Electromagnetic Waves Test

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Electromagnetic Waves Test - 34

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Weekly Quiz Competition

Question 1
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Choose the correct answer from the alternatives given.
Assume a bulb of efficiency 2.5% as a point source. The peak values of electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m is respectively

A
$$2.5 \, V \, m^1, \, 3.6 \, \times \, 10^{-8} T$$

B
$$4.2 \, V \, m^1, \, 2.8 \, \times \, 10^{-8} T$$

C
$$4.08 \, V \, m^1, \, 1.36 \, \times \, 10^{-8} T$$

D
$$3.6 \, V \, m^1, \, 3.6 \, \times \, 10^{-8} T$$

Solution

Here intensity, $$ I = \dfrac{power}{area}$$

= $$\dfrac{100 \times 2.5 }{4\pi (3)^2 \times 100} = \dfrac{2.5}{36\pi }Wm^{-2}$$

Half of this intensity belongs to electronic field and half of that to magnetic field.

$$\therefore \dfrac{I}{2} \,= \, \dfrac{1}{4}\varepsilon_0E^2_0c$$

or
$$E_o = \sqrt{\dfrac{2I}{\epsilon _0c}}\,\\\Rightarrow \, \sqrt{\dfrac{2 \times \dfrac{2.5}{36\pi }}{\dfrac{1}{4\pi \times 9\times 10^9 }}\times 3\times 10^8}\\\Rightarrow 4.08 V m^{-i}$$

$$\therefore B_o = \dfrac{E_0}{c} \,=\, \dfrac{4.08}{3\times 10^8}\\ = 1.36 \times 10^{-8}T$$
Question 2
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A plane EM wave travelling alone z-direction is described by $${ \overrightarrow { E } \ =\ { E }_{ 0 } }\sin { (kz\ -\ \omega t)\hat{ i } }$$ and $${\vec B\ =\ { B }_{ 0 } }\sin { (kz\ -\ \omega t)\hat{ j } }$$.

A
The average energy density of the wave is given by $$\displaystyle { u }_{ av }\ =\ \dfrac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }\ +\ \dfrac { 1 }{ 4 } \dfrac { { B }_{ 0 }^{ 2 } }{ { \mu }_{ 0 } } .$$

B
The time averaged intensity of the wave is given by $$ { I }_{ av }\ =\dfrac{1}{2}\in_0cE_0^2$$

C
Both (a) and (b)

D
None of these

Solution

(i) The e.m. wave carries energy which is due to electric field vector and magnetic field vector. In e.m. wave $$E$$ and $$B$$ vary from point to point and from moment to moment. Let $$E$$ and $$B$$ be their time average. The energy density due to the electric field $$E$$ is

$$u_E = \dfrac{1}{2} \in_0 E^2$$

The energy density due to magnetic field $$B$$ is $$u B = \dfrac{1}{2}\dfrac{B^2}{\mu_0}$$
Total average energy density of em wave
$$u_{av}=uE+uB=\dfrac{1}{2}\in_0E^2+\dfrac{1}{2}\dfrac{B^2}{\mu_0}$$ ...(i)

Let the em wave be propagation along z-direction. The electric field vector and magnetic field vectors be represented by
$$E=E_0\sin (kz-wt)$$
$$B= B_0\sin (kz-wt)$$

The time average value of $$E^2$$ over complete cycle $$= E_0^2/2$$

and time average value of $$B^2$$ over complete cycle $$= B_0^2/2$$

$$\therefore u_{av} = \dfrac{1}{2} \in_0 \dfrac{E_0^2}{2} + \dfrac{1}{2}\mu_0\left(\dfrac{B_0^2}{2}\right) = \dfrac{1}{4} \in_0 E_0^2 + \dfrac{B_0^2}{4\mu_0}$$

(ii) we know that $$E_0 = cB_0$$ and $$c = \dfrac{1}{\sqrt{\mu_0\in_0}}$$.

$$\therefore \dfrac{1}{4} \dfrac{B_0^2}{\mu_0} = \dfrac{1}{4} \dfrac{E_0^2/c^2}{\mu_0} = \dfrac{E_0^2}{4\mu_0} \times \mu_0 \in_0 = \dfrac{1}{4} \in_0 E_0^2$$

$$\therefore uB = uE$$

Hence, $$u_{av} = \dfrac{1}{4} \in_0 E^2_0 + \dfrac{1}{4} \dfrac{B_0^2}{\mu_0} = \dfrac{1}{4} \in_0 E_0^2 + \dfrac{1}{4}\in_0 E^2_0=\dfrac{1}{4}\in_0E_0^2=\dfrac{1}{2}\dfrac{B^2_0}{\mu_0}$$

Time average intensity of the wave

$$I_{av} = u_{av}c=\dfrac{1}{2} \in E_0^2c=\dfrac{1}{2}\in_0cE_0^2$$
Hence, the correct option is $$(C)$$
Question 3
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Suppose that the electric field amplitude of an electromagnetic wave propagating along x-direction is $${ E }_{ 0}$$ = 120 N $${ C }^{ -1 }$$ and that its frequency is $$\upsilon$$ = 50.0 MHz.

A
The expression of electric field is $${ \overrightarrow { E } \ =\ 120\sin { (\dfrac { \pi }{ 3 } } x\ -\ 100\pi \ \times \ { 10 }^{ 6 } }t)\hat{ j }$$

B
The expression of electric field is $${ \overrightarrow { E } \ =\ 60\sin { (\dfrac { \pi }{ 3 } } x\ -\ 100\pi \ \times \ { 10 }^{ 6 } }t)\hat{ j }$$

C
The expression of magnetic field is $${ \overrightarrow { B } \ = 40 \times { 10 }^{ -8 }\sin { (\dfrac { \pi }{ 3 } } x\ -\ \pi \ \times \ { 10 }^{ 8 } }t)\hat{ k }$$

D
Both (a) and (c)

Solution

$$E_0=120\,NC$$

$$\omega=2\pi f$$
$$\omega=2\pi\times 50\times 10^{6}=100\pi\times 10^{6}$$

$$\cfrac{\omega}{k}=c$$
$$\Rightarrow \cfrac{100\pi\times 10^{6}}{k}=3\times 10^{8}$$
$$\Rightarrow k=\cfrac{\pi}{3}$$

The standard equaton for electric field is:
$$E=E_0\ sin(kx-\omega t)$$
$$\implies E=120\sin (\cfrac{\pi}{3}x-100\pi\times 10^{6}t)\hat j$$

The magnetic field intensity is given by:
$$B_0=\dfrac{E_0}{c}$$

$$\Rightarrow B_0=\dfrac{120}{3\times10^8}=40\times 10^{-8}\ T$$

The standard equaton for magnetic field is:
$$B=B_0\ sin(kx-\omega t)$$
$$\implies B=40\times 10^{-8}\sin(\cfrac{\pi}{3}x-\pi\times 10^{8}t)\hat k$$

Option $$(D)$$ is correct.
Question 4
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A plane e.m wave of frequency $$30MHz$$ travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time $$E=6V/m$$ along $$y$$-direction. Its magnetic field component $$B$$ at this point would be

A
$$2\times { 10 }^{ -8 }T$$ along $$z$$-direction

B
$$6\times { 10 }^{ -8 }T$$ along $$x$$-direction

C
$$2\times { 10 }^{ -8 }T$$ along $$y$$-direction

D
$$6\times { 10 }^{ -8 }T$$ along $$z$$-direction

Solution

Given data
frequency$$ = 30 MHz$$
Electric field $$\varepsilon=\cfrac { 6v }{ m } \hat { J } $$ (along + y axis)
Now we know that
$$c=\cfrac { { \varepsilon }_{ 0 } }{ { B }_{ 0 } } $$
c = speed of EM wave$$=3\times { 10 }^{ 8 }m/s$$
$${ \varepsilon }_{ 0 }=$$peak value of electric field
$${ B }_{ 0 }=$$peak value of magnetic field
$$c=\cfrac { \varepsilon }{ B } $$
B = magnetic field
$$3\times { 10 }^{ 8 }=\cfrac { 6 }{ B } $$
$$B=\cfrac { 6 }{ 3\times { 10 }^{ 8 } } $$ $$B=2\times { 10 }^{ -8 }$$ Tesla
Now we have to find direction of magnetic field
We know that wave propagates in the direction perpendicular to oscillating electric field and magnetic field.
$$c=\vec { \varepsilon } \times \vec { B } $$ (only for direction)
c = along x axis
$$\varepsilon =$$along y axis
So B should be along z axis
$$B=2\times { 10 }^{ -8 }T$$ along z axis
Question 5
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The electric field part of an electromagnetic wave in vacuum is
$$E=3.1\cos { \left[ \left( 1.8\dfrac { rad }{ m } \right) y+\left( 5.4\times { 10 }^{ 8 }\dfrac { rad }{ s } \right) t \right] \hat { i } } $$.The frequency corresponding to the given part of the electromagnetic wave is?

A
$$5.4 10^{8} Hz$$

B
$$8.6 10^{7} Hz$$

C
$$3.2 10^{8} Hz$$

D
$$4.8 10^{7} Hz$$

Solution

Given equation
$$E=3.1\cos { \left[ \left( 1.8\dfrac { rad }{ m } \right) y+\left( 5.4\times { 10 }^{ 8 }\dfrac { rad }{ s } \right) t \right] \hat { i } } $$...........(i)

Standard equation at electromagnetic wave in terms of electric field,

$$E=E_0\cos(ky+wt)$$.........(ii)

If the wave is moving in $$-y$$ direction

Wave length, $$\lambda=\dfrac{2\pi}{Coefficient\,of\,y}$$

$$=\dfrac{2\pi}{1.8}=\dfrac{2\times 22/7}{1.8}$$

$$\lambda=\dfrac{44}{1.8\times 7}=3.49=3.5m$$

velocity of wave, $$v=\dfrac{\text{coefficient of l}}{\text{coefficient of y}}$$

$$v=\dfrac{5.4\times 10^8}{1.8}=3\times 10^8$$

frequency, $$f=\dfrac{v}{\lambda}=\dfrac{3\times 10^8}{3.5}=8.6\times 10^7$$ Hz
Hence, the correct option is $$(B)$$
Question 6
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In an electromagnetic wave, the amplitude of electric and magnetizing fields are 100 V/m and 0.265 A/m/ The maximum energy flow is -

A
$$26.5\ W/{ m }^{ 2 }$$

B
$$36.5\ W/{ m }^{ 2 }$$

C
$$46.7\ W/{ m }^{ 2 }$$

D
$$765\ W/{ m }^{ 2 }$$

Solution
Question 7
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Potential difference across plates of a capacitor $$6\mu F$$ is changing at the rate of $$72V{ s }^{ -1 }$$. Displacement current at that instant will be

A
$$1.2A$$

B
12 x $${ 10 }^{ -6 }$$A

C
7.2 x $${ 10 }^{ -6 }$$A

D
none

Solution

Given,
$$C=\dfrac{A\varepsilon_0}{d}=6\mu C$$
$$\dfrac{dV}{dt}=72V/s$$
Displacement current, $$I_D=\varepsilon_0 \dfrac{d\phi_E}{dt}$$
$$I_D=\varepsilon_0\dfrac{d(EA)}{dt}$$

$$I_D=\varepsilon_0 A\dfrac{d(V/d)}{dt}$$

$$I_D=\dfrac{\varepsilon_0 A}{d}.\dfrac{dV}{dt}$$

$$I_D=C\dfrac{dV}{dt}$$

$$I_D=6\times 10^{-6}\times 72$$

$$I_D=432\times 10^{-6}A$$

The correct option is D.
Question 8
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A plane electromagnetic wave with an intensity of $$200 W/m^2$$ is incident normal to a flat plate of radius 30 cm. If the plate absorbs $$60\%$$ and reflects $$40\%$$ of the incident radiation, what is the momentum transferred to it in 5 min?

A
$$1.1 \times 10^{-4} kg ms^{-1}$$

B
$$2.7 \times 10^{-4} kg ms^{-1}$$

C
$$3.7 \times 10^{-4} kg ms^{-1}$$

D
$$3.7 \times 10^{-3} kg ms^{-1}$$

Solution
Question 9
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The magnetic field in a plane electromagnetic wave is given by$$B=2.01\times10^{-7} \sin(6.28\times10^2x+2.2\times10^{10}t)T$$. [where x in cm and t in second]. The wavelength of the given wave is

A
1 cm

B
628 cm

C
1.29 cm

D
314 cm

Solution
Question 10
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A parallel late condenser consists of two circular plate each of radius $$2cm$$ separated by a distance of $$0.1mm$$. A time varying potential difference of $$5\times 10^{13}v/s$$ is applied across the plated of the condenser: The displacement current is

A
$$550A$$

B
$$5.56 \times 10^2A$$

C
$$5.56 \times 10^3A$$

D
$$2.28 \times 10^4A$$

Solution

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Electromagnetic Waves Test - 34

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Electromagnetic Waves Test - 34

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Question 1

$$2.5 \, V \, m^1, \, 3.6 \, \times \, 10^{-8} T$$

$$4.2 \, V \, m^1, \, 2.8 \, \times \, 10^{-8} T$$

$$4.08 \, V \, m^1, \, 1.36 \, \times \, 10^{-8} T$$

$$3.6 \, V \, m^1, \, 3.6 \, \times \, 10^{-8} T$$

Solution

Question 2

The average energy density of the wave is given by $$\displaystyle { u }_{ av }\ =\ \dfrac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }\ +\ \dfrac { 1 }{ 4 } \dfrac { { B }_{ 0 }^{ 2 } }{ { \mu }_{ 0 } } .$$

The time averaged intensity of the wave is given by $$ { I }_{ av }\ =\dfrac{1}{2}\in_0cE_0^2$$

Both (a) and (b)

None of these

Solution

Question 3

The expression of electric field is $${ \overrightarrow { E } \ =\ 120\sin { (\dfrac { \pi }{ 3 } } x\ -\ 100\pi \ \times \ { 10 }^{ 6 } }t)\hat{ j }$$

The expression of electric field is $${ \overrightarrow { E } \ =\ 60\sin { (\dfrac { \pi }{ 3 } } x\ -\ 100\pi \ \times \ { 10 }^{ 6 } }t)\hat{ j }$$

The expression of magnetic field is $${ \overrightarrow { B } \ = 40 \times { 10 }^{ -8 }\sin { (\dfrac { \pi }{ 3 } } x\ -\ \pi \ \times \ { 10 }^{ 8 } }t)\hat{ k }$$

Both (a) and (c)

Solution

Question 4

$$2\times { 10 }^{ -8 }T$$ along $$z$$-direction

$$6\times { 10 }^{ -8 }T$$ along $$x$$-direction

$$2\times { 10 }^{ -8 }T$$ along $$y$$-direction

$$6\times { 10 }^{ -8 }T$$ along $$z$$-direction

Solution

Question 5

$$5.4 10^{8} Hz$$

$$8.6 10^{7} Hz$$

$$3.2 10^{8} Hz$$

$$4.8 10^{7} Hz$$

Solution

Question 6

$$26.5\ W/{ m }^{ 2 }$$

$$36.5\ W/{ m }^{ 2 }$$

$$46.7\ W/{ m }^{ 2 }$$

$$765\ W/{ m }^{ 2 }$$

Solution

Question 7

$$1.2A$$

12 x $${ 10 }^{ -6 }$$A

7.2 x $${ 10 }^{ -6 }$$A

none

Solution

Question 8

$$1.1 \times 10^{-4} kg ms^{-1}$$

$$2.7 \times 10^{-4} kg ms^{-1}$$

$$3.7 \times 10^{-4} kg ms^{-1}$$

$$3.7 \times 10^{-3} kg ms^{-1}$$

Solution

Question 9

1 cm

628 cm

1.29 cm

314 cm

Solution

Question 10

$$550A$$

$$5.56 \times 10^2A$$

$$5.56 \times 10^3A$$

$$2.28 \times 10^4A$$

Solution

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