Electromagnetic Waves Test

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Electromagnetic Waves Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

A light of wavelength $$6000 A^0$$ in air, enters a medium with refractive index 1.5. What will be the wavelength of light in the medium?

A
$$4000A^0$$

B
$$6000A^0$$

C
$$9000A^0$$

D
None of the above

Solution

Here, wavelength of light in air,
$$\lambda=6000 A^0=6\times 10^{-7}m;$$
Refractive index of the medium, $$\mu=1.5$$
The frequency of the light does not change, when light travels from air to a refracting medium.
$$\therefore v'=v=\dfrac {c}{\lambda}=\dfrac {3\times 10^8}{6\times 10^{-7}}=5\times 10^{14}S^{-1}$$
The wavelength of light in the medium,
$$\lambda'=\dfrac {\lambda}{\mu}=\dfrac {6000}{1.5}=4000 A^0$$
Question 2
1 / -0

A pulse of light of duration $$100\ ns$$ is absorbed completely by a small object initially at rest. Power of the pulse is $$30\ mW$$ and the speed of light $$3\times 10^8 m/s.$$ The final momentum of the object is

A
$$0.3\times 10^{-17} kg\ ms^{-1}$$

B
$$1.0\times 10^{-17} kg\ ms^{-1}$$

C
$$3\times 10^{-17} kg\ ms^{-1}$$

D
$$9.0\times 10^{-17} kg ms^{-1}$$

Solution

$$answer:-$$ B option
using Einstein mass energy equation
$$E=mc^2$$
$$Energy(E)=power\times time$$
$$E=30\times { 10 }^{ -3 }\times 100\times { 10 }^{ -9 }=m{ c }^{ 2 }=m\times { (3\times { 10 }^{ 8 }) }^{ 2 }$$
$$ m=\dfrac { 1 }{ 3 } \times { 10 }^{ -25 }$$
using
$$p=mv$$
$$p=\dfrac { 1 }{ 3 } \times { 10 }^{ -25 }\times 3\times { 10 }^{ 8 }={ 10 }^{ -17 } kgm/s$$
Question 3
1 / -0

Light of intensity$$ = 3 W/m^{2}$$ is incident on a perfectly absorbing metal surface of area $$1 m^{2}$$ making an angle of $$60^0$$ with the normal. If the force exerted by the photons on the surface is $$p \times 10^{-9}$$ (in Newton), then the value of p is :

A
5

B
2

C
1

D
3

Solution

We will be calculating everything per unit time.

Effective area for the light on the surface is $$ = 1 \times cos 60^{0}$$
Energy falling on the surface$$=$$ intensity x Effective area $$ = 3 \times 1 \times cos 60^{0} = \dfrac{3}{2} watt$$
Momentum carried by the light per sec $$ = 3/(2c) = 5\times 10^{-9}$$
So, $$p= 5\times 10^{-9}$$
Question 4
1 / -0

If $$\vec{E}$$ and $$\vec{B}$$ are the electric and magnetic field vectors of electromagnetic waves, then the direction of propagation of the electromagnetic wave is along the direction of

A
$$\vec{E}$$

B
$$\vec{B}$$

C
$$\vec{E} \times \vec{B}$$

D
None of these

Solution

we know that;
The direction of propagation of the electromagnetic wave is perpendicular to the plane of oscillation of electric and magnetic field of the EM wave.

Thus,
If, $$\overrightarrow E$$ and $$\overrightarrow B$$ are electric and magnetic field vectors of the EM wave, the direction of its propagation will be given by $$\overrightarrow E \times \overrightarrow B$$

As in an EM wave, the vectors $$\overrightarrow E$$ and $$\overrightarrow B$$ are perpendicular to each other, the direction of the wave propagation will be perpendicular to the plane containing the electric and magnetic field vectors.
Question 5
1 / -0

Directions For Questions

A parallel plate capacitor made of circular plates each of radius $$R=6\ cm$$ has capacitance $$C=100\ pF$$. The capacitor is connected to a $$230\ V$$ AC supply with an angular frequency of $$300\ rad/s$$.

...view full instructions

The displacement current will be

A
$$6.9\mu A$$

B
$$9.6\mu A$$

C
$$6.3\mu A$$

D
$$5.7\mu A$$

Solution

Capacitance, $$C=100\times {10}^{-12}F={10}^{-10}F$$
Voltage, $$V=230V$$
Angular frequency, $$\omega=300rad/s$$

Since we know that,
$$I_D = \epsilon_o \dfrac{d( \phi _E)}{dt} = \epsilon_o \dfrac{d(EA)}{dt} $$ ( $$\because \phi_E = EA$$ where $$\phi_E$$ isthe electromagnetic flux and $$E$$ and $$A$$ are electric field and area respectively)
$$\Rightarrow I_D = \epsilon_o A \dfrac{dE}{dt}$$
$$\Rightarrow I_D = \epsilon_o A \dfrac{d}{dt} \left(\dfrac{Q}{ \epsilon_o A} \right)$$ $$\left( \because E = \dfrac{ \sigma}{ \epsilon_o} = \dfrac{Q}{ \epsilon_o A} \right)$$
$$\Rightarrow I_D = \epsilon_o A \times \dfrac{1}{\epsilon_o A} \dfrac{dQ}{dt} = \dfrac{dQ}{dt} = I_{conduction \ current}$$

Now, Conduction current is given by, $$I=\dfrac{V}{{X}_{c}}$$
Capacitive reactance is given by, $${X}_{c}=\dfrac{1}{\omega C}$$
Current, $$I=V\omega C=230\times300\times{10}^{-10}$$
$$I=6.9\times {10}^{-6}A=6.9\mu A$$

Displacement current is equal to conduction current here. Hence, $$I_D = 6.9 \mu A$$
Question 6
1 / -0

The incident intensity on a horizontal surface at sea level from the sun is about $$1 kW m^{-2}$$. Assuming that 50 per cent of this intensity is reflected and 50 per cent is absorbed, determine the radiation pressure on this horizontal surface (in pascals).

A
$$8.2\times 10^{-2}$$

B
$$5\times 10^{-6}$$

C
$$3\times 10^{-5}$$

D
$$6\times 10^{-5}$$

Solution

Given: $$I = 1000$$ $$Wm^{-2}$$

As 50 percent of light is reflected, thus $$e=0.5$$

Radiation pressure, $$P = \dfrac{(1+e)I}{c}$$

$$\therefore$$ $$P = \dfrac{(1+0.5)\times 1000}{3\times 10^8} = 5\times 10^{-6}$$ Pa
Question 7
1 / -0

A parallel beam of light is incident normally on a plane surface absorbing 40 % of the light and reflecting the rest. If the incident beam carries 60 watt of power, the force exerted by it on the surface is:

A
$$3.2 \times 10^{-8}\, N$$

B
$$3.2 \times 10^{-7}\, N$$

C
$$5.2 \times 10^{-7}\, N$$

D
$$5.12 \times 10^{-8}\, N$$

Solution

momentum of incident ray per second $$P_1=\dfrac{E}{c}=\dfrac{60}{3\times 10^8}=2\times 10^{-7} N$$
momentum of reflected ray per second $$P_2=0.6\times\dfrac{E}{c}=\dfrac{0.6\times 60}{3\times 10^8}=1.2\times 10^{-7} N$$
now rate of change of momentum $$F=P_2-(-P1)=3.2\times 10^{-7} N$$
Question 8
1 / -0

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 wavelength of this part of electromagnetic wave is

A
$$1.5 m$$

B
$$2 m$$

C
$$2.5 m$$

D
$$3.5 m$$

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

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

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

Option - $$D$$
Question 9
1 / -0

Sea water at frequency $$\nu \ =\ 4\ x\ { 10 }^{ 8 }$$ Hz has permittivity $$\varepsilon \ \approx \ 80\ { \varepsilon }_{ 0 }$$, permeability $$\mu \ \approx \ { \mu }_{ 0 }$$ and resistivity $$\rho \ =\ 0.25\ \Omega m$$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t) = $${ V }_{ 0 }\ \sin { \ (2\pi \nu t) }$$. The of amplitude of the displacement current density to the conduction current density is

A
$$\dfrac { 2 }{ 3 }$$

B
$$\dfrac { 4 }{ 9 }$$

C
$$\dfrac { 9 }{ 4 }$$

D
2

Solution

Suppose distance between the parallel plates is $$D$$ and applied voltage $$V_{(t)} = V_02\pi vt$$.
thus electric field
$$E = \dfrac{V_0}{d} \sin (2\pi vt)$$
Now using Ohm's law
$$J_c = \dfrac{1}{\phi} \dfrac{V_0}{d}\sin (2\pi vt)$$

$$\dfrac{V_0}{\phi d}\sin (2 \pi vt) = J_0^c \sin 2 \pi vt$$

Here $$J_0^c = \dfrac{V_0}{pd}$$
Now the displacement current density is given as
$$Jd = \in \dfrac{\delta E}{dt} =\dfrac{\in \delta}{dt}$$ $$\left[\dfrac{V_0}{dt} \sin (2\pi vt)\right]$$

$$= \dfrac{\in 2\pi v V_0}{d} \cos (2\pi vt)$$

$$\Rightarrow = J^d_0 \cos (2\pi vt)$$

Where $$J_0^d = \dfrac{2\pi V\in V_0}{d}$$

$$\Rightarrow \dfrac{J^d_0}{J^c_0} = \dfrac{2\pi v \in V_0}{d}. \dfrac{pd}{V_0} = 2\pi v \in \rho$$

$$= 2\pi \times 80\in_0v\times 0.25 = 4\pi \in_0v \times 10$$

$$= \dfrac{10v}{9\times 10^9} = \dfrac{4}{9}$$
Question 10
1 / -0

Choose the correct answer from the alternatives given.
A parallel plate capacitor with plate area $$A$$ and separation between the plates $$d$$ is charged by a constant current $$I$$. Consider a plane surface of area $$\dfrac{A}{2}$$ parallel to the plate and drawn between the plates. The displacement current through the area is :

A
I

B
$$\dfrac{I}{2}$$

C
$$\dfrac{I}{4}$$

D
$$\dfrac{I}{8}$$

Solution

Charge on capacitor plates at time $$t$$ is , $$q = It$$.

Electric field between the plates at this instant is
$$E = \dfrac{Q}{A\varepsilon _0}\, =\, \dfrac{It}{A\varepsilon _0} $$

The Electric flux through the given area is:
$$\phi=A.E$$

The Electric flux through the area $$\dfrac{A}{2}$$ is:
$$\phi _E \, = \, \left(\dfrac{A}{2}\right)E$$

$$\phi_E=\dfrac{A}{2}\dfrac{It}{A\varepsilon _0}=\dfrac{It}{2\varepsilon_0}$$

The displacement current through the area is given as:
$$I_D \, = \, \varepsilon_0 \dfrac{d \phi _E}{dt}$$

$$\,\,\,\implies\varepsilon_0 \dfrac{d}{dt} \left(\dfrac{It}{2 \varepsilon_0} \right) = \dfrac{I}{2}$$

Option $$(B)$$ is correct.

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Electromagnetic Waves Test - 33

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Electromagnetic Waves Test - 33

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Question 1

$$4000A^0$$

$$6000A^0$$

$$9000A^0$$

None of the above

Solution

Question 2

$$0.3\times 10^{-17} kg\ ms^{-1}$$

$$1.0\times 10^{-17} kg\ ms^{-1}$$

$$3\times 10^{-17} kg\ ms^{-1}$$

$$9.0\times 10^{-17} kg ms^{-1}$$

Solution

Question 3

5

2

1

3

Solution

Question 4

$$\vec{E}$$

$$\vec{B}$$

$$\vec{E} \times \vec{B}$$

None of these

Solution

Question 5

$$6.9\mu A$$

$$9.6\mu A$$

$$6.3\mu A$$

$$5.7\mu A$$

Solution

Question 6

$$8.2\times 10^{-2}$$

$$5\times 10^{-6}$$

$$3\times 10^{-5}$$

$$6\times 10^{-5}$$

Solution

Question 7

$$3.2 \times 10^{-8}\, N$$

$$3.2 \times 10^{-7}\, N$$

$$5.2 \times 10^{-7}\, N$$

$$5.12 \times 10^{-8}\, N$$

Solution

Question 8

$$1.5 m$$

$$2 m$$

$$2.5 m$$

$$3.5 m$$

Solution

Question 9

$$\dfrac { 2 }{ 3 }$$

$$\dfrac { 4 }{ 9 }$$

$$\dfrac { 9 }{ 4 }$$

2

Solution

Question 10

I

$$\dfrac{I}{2}$$

$$\dfrac{I}{4}$$

$$\dfrac{I}{8}$$

Solution

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