}cos2vt$$. The plates are very large and close together (area = A, separation = d ) , the displacement current through the capacitor is

A
$${q}_{0}2v sinvt$$

B
$$-{q}_{0}2v sinvt$$

C
$${q}_{0}2 sinvt$$

D
$${q}_{0}v sinvt$$

Solution

Given : $$q = q_0 \cos 2vt$$
To find : $$I_d = ?$$
Formula : $$I_d = \dfrac{dq}{dt}$$
Sol : From formula,
The displacement current through the capacitor
$$I_d= \dfrac{dq}{dt} = \dfrac{d}{dt} (q_0\cos 2vt) = -q_0 \sin 2vt \times 2v$$
$$\therefore I_d = -q_02v \sin 2vt$$
$$\therefore $$ Correct option is (B)
Question 2
1 / -0

Two opposite charged particles oscillate about their mean equilibrium position in free space, with a frequency of $${10}^{9}Hz$$. The wavelength of the corresponding electromagnetic wave produced is ______

A
$$0.3m$$

B
$${10}^{9}m$$

C
$$3\times { 10 }^{ 17 }m$$

D
$$3.3m$$

Solution

Electromagnetic wave $$V=3\times10^8m/s$$
Given frequency $$(f)=10^9 Hz$$
$$V=f\lambda\\ \lambda=\cfrac{V}{f}=\cfrac{3\times 10^8}{10^9}=0.3m$$
Question 3
1 / -0

The rms value of the electric field of the light coming from sun is 720 N $$C^{-1}$$. The average total energy density of the electromagnetic wave is

A
$$3.3 10^{-3} J m^{-3}$$

B
$$4.58 10^{-6} J m^{-3}$$

C
$$6.37 10^{-9} J m^{-3}$$

D
$$81.35 10^{-12} J m^{-3}$$

Solution

Given: The RMS value of the electric field of light is $$720\ N/C$$.

Energy density of electromagnetic wave is given by:
$$E=\epsilon_0 E_{rms}^2$$

$$=8.85\times 10^{-12}\times (720)^2$$

$$E\approx 4.58\times 10^{-6}J/m^3$$

Option $$(B)$$ is corret.
Question 4
1 / -0

An electromagnetic wave radiates outwards from a dipole antenna, with $${ E }_{ 0 }$$ as the amplitude of its electric field vector. The electric field $${ E }_{ 0 }$$ which transports significant energy from the source falls off as

A
$$\dfrac { 1 }{ { r }^{ 3 } }$$

B
$$\dfrac { 1 }{ { r }^{ 2 } }$$

C
$$\dfrac { 1 }{ { r }}$$

D
remains constant.

Solution

A diode antenna radiates the electromagnetic wave outwards. The amplitude of electric field vector $$(E_0)$$ which transports significant energy from the source falls intensity inversely as the distance (r) from the antenna,

i.e,$$ E_{0}\propto \dfrac{1}{r}$$
Question 5
1 / -0

One requires 11eV of energy to dissociate of carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in:

A
visible region.

B
infrared region.

C
ultraviolet region.

D
microwave region.

Solution

The energy required for the dissociation is given by:
$$E=h\nu$$

$$\Rightarrow 11\times{1.6\times10^{-19}}=6.626\times10^{-34}\times\nu$$

$$\Rightarrow \nu=2.656\times 10^{15}\ Hz$$

Hence, the frequency required for the dissociation is $$2.656\times 10^{15}\ Hz$$ and it lies in the range of UV region.
Question 6
1 / -0

A parallel plate capacitor with circular plates of radius $$R$$ is being charged as shown. At the instant shown, the displacement current in the region between the plates enclosed between $$\cfrac{R}{2}$$ and $$R$$ is given by

A
$$\cfrac{3}{4}i$$

B
$$\cfrac{1}{4}i$$

C
$$3i$$

D
$$\cfrac{4}{3}i$$

Solution

Displacement current is given by

$$ I={{\varepsilon }_{0}}\dfrac{d{{\phi }_{e}}}{dt} $$

$$ =\,{{\varepsilon }_{0}}A\dfrac{dE}{dt}\,\,\,\,(\because \,\phi =A.E) $$

$$ =A\dfrac{d}{dt}\left(\dfrac{q}{{{A}^{'}}}\right) $$

$$ =\dfrac{A}{{{A}^{'}}}i\,\,(\because \dfrac{dq}{dt}=i) $$

$$ =\dfrac{\pi {{\left(\dfrac{R}{2}\right)}^{2}}}{\pi {{R}^{2}}} $$

$$ =\dfrac{1}{4}i $$
Question 7
1 / -0

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is:

A
$$\dfrac{E}{c}$$

B
$$\dfrac{2E}{c}$$

C
$$Ec$$

D
$$\dfrac{E}{c^2}$$

Solution

Hint:-Momentum transfered to surface is equal to change in momentum of radiation.

Step 1: WRITE INITIAL AND FINAL MOMENTUM OF RADIATION

$$P_i = \dfrac{E}{c}$$ and $$P_f = \dfrac{-E}{c}$$ too.

Step 2: CALCULATE MOMENTUM TRANSFERED TO SURFACE
Momentum transferred to the surface $$\Rightarrow P_{final} – P_{initial}.$$
So,

Net Momentum $$= \dfrac{E}{c}-(-\dfrac{E}{c}) \Rightarrow \dfrac{2E}{c}$$ .
Question 8
1 / -0

A plane polarized monochromatic $$EM$$ wave is traveling in vacuum along $$z$$ direction such that at $$t={t}_{1}$$ it is found that the electric field is zero at a spatial point $${z}_{1}$$. The next zero that occurs in its neighborhood is at $${z}_{2}$$. The frequency of the electromagnetic wave is:

A
$$\dfrac { 3\times { 10 }^{ 8 } }{ \left| { z }_{ 2 }-{ z }_{ 1 } \right| }$$

B
$$\dfrac { 1.5\times { 10 }^{ 8 } }{ \left| { z }_{ 2 }-{ z }_{ 1 } \right| }$$

C
$$\dfrac { 6\times { 10 }^{ 8 } }{ \left| { z }_{ 2 }-{ z }_{ 1 } \right| }$$

D
$$\dfrac { 1 }{ { t }_{ 1 }+\dfrac { \left| { z }_{ 2 }-{ z }_{ 1 } \right| }{ 3\times { 10 }^{ 8 } } }$$

Solution

From question:
$$\begin{array}{l} \frac { \lambda }{ 2 } =\left| { { z_{ 2 } }-{ z_{ 1 } } } \right| \Rightarrow \lambda =2\left| { { z_{ 2 } }-{ z_{ 1 } } } \right| \\ c=f\times \lambda \Rightarrow f=\frac { c }{ \lambda } =\frac { { 3\times { { 10 }^{ 8 } } } }{ { 2\left| { { z_{ 2 } }-{ z_{ 1 } } } \right| } } =\frac { { 1.5\times { { 10 }^{ 8 } } } }{ { \left| { { z_{ 2 } }-{ z_{ 1 } } } \right| } } \end{array}\begin{array}{l} \frac { \lambda }{ 2 } =\left| { { z_{ 2 } }-{ z_{ 1 } } } \right| \Rightarrow \lambda =2\left| { { z_{ 2 } }-{ z_{ 1 } } } \right| \\ c=f\times \lambda \Rightarrow f=\frac { c }{ \lambda } =\frac { { 3\times { { 10 }^{ 8 } } } }{ { 2\left| { { z_{ 2 } }-{ z_{ 1 } } } \right| } } =\frac { { 1.5\times { { 10 }^{ 8 } } } }{ { \left| { { z_{ 2 } }-{ z_{ 1 } } } \right| } } \end{array}$$
zero electric field occurs at half of wavelength
Question 9
1 / -0

The intensity of sun on earth is $$1400 W/m^2$$. Assuming earth to be a black body. Calculate radiation pressure?

A
$$4.66 \times 10^6 N/m^2$$

B
$$1.4 \times 10^{-6} N/m^2$$

C
$$4.66 \times 10^{-6} N/m^2$$

D
$$7 \times 10^{-6} N/m^2$$

Solution
Question 10
1 / -0

The electric field associated with an e.m. wave in vacuum is given by $$\vec {E} = 40\cos (kz - 6\times 10^{8}t)\hat {i}$$, where $$E, z$$ and $$t$$ in $$volt/m$$, meter and seconds respectively. The value of wave vector $$k$$ is

A
$$6m^{-1}$$

B
$$3m^{-1}$$

C
$$2m^{-1}$$

D
$$0.5m^{-1}$$

Solution

Given: The electric field associated with an electromagnetic wave in vacuum is given by $$\vec E =40 \cos(kz−6\times 10^8t)\hat i$$ , where E, z and t are in volt per meter, meter and second respectively.
To find the value of wave vector k
Solution:
We know electromagnetic wave eqution is
$$E=E_0\cos(kz-\omega t)$$
And given equation is
$$\vec E =40 \cos(kz−6\times 10^8t)\hat i$$
By comparing these two, we get
$$\omega=6\times10^8$$ and
$$E_0=40\hat i$$
we also know,
Speed of electromagnetic wave, $$v=\dfrac \omega k$$
where v is the speed of the light
Hence, $$k=\dfrac \omega v\\\implies k=\dfrac {6\times 10^8}{3\times 10^8}\\\implies k=2m^{-1}$$
is the required value

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Electromagnetic Waves Test - 27

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Electromagnetic Waves Test - 27

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Question 1

$${q}_{0}2v sinvt$$

$$-{q}_{0}2v sinvt$$

$${q}_{0}2 sinvt$$

$${q}_{0}v sinvt$$

Solution

Question 2

$$0.3m$$

$${10}^{9}m$$

$$3\times { 10 }^{ 17 }m$$

$$3.3m$$

Solution

Question 3

$$3.3 10^{-3} J m^{-3}$$

$$4.58 10^{-6} J m^{-3}$$

$$6.37 10^{-9} J m^{-3}$$

$$81.35 10^{-12} J m^{-3}$$

Solution

Question 4

$$\dfrac { 1 }{ { r }^{ 3 } }$$

$$\dfrac { 1 }{ { r }^{ 2 } }$$

$$\dfrac { 1 }{ { r }}$$

remains constant.

Solution

Question 5

visible region.

infrared region.

ultraviolet region.

microwave region.

Solution

Question 6

$$\cfrac{3}{4}i$$

$$\cfrac{1}{4}i$$

$$3i$$

$$\cfrac{4}{3}i$$

Solution

Question 7

$$\dfrac{E}{c}$$

$$\dfrac{2E}{c}$$

$$Ec$$

$$\dfrac{E}{c^2}$$

Solution

Question 8

$$\dfrac { 3\times { 10 }^{ 8 } }{ \left| { z }_{ 2 }-{ z }_{ 1 } \right| }$$

$$\dfrac { 1.5\times { 10 }^{ 8 } }{ \left| { z }_{ 2 }-{ z }_{ 1 } \right| }$$

$$\dfrac { 6\times { 10 }^{ 8 } }{ \left| { z }_{ 2 }-{ z }_{ 1 } \right| }$$

$$\dfrac { 1 }{ { t }_{ 1 }+\dfrac { \left| { z }_{ 2 }-{ z }_{ 1 } \right| }{ 3\times { 10 }^{ 8 } } }$$

Solution

Question 9

$$4.66 \times 10^6 N/m^2$$

$$1.4 \times 10^{-6} N/m^2$$

$$4.66 \times 10^{-6} N/m^2$$

$$7 \times 10^{-6} N/m^2$$

Solution

Question 10

$$6m^{-1}$$

$$3m^{-1}$$

$$2m^{-1}$$

$$0.5m^{-1}$$

Solution

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