Electromagnetic Waves Test

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Electromagnetic Waves Test - 13

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Question 1
1 / -0

Magnetic field in a plane electromagnetic wave is given by
$$ \overrightarrow{B} = B_0 sin (kx + \omega t) \hat{j} T $$
Expression for corresponding electric field will be :
Where c is speed of light.

A
$$ \overrightarrow {E} = B_0 c\sin (kx + \omega t) \hat{k} V/m $$

B
$$ \overrightarrow{E} = \dfrac{B_0}{c} \sin (kx + \omega t) \hat{k} V/m $$

C
$$ \overrightarrow {E} = - B_0 c \sin (kx + \omega t) \hat{k} V/m $$

D
$$ \overrightarrow{E} = B_0 c \sin (kx - \omega t) \hat{k} V/m $$

Solution

$$\dfrac{\overrightarrow{E}}{\overrightarrow{B}}=C$$
$$\overrightarrow{B} = B_0\sin(kx + \omega t)$$
$$\overrightarrow {E} = B_0C\sin(kx+\omega t)\hat k$$ $$V/m$$
Question 2
1 / -0

Consider an electromagnetic wave propagating in vacuum. Choose the correct statement :

A
For an electromagnetic wave propagating in +y direction the 1 A electric field is $$\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (x, t) \hat{z}$$ and the magnetic field is $$\overset{\rightarrow}{B} =\frac{1}{\sqrt{2}} B_z (x, t) \hat{y}$$

B
For an electromagnetic wave propagating in +y direction the electric field is $$\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (x, t) \hat{y}$$ and the magnetic field is $$\overset{\rightarrow}{B} =\frac{1}{\sqrt{2}} B_z (x, t) \hat{z}$$

C
For an electromagnetic wave propagating in +x direction the electric field is $$\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (y, z, t) (\hat{y}+\hat{z})$$ and the magnetic field is $$\overset{\rightarrow}{B} = \frac{1}{\sqrt{2}}E_{yz} (y, z, t) (\hat{y}+\hat{z})$$

D
For an electromagnetic wave propagating in +x direction the electric field is $$\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}-\hat{z})$$ and the magnetic field is $$\overset{\rightarrow}{B} = \frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}+\hat{z})$$

Solution

Electromagnetic waves travel in the direction perpendicular to electric as well as magnetic field. Cross product of electric and magnetic field should give the direction of electromagnetic wave.
$$ \overset{\rightarrow}{ E } \times \overset{\rightarrow}{B} $$ = $$\frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}-\hat{z})$$$$ \times \frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}+\hat{z})$$ = $$E_{yz}^{2} (x, t)$$ $$\hat{x}$$
Question 3
1 / -0

A plane electromagnetic wave of frequency $$50MHz$$ travels in free space along the positive x-direction. At a particular point in space and time, $$\overrightarrow{E} = 6.3\hat{j}V/m$$. The corresponding magnetic field $$\overrightarrow{B}$$, at that point will be:

A
$$18.9\times 10^{-8}\hat{k}T$$

B
$$6.3\times 10^{-8}\hat{k}T$$

C
$$2.1\times 10^{-8}\hat{k}T$$

D
$$18.9\times 10^{8}\hat{k}T$$

Solution

$$|B| = \dfrac{|E|}{[C} = \dfrac{6.3}{3\times 10^8} = 2.1 \times 10^{-8}T$$
and $$\hat{E} \times \hat{B} = \hat{C}$$
$$\hat{j}\times {B} = \hat{i}$$
$$\hat{B} = \hat{k}$$
$$\overrightarrow{B} = |B| \hat{B} = 2.1\times 10^{-8}\hat{k}T$$
Question 4
1 / -0

A plane electromagnetic wave of wavelength $$\lambda$$ has an intensity $$I$$. It is propagating along the positive $$Y-direction$$. The allowed expressions for the electric and magnetic fields are given by

A
$$\vec {E} = \sqrt {\dfrac {I}{\epsilon_{0}C}}\cos \left [\dfrac {2\pi}{\lambda} (y - ct)\right ]\hat {i}; \vec {B} = \dfrac {1}{c} E \hat {k}$$

B
$$\vec {E} = \sqrt {\dfrac {I}{\epsilon_{0}C}}\cos \left [\dfrac {2\pi}{\lambda} (y - ct)\right ]\hat {k}; \vec {B} = -\dfrac {1}{c} E \hat {i}$$

C
$$\vec {E} = \sqrt {\dfrac {2I}{\epsilon_{0}C}}\cos \left [\dfrac {2\pi}{\lambda} (y - ct)\right ]\hat {k}; \vec {B} = +\dfrac {1}{c} E \hat {i}$$

D
$$\vec {E} = \sqrt {\dfrac {2I}{\epsilon_{0}C}}\cos \left [\dfrac {2\pi}{\lambda} (y + ct)\right ]\hat {k}; \vec {B} = \dfrac {1}{c} E \hat {i}$$

Solution

E is the electric field vector, and B is the magnetic field vector of the EM wave. For electromagnetic waves E and B are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of E x B.
So, if the wave propagates in the +Y direction then the direction of E and B should be in +X and +Z or vice versa i.e +Z and +Xrespectively.

$$Case 1.$$
Let us suppose $$\vec E$$ is in $$\hat i $$ and $$ \vec B $$ is in $$\hat k$$

Then $$\vec E \times \vec B$$ will be in $$- \hat j $$

Not Possible.

$$Case 2. $$
Let us suppose $$\vec E$$ is in $$\hat k $$ and $$ \vec B $$ is in $$\hat i$$

Then $$\vec E \times \vec B$$ will be in $$ \hat j $$

This is satisfying option (3)

as the electric and magnetic field also propagate in positive y direction with time so $$(y - ct)$$ should be there in wave equation.

Also $$I = \dfrac{c \epsilon_o}{2} |E_o|^2$$

$$|E_o| = \sqrt {\dfrac{2I}{c \epsilon_o}}$$

From these, we can say that option (c) would be the best option.
Question 5
1 / -0

A monochromatic beam of light has a frequency $$v=\cfrac { 3 }{ 2\pi } \times { 10 }^{ 12 }Hz$$ and is propagating along the direction $$\cfrac { \hat { i } +\hat { j } }{ \sqrt { 2 } } $$. It is polarized along the $$\hat { k } $$ direction. The acceptable form for the magnetic field is:

A
$$\hat { k } \cfrac { { E }_{ 0 } }{ C } \left( \cfrac { \hat { i } -\hat { j } }{ \sqrt { 2 } } \right) \cos { \left[ { 10 }^{ 4 }\left( \cfrac { \hat { i } -\hat { j } }{ \sqrt { 2 } } \right) .\vec { r } -\left( 3\times { 10 }^{ 12 } \right) t \right] } $$

B
$$\cfrac { { E }_{ 0 } }{ C } \left( \cfrac { \hat { i } -\hat { j } }{ \sqrt { 2 } } \right) \cos { \left[ { 10 }^{ 4 }\left( \cfrac { \hat { i } +\hat { j } }{ \sqrt { 2 } } \right) .\vec { r } -\left( 3\times { 10 }^{ 12 } \right) t \right] } $$

C
$$\cfrac { { E }_{ 0 } }{ C } \hat { k } \cos { \left[ { 10 }^{ 4 }\left( \cfrac { \hat { i } +\hat { j } }{ \sqrt { 2 } } \right) .\vec { r } +\left( 3\times { 10 }^{ 12 } \right) t \right] } $$

D
$$\cfrac { { E }_{ 0 } }{ C } \cfrac { \left( \hat { i } +\hat { j } +\hat { k } \right) }{ \sqrt { 3 } } \cos { \left[ { 10 }^{ 4 }\left( \cfrac { \hat { i } +\hat { j } }{ \sqrt { 2 } } \right) .\vec { r } +\left( 3\times { 10 }^{ 12 } \right) t \right] } $$

Solution

Direction of B is ,
$$= \hat{K} \times \hat{E}$$
$$= (\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}) \times \hat{K}$$
$$=\dfrac {\hat{i}+ \hat{k}}{\sqrt {2}} + \dfrac{\hat{j} \times \hat{k}}{\sqrt{2}}$$
$$=\dfrac{\hat{j}}{\sqrt{2}}+ \dfrac{\hat{i}}{\sqrt{2}} = \dfrac {\hat {i} - \hat {j}}{\sqrt{2}}$$
So, ans is between (i) and (ii)
Propagation direction , $$\hat{k} = \dfrac {\hat{i} + \hat {j}}{\sqrt{2}}$$
$$\bar { B } $$ wave $$eq^n$$ will be
$$\Rightarrow \dfrac {E_0}{c} (\hat{B}) $$ $$ Cos [|k| \hat{k} - wt]$$
$$\downarrow$$ $$\downarrow$$
we know it is we know it is
$$\dfrac{\hat{i}-\hat{j}}{\sqrt{2}}$$ $$ \dfrac{\hat{i}+\hat{j}}{\sqrt{2}}$$
Only (i) satisfies
Hence , correct answer is (i)
Question 6
1 / -0

Light with an energy flux of $$18 W/cm^2$$ falls on a non-reflecting surface at normal incidence. The pressure exerted on the surface is:

A
$$2N/m^2$$

B
$$2\times 10^{-4}N/m^2$$

C
$$6N/m^2$$

D
$$6\times 10^{-4}N/m^2$$

Solution

Pressure exerted on the surface$$=P=\dfrac{F}{A}$$$$=\dfrac{\dfrac{dp}{dt}}{A}$$
Momentum $$p=\dfrac{E}{c}$$
where E is the energy of incident light, c is the speed of light.
Thus $$P=\dfrac{1}{c}(\dfrac{1}{A}\dfrac{dE}{dt})$$$$=\dfrac{Flux}{c}$$$$=\dfrac{18\times 10^{4}}{3\times 10^8}=6\times 10^{-4}N/m^2$$
Question 7
1 / -0

Radiations of intensity $$0.5\ W/m^{2}$$ are striking a metal plate. The pressure on the plate is

A
$$0.166\times 10^{-8} N/m^{2}$$

B
$$0.332\times 10^{-8}N/m^{2}$$

C
$$0.111\times 10^{-8}N/m^{2}$$

D
$$0.083\times 10^{-8}N/m^{2}$$

Solution

Intensity or power per unit area of the radiations,
$$P = pv$$
$$\Rightarrow p = \dfrac {P}{v} = \dfrac {0.5}{3\times 10^{8}} = .166\times 10^{-8} N/m$$
Question 8
1 / -0

Displacement current is continuous:

A
when electric field is changing in the circuit

B
when magnetic field is changing in the circuit

C
in both types of fields

D
through wires and resistance only

Solution

Displacement current is a quantity appearing in Maxwell's equations that is defined in terms of the rate of change of electric displacement field.
When electric field is changing with time continuously, the displacement current is constant.
Question 9
1 / -0

The displacement current flows in the dielectric of a capacitor when the potential difference across its plates

A
becomes zero

B
has assumed a constant value

C
is increasing with time

D
is decreasing with time

Solution

According to Maxwell's hypothesis, a displacement current will flow through a capacitor when the potential difference across its plates is varying. Thus a varying electric field will exist between the plates and this displacement current is same in magnitude to the current flowing in outer circuit. When a D.C voltage applied across its plates, constant voltage appears across its plates and so there will be no displacement current flowing through the capacitor. Thus the displacement current will flow when the potential is increasing with time.
Question 10
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Choose the correct answer from the alternatives given.
The conduction current is the same as displacement current when the source is

A
AC only

B
DC only

C
both AC and DC

D
neither AC nor for DC

Solution

For a capacitor, we have:
$$Q=CV$$
If $$Q$$ is changing, there will be a current in capacitor plates,
$$I=dQ/dt=CdV/dt$$
when voltage across the capacitor is constant, $$dV/dt=0$$
therefore, $$I=0$$
It implies that, for a DC (constant ) voltage, the capacitor current is zero. Hence, for a DC source the conduction current and displacement current (capacitor current) are not same.
Whereas, by Maxwell's equation for a time varying voltage (AC voltage), both conduction and displacement currents are same.

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Electromagnetic Waves Test - 13

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Electromagnetic Waves Test - 13

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Question 1

$$ \overrightarrow {E} = B_0 c\sin (kx + \omega t) \hat{k} V/m $$

$$ \overrightarrow{E} = \dfrac{B_0}{c} \sin (kx + \omega t) \hat{k} V/m $$

$$ \overrightarrow {E} = - B_0 c \sin (kx + \omega t) \hat{k} V/m $$

$$ \overrightarrow{E} = B_0 c \sin (kx - \omega t) \hat{k} V/m $$

Solution

Question 2

For an electromagnetic wave propagating in +y direction the 1 A electric field is $$\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (x, t) \hat{z}$$ and the magnetic field is $$\overset{\rightarrow}{B} =\frac{1}{\sqrt{2}} B_z (x, t) \hat{y}$$

For an electromagnetic wave propagating in +y direction the electric field is $$\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (x, t) \hat{y}$$ and the magnetic field is $$\overset{\rightarrow}{B} =\frac{1}{\sqrt{2}} B_z (x, t) \hat{z}$$

For an electromagnetic wave propagating in +x direction the electric field is $$\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (y, z, t) (\hat{y}+\hat{z})$$ and the magnetic field is $$\overset{\rightarrow}{B} = \frac{1}{\sqrt{2}}E_{yz} (y, z, t) (\hat{y}+\hat{z})$$

For an electromagnetic wave propagating in +x direction the electric field is $$\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}-\hat{z})$$ and the magnetic field is $$\overset{\rightarrow}{B} = \frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}+\hat{z})$$

Solution

Question 3

$$18.9\times 10^{-8}\hat{k}T$$

$$6.3\times 10^{-8}\hat{k}T$$

$$2.1\times 10^{-8}\hat{k}T$$

$$18.9\times 10^{8}\hat{k}T$$

Solution

Question 4

$$\vec {E} = \sqrt {\dfrac {I}{\epsilon_{0}C}}\cos \left [\dfrac {2\pi}{\lambda} (y - ct)\right ]\hat {i}; \vec {B} = \dfrac {1}{c} E \hat {k}$$

$$\vec {E} = \sqrt {\dfrac {I}{\epsilon_{0}C}}\cos \left [\dfrac {2\pi}{\lambda} (y - ct)\right ]\hat {k}; \vec {B} = -\dfrac {1}{c} E \hat {i}$$

$$\vec {E} = \sqrt {\dfrac {2I}{\epsilon_{0}C}}\cos \left [\dfrac {2\pi}{\lambda} (y - ct)\right ]\hat {k}; \vec {B} = +\dfrac {1}{c} E \hat {i}$$

$$\vec {E} = \sqrt {\dfrac {2I}{\epsilon_{0}C}}\cos \left [\dfrac {2\pi}{\lambda} (y + ct)\right ]\hat {k}; \vec {B} = \dfrac {1}{c} E \hat {i}$$

Solution

Question 5

$$\hat { k } \cfrac { { E }_{ 0 } }{ C } \left( \cfrac { \hat { i } -\hat { j } }{ \sqrt { 2 } } \right) \cos { \left[ { 10 }^{ 4 }\left( \cfrac { \hat { i } -\hat { j } }{ \sqrt { 2 } } \right) .\vec { r } -\left( 3\times { 10 }^{ 12 } \right) t \right] } $$

$$\cfrac { { E }_{ 0 } }{ C } \left( \cfrac { \hat { i } -\hat { j } }{ \sqrt { 2 } } \right) \cos { \left[ { 10 }^{ 4 }\left( \cfrac { \hat { i } +\hat { j } }{ \sqrt { 2 } } \right) .\vec { r } -\left( 3\times { 10 }^{ 12 } \right) t \right] } $$

$$\cfrac { { E }_{ 0 } }{ C } \hat { k } \cos { \left[ { 10 }^{ 4 }\left( \cfrac { \hat { i } +\hat { j } }{ \sqrt { 2 } } \right) .\vec { r } +\left( 3\times { 10 }^{ 12 } \right) t \right] } $$

$$\cfrac { { E }_{ 0 } }{ C } \cfrac { \left( \hat { i } +\hat { j } +\hat { k } \right) }{ \sqrt { 3 } } \cos { \left[ { 10 }^{ 4 }\left( \cfrac { \hat { i } +\hat { j } }{ \sqrt { 2 } } \right) .\vec { r } +\left( 3\times { 10 }^{ 12 } \right) t \right] } $$

Solution

Question 6

$$2N/m^2$$

$$2\times 10^{-4}N/m^2$$

$$6N/m^2$$

$$6\times 10^{-4}N/m^2$$

Solution

Question 7

$$0.166\times 10^{-8} N/m^{2}$$

$$0.332\times 10^{-8}N/m^{2}$$

$$0.111\times 10^{-8}N/m^{2}$$

$$0.083\times 10^{-8}N/m^{2}$$

Solution

Question 8

when electric field is changing in the circuit

when magnetic field is changing in the circuit

in both types of fields

through wires and resistance only

Solution

Question 9

becomes zero

has assumed a constant value

is increasing with time

is decreasing with time

Solution

Question 10

AC only

DC only

both AC and DC

neither AC nor for DC

Solution

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