Electromagnetic Waves — Page 9 | JEE Physics

Q81

Match List I with List II : .tg .tg List I List II A. Microwaves I. Radio active decay of the nucleus B. Gamma rays II. Rapid acceleration and deceleration of electron in aerials C. Radio waves III. Inner shell electrons D. X-rays IV. Klystron valve Choose the correct answer from the options given below :

A A-I, B-III, C-IV, D-II

B A-IV, B-III, C-II, D-I

C A-IV, B-I, C-II, D-III

D A-I, B-II, C-III, D-IV

Correct Answer

Option C

Solution

1.

Klystron valve used to produce Microwave 2.

Gamma ray $\rightarrow$ Radioactive decay 3.

Radio wave $\rightarrow$ Rapid acceleration and deacceleration of electrons in aerials 4.

X-ray $\rightarrow$ Inner shell electrons

Q82

Match List I with List II .tg .tg List I List II A. Gauss's Law in Electrostatics I.

\oint {\overrightarrow E \,.\,d\overrightarrow l = - {{d{\phi _B}} \over {dt}}}

B. Faraday's Law II.

\oint {\overrightarrow B \,.\,d\overrightarrow A = 0}

C. Gauss's Law in Magnetism III.

\oint {\overrightarrow B \,.\,d\overrightarrow l = {\mu _0}{i_c} + {\mu _0}{ \in _0}{{d{\phi _E}} \over {dt}}}

D. Ampere-Maxwell Law IV.

\oint {\overrightarrow E \,.\,d\overrightarrow s = {q \over {{ \in _0}}}}

Choose the correct answer from the options given below :

A A-I, B-II, C-III, D-IV

B A-III, B-IV, C-I, D-II

C A-IV, B-I, C-II, D-III

D A-II, B-III, C-IV, D-I

Correct Answer

Option C

Solution

Gauss's law $\oint \vec{E} \cdot \overrightarrow{d s}=\dfrac{q}{\epsilon_{0}} \quad(\mathrm{~A} \rightarrow \mathrm{IV})$ Faraday's law $\oint \vec{E} \cdot \overrightarrow{d l}=-\dfrac{d \phi_{B}}{d t} \quad(\mathrm{~B} \rightarrow \mathrm{I})$ Gauss's law in magnetism $\oint \vec{B} \cdot \overrightarrow{d A}=0 \quad(\mathrm{C} \rightarrow \mathrm{II})$ Ampere's-Maxwell law $\oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} i_{c}+\mu_{0} \in_{0} \dfrac{d \phi_{E}}{d t}$

\text{ (D } \rightarrow \text{ III) }

Q83

Match with of Electromagnetic waves with corresponding wavelength range : $to$ 1 \mathrm{~nm}$

List - I		List - II
(B)	Ultraviolet	(II)	$1 \mathrm{~nm}$ to $10^{-3} \mathrm{~nm}$
(C)	X-Ray	(III)	$1 \mathrm{~mm}$ to $700 \mathrm{~nm}$
(D)	Infra-red	(IV)	$0.1 \mathrm{~m}$ to $1 \mathrm{~mm}$

A (A)-(I), (B)-(IV), (C)-(II), (D)-(III)

B (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

C (A)-(IV), (B)-(I), (C)-(III), (D) -(II)

D (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

Correct Answer

Option B

Solution

The correct matching of the electromagnetic waves with their corresponding wavelength ranges is: (A) Microwave --> (IV) 0.1 m to 1 mm (B) Ultraviolet --> (I) 400 nm to 1 nm (C) X-Ray --> (II) 1 nm to $10^{-3}$ nm (D) Infra-red --> (III) 1 mm to 700 nm Therefore, the correct option is (A-IV, B-I, C-II, D-III).

Q84

An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:

A

\left( {{1 \over {\sqrt n }},{1 \over {\sqrt n }}} \right)

B

\left( {\sqrt n ,\sqrt n } \right)

C

\left( {\sqrt n ,{1 \over {\sqrt n }}} \right)

D

\left( {{1 \over {\sqrt n }},\sqrt n } \right)

Correct Answer

Option C

Solution

C

= {1 \over {\sqrt {{\mu _0}{ \in _0}} }}

V =

= {1 \over {\sqrt {k{ \in _0}{\mu _0}} }}

[For transparent medium $\mu$ r $\approx$ $\mu$ 0] $\therefore$

{C \over V}

$=$

\sqrt k =

n

{1 \over 2} \in {}_0\,E_0^2

C $=$ intensity $=$

{1 \over 2}

\in

0 kE2v $\therefore$ E

_0^2

C $=$ kE2v $\Rightarrow$

{{E_0^2} \over {{E^2}}} = {{kV} \over C} = {{{n^2}} \over n} \Rightarrow {{{E_0}} \over E} = \sqrt n

similarly

{{B_0^2C} \over {2{\mu _0}}} = {{{B^2}v} \over {2{\mu _0}}} \Rightarrow {{{B_0}} \over B} = {1 \over {\sqrt n }}

Q85

An EM wave from air enters a medium. The electric fields are

\overrightarrow {{E_1}}

=

{E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]

in air and

\overrightarrow {{E_2}}

=

{E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]

in medium, where the wave number k and frequency

\nu

refer to their values in air. The medium is non-magnetic. If

{\varepsilon _{{r_1}}}

and

{\varepsilon _{{r_2}}}

refer to relative permittivities of air and medium respectively, which of the following options is correct ?

A

{{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 4

B

{{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 2

C

{{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 4}

D

{{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 2}

Correct Answer

Option C

Solution

Electric field in air,

\overrightarrow {{E_1}}

= E01

\widehat x

cos (

{{2\pi vz} \over c}

$-$ 2 $\pi$ vt )

\therefore\,\,\,

Velocity in air =

{{2\pi v} \over {{{2\pi v} \over c}}}

= c Also, c =

{1 \over {\sqrt {\mu \varepsilon {r_1}{\varepsilon _0}} }}

. . . . . . (1)

\overrightarrow {{E_2}}

= E02

\widehat x

cos(2kz $-$ kct)

\therefore\,\,\,

Velocity in medium =

{{kc} \over {2k}}

=

{c \over 2}

Also,

{c \over 2}

=

{1 \over {\sqrt {\mu {\varepsilon _{r2}}\,{\varepsilon _0}} }}

. . . . . (2) As, medium is non magnetic, So, $\mu$ medium = $\mu$ air = $\mu$ Dividing (1) by (2), we get 2 =

\sqrt {{{{\varepsilon _{r2}}} \over {{\varepsilon _{r1}}}}}

$\Rightarrow$

\,\,\,

{{{{\varepsilon _{r1}}} \over {{\varepsilon _{r2}}}}}

=

{1 \over 4}

Q86

The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression

\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].

The magnetic field

\overrightarrow B

(x,z, t) is given by

-

(c is the velocity of light)

A

{1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]

B

{1 \over c}\left( {6\widehat k - 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]

C

{1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]

D

{1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]

Correct Answer

Option B

Solution

\overrightarrow E = 10\widehat j\cos \left[ {\left( {6\widehat i + 8\widehat k} \right).\left( {x\widehat i + z\widehat k} \right)} \right]

= 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]

$\therefore$

\overrightarrow K = 6\widehat i + 8\widehat k;

direction of waves travel. i.e., direction of 'c'. $\therefore$ Direction of

\widehat B

will be along

\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}

Mag. of

\overrightarrow B

will be along

\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}

Mag. of

\overrightarrow B = {E \over C} = {{10} \over C}

$\therefore$

\overrightarrow B = {{10} \over C}\left( {{{ - 4\widehat i + 3\widehat k} \over 5}} \right) = {{\left( { - 8\widehat i + 6\widehat k} \right)} \over C}

Q87

The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :

A

{U_E} = {{{U_B}} \over 2}

B

{U_E} > {U_B}

C

{U_E} < {U_B}

D

{U_E} = {U_B}

Correct Answer

Option D

Solution

Energy density of magnetic field (UB) =

{{{B^2}} \over {2{\mu _0}}}

Also, Energy density of electric field UE =

{1 \over 2}

\varepsilon

0E2 =

{1 \over 2}

\varepsilon

0 B2 C2 [as

{E \over B}

= C ] =

{1 \over 2}

\varepsilon

0 B2 $\times$

\left( {{1 \over {{\varepsilon _0}{\mu _0}}}} \right)

[as C2 =

{{1 \over {{\varepsilon _0}{\mu _0}}}}

] =

{{{B^2}} \over {2{\mu _0}}}

$\therefore$ UB = UE

Q88

Given below are two statements : Statement I : A time varying electric field is a source of changing magnetic field and vice-versa. Thus a disturbance in electric or magnetic field creates EM waves. Statement II : In a material medium, the EM wave travels with speed

v = {1 \over {\sqrt {{\mu _0}{ \in _0}} }}

. In the light of the above statements, choose the correct answer from the options given below.

A Both Statement I and Statement II are true

B Both Statement I and Statement II are false

C Statement I is correct but Statement II is false

D Statement I is incorrect but Statement II is true

Correct Answer

Option C

Solution

In a material medium speed of light is given by

v = {1 \over {\sqrt {{\varepsilon _0}{\varepsilon _r}{\mu _0}{\mu _r}} }}

. So statement 2 is false.

Q89

The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by

\mathrm{{E_x} = {E_o}\sin (kz - \omega t)}

\mathrm{{B_y} = {B_o}\sin (kz - \omega t)}

Then the correct relation between E

_0

and B

_0

is given by

A

\mathrm{{E_o}{B_o} = \omega k}

B

\mathrm{{E_0} = k{B_0}}

C

\mathrm{k{E_0} = \omega {B_0}}

D

\mathrm{\omega {E_0} = k{B_0}}

Correct Answer

Option C

Solution

$E_{x}=E_{0} \sin (k z-\omega t)$ $B_{y}=B_{0} \sin (k z-\omega t)$ $\because$ Velocity $=\dfrac{E_{0}}{B_{0}}$ $\dfrac{\omega}{k}=\dfrac{E_{0}}{B_{0}}$ $\omega B_{0}=k E_{0}$

Q90

Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by E = 20cos(2

\times

1010 t

-

200x) V/m. The dielectric constant of the medium is equal to : (Take

\mu

r = 1)

A 9

B 2

C

{1 \over 3}

D 3

Correct Answer

Option A

Solution

Given, electric field, E = 20 cos(2 $\times$ 1010t $-$ 200 x) V/m Comparing with the standard equation, E = E0 cos( $\omega$ t $-$ kx) V/m, we get Wave constant, k = 200 Angular frequency, $\omega$ = 2 $\times$ 1010 rad/s Speed of the wave,

v = {\omega \over k} = {{2 \times {{10}^{10}}} \over {200}} = {10^8}

m/s Refractive index,

\mu = {c \over v} = {{3 \times {{10}^8}} \over {{{10}^8}}} = 3

As we know the relation between the refractive index and dielectric constant,

\mu = \sqrt {{\varepsilon _r}{\mu _r}}

Substituting the value in the above equations, we get

3 = \sqrt {{\varepsilon _r}(1)}

{\varepsilon _r} = 9

Thus, the dielectric constant of the medium is 9.