Electromagnetic Waves Test

Result

Electromagnetic Waves Test - 23

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In electromagnetic wave, according to Maxwell, changing electric field gives.

A
Stationary magnetic field

B
Conduction current

C
Eddy current

D
Displacement current

Solution

By Maxwell
$${ I }_{ d }=\dfrac { \varepsilon dE }{ dt } $$
dE is electric field
$${ I }_{ d }$$ is displacement current per unit area.
Hence changing electric field gives displacement current.
Question 2
1 / -0

The capacity of a parallel plate air capacitor is $$2 \mu F$$ and voltage between the plates is changing at the rate of $$3 {V}/{S}$$. The displacement current in the capacitor is:

A
$$2 \mu F$$

B
$$3 \mu F$$

C
$$5 \mu F$$

D
$$6 \mu F$$

Solution

Displacement current in capacitor $$i = A\epsilon_o \dfrac{dE}{dt}$$
Or $$i = \dfrac{A\epsilon_o}{l} \dfrac{d(El)}{dt}$$ where $$l$$ is the separation between the two plates.
Potential difference between the two plates $$V = El$$
$$\implies$$ $$i = C \dfrac{dV}{dt}$$ where capacitance of the capacitor is $$C = \dfrac{A\epsilon_o }{l}$$
$$\therefore$$ $$i = 2\mu F \times 3 V/s = 6\mu A$$
Question 3
1 / -0

Unpolarized light falls first on polarizer $$\left( P \right) $$ and then on analyzer $$\left( A \right) $$. If the intensity of the transmitted light from the analyser is $$\dfrac { 1 }{ 8 }$$th of the incident unpolarized light. What will be the angle between optic axes of $$P$$ and $$A$$?

A
$${ 45 }^{ o }$$

B
$${ 30 }^{ o }$$

C
Zero

D
$${ 60 }^{ o }$$

Solution

Given,
$$I=\dfrac { { I }_{ 0 } }{ 2 } $$ ....(i)
$${ I }^{ ' }=I\cos ^{ 2 }{ \theta } $$ $$\left( \because { I }^{ ' }=\dfrac { { I }_{ 0 } }{ 8 } \right) $$
$$\therefore \dfrac { { I }_{ 0 } }{ 8 } =\dfrac { { I }_{ 0 } }{ 2 } \cos ^{ 2 }{ \theta } $$
From the equation (i), we have
$$\dfrac { 1 }{ 4 } =\cos ^{ 2 }{ \theta } \Rightarrow \cos { \theta } ={ 1 }/{ 2 }$$
$$\Rightarrow \cos { \theta } =\cos { { 60 }^{ o } } $$
$$\Rightarrow \theta ={ 60 }^{ o }$$
Question 4
1 / -0

Light is an electromagnetic wave. Its speed in vacuum is given by the expression

A
$$\displaystyle \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } $$

B
$$\displaystyle \sqrt { \frac { { \mu }_{ 0 } }{ { \varepsilon }_{ 0 } } } $$

C
$$\displaystyle \sqrt { \frac { { \varepsilon }_{ 0 } }{ { \mu }_{ 0 } } } $$

D
$$\displaystyle \frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } $$

Solution

We know that,
$$\displaystyle { \mu }_{ 0 }=4\pi \times { 10 }^{ -7 }Wb{ A }^{ -1 }{ m }^{ -1 }$$
$$\displaystyle { \varepsilon }_{ 0 }=8.85\times { 10 }^{ -12 }\frac { N-{ m }^{ 2 } }{ { C }^{ 2 } } $$
So, speed of light $$\displaystyle c=\frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } =3\times { 10 }^{ 8 }{ m }/{ s }$$
Question 5
1 / -0

The Electromagnetic wave travelling in a medium has relative permeability $$1.3$$ and relative premittivity $$2.14$$. The speed of EM wave in that medium will be.

A
$$13.6\times 10^6\ m/s$$

B
$$1.8\times 10^6\ m/s$$

C
$$3.6\times 10^8\ m/s$$

D
$$1.8\times 10^8\ m/s$$

Solution

Velocity of EM wave in vacuum is given by
$$\displaystyle c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$$
Velocity of EM wave in medium is given by
$$\displaystyle v=\frac{1}{\sqrt{\mu \varepsilon}}$$
$$\therefore \displaystyle\frac{c}{v}=\sqrt{\displaystyle\frac{\mu\varepsilon}{\mu_0\varepsilon_0}}=\sqrt{\displaystyle\frac{\mu_0\mu_r\varepsilon_0\cdot \varepsilon_r}{\mu_0\varepsilon}}$$
$$v=c/\sqrt{\mu_r\varepsilon_r}$$
$$=3\times 10^8/\sqrt{1.3\times 2.14}$$
$$=1.8\times 10^8$$m.
Question 6
1 / -0

In an electromagnetic wave, the electric magnetising fields are $$100V/m$$ and $$0.265 A/m$$. The maximum energy flow is?

A
$$26.5W/m^2$$

B
$$36.5W/m^2$$

C
$$46.7W/m^2$$

D
$$765W/m^2$$

Solution

Maximum rate of energy flow, $$S=E_0\times H_0$$
Given, $$E_0=100V/m, H_0=0.265 A/m$$
$$\therefore S=100\times 0.265=26.5W/m^2$$.
Question 7
1 / -0

Electromagnetic waves travel only through.

A
material medium

B
vacuum

C
oscillating electric and magnetic fields

D
oscillating electric and magnetic fields whose directions are perpendicular to each other

Solution

Electromagnetic waves travel through oscillating electric and magnetic fields whose directions are perpendicular to each other.
Question 8
1 / -0

The time varying electric and magentic fields are $${E}=E_x\hat{i}+E_x\hat{j}+E_z\hat{k}$$ and $$B=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$$ respectively. They generate a plane electromagnetic wave travelling towards x-direction in pair of space.

A
$$E_y, B_z$$

B
$$E_x, B_y$$

C
$$E_z, B_x$$

D
$$E_z, B_z$$

Solution

As the electromagnetic wave is travelling towards x-direction, so the electric and magnetic fields will oscillate perpendicular to x-axis. Also, $$E\perp B$$, so if B is towards y-direction B must be towards z-direction. i.e., $$E_y, B_z$$.
Question 9
1 / -0

If the magnetic field of an electromagnetic wave is given as $$B_y=2\times 10^{-7}\sin(10^3x+1.5\times 10^{12}t)$$ tesla, the wavelength of the electromagnetic wave is

A
$$0.314 mm$$

B
$$0.628 mm$$

C
$$6.28 mm$$

D
$$1.26 mm$$

E
$$0.0628 mm$$

Solution

The general equation of an electromagnetic wave is $$B=A\sin(kx+\omega t)$$
Comparing this equation with the given equation, $$A=2\times 10^{-7}$$, $$k=10^3 $$ and $$\omega=1.5\times 10^{12}$$
We know that , wave number $$k=\dfrac{2\pi}{\lambda}$$
So, $$10^3=\dfrac{2\pi}{\lambda}$$
or $$\lambda=6.28\times 10^{-3} m=6.28 \,mm$$
Question 10
1 / -0

If a source is transmitting electro-magnetic waves of frequency $$8.196\times 10^{6}Hz$$, then the wavelength of the electro-magnetic waves transmitted from the source will be

A
$$5090\ cm$$

B
$$4050\ cm$$

C
$$4230\ cm$$

D
$$3660\ cm$$

Solution

Given, frequency of $$EM$$ waves
$$v = 8.196\times 10^{6}Hz$$
velocity of $$EM$$ waves $$(v) = 3\times 10^{8} m/s$$
$$\therefore$$ Wavelength of $$EM$$ waves $$\lambda = \dfrac {v}{v}$$
$$= \dfrac {3\times 10^{8}}{8.196\times 10^{6}} = 36.60\ m$$
$$= 3660\ cm$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Electromagnetic Waves Test - 23

Result

Electromagnetic Waves Test - 23

Score

-

Rank

-

SHARING IS CARING

Question 1

Stationary magnetic field

Conduction current

Eddy current

Displacement current

Solution

Question 2

$$2 \mu F$$

$$3 \mu F$$

$$5 \mu F$$

$$6 \mu F$$

Solution

Question 3

$${ 45 }^{ o }$$

$${ 30 }^{ o }$$

Zero

$${ 60 }^{ o }$$

Solution

Question 4

$$\displaystyle \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } $$

$$\displaystyle \sqrt { \frac { { \mu }_{ 0 } }{ { \varepsilon }_{ 0 } } } $$

$$\displaystyle \sqrt { \frac { { \varepsilon }_{ 0 } }{ { \mu }_{ 0 } } } $$

$$\displaystyle \frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } $$

Solution

Question 5

$$13.6\times 10^6\ m/s$$

$$1.8\times 10^6\ m/s$$

$$3.6\times 10^8\ m/s$$

$$1.8\times 10^8\ m/s$$

Solution

Question 6

$$26.5W/m^2$$

$$36.5W/m^2$$

$$46.7W/m^2$$

$$765W/m^2$$

Solution

Question 7

material medium

vacuum

oscillating electric and magnetic fields

oscillating electric and magnetic fields whose directions are perpendicular to each other

Solution

Question 8

$$E_y, B_z$$

$$E_x, B_y$$

$$E_z, B_x$$

$$E_z, B_z$$

Solution

Question 9

$$0.314 mm$$

$$0.628 mm$$

$$6.28 mm$$

$$1.26 mm$$

$$0.0628 mm$$

Solution

Question 10

$$5090\ cm$$

$$4050\ cm$$

$$4230\ cm$$

$$3660\ cm$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.