Electromagnetic Waves — Page 6 | NEET Physics

Q51

The magnetic field in a plane electromagnetic wave is

\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{T}

. The corresponding electric field will be :

A

E_z=105 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

B

E_y=10.5 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

C

E_y=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

D

E_z=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

Correct Answer

Option A

Solution

For an electromagnetic wave propagating in free space, the relationship between the magnitudes of the electric field (

E

) and the magnetic field (

B

) can be described using the equation: $E = cB$ where $c$ is the speed of light in vacuum, approximately $3.0 \times 10^8 \, \text{m/s}$ . $B$ is the magnitude of the magnetic field.

Given the magnetic field $B_y = (3.5 \times 10^{-7}) \sin (1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{T}$ , we can calculate the corresponding electric field magnitude using the formula above: $E = (3.0 \times 10^8) \times (3.5 \times 10^{-7})$ $= 105 \, \text{Vm}^{-1}$ Thus, the magnitude of the electric field associated with the given magnetic field is $105 \, \text{Vm}^{-1}$ .

The direction of the electric field is perpendicular to both the magnetic field and the direction of propagation.

Given $B_y$ , this means $E$ will have components in the $x-z$ plane.

Since electromagnetic waves are transverse, and given that the magnetic field is specified to be in the $y$ -direction, the corresponding electric field component must lie in a plane perpendicular to the $y$ -axis, which could be either the $x$ or the $z$ direction.

However, knowing electromagnetic wave properties, if the wave is propagating along the $x$ -axis and the magnetic field ( $B_y$ ) is along the $y$ -axis, then by right-hand rule, the electric field ( $E$ ) must be along the $z$ -axis to maintain the orthogonal relationship among the direction of propagation, electric field, and magnetic field vector directions.

Therefore, the correct option is: $\text{Option A: } E_z = 105 \sin \left(1.5 \times 10^3 x + 0.5 \times 10^{11} t\right) \, \text{Vm}^{-1}$

Q52

A plane electromagnetic wave propagates along the + x direction in free space. The components of the electric field,

\vec{E}

and magnetic field,

\vec{B}

vectors associated with the wave in Cartesian frame are

A

E_x, B_y

B

E_y, B_x

C

E_y, B_z

D

E_z, B_y

Correct Answer

Option C

Solution

We know, for a plane electromagnetic wave,

\widehat E \times \widehat B = \widehat C

Given,

\widehat C = \widehat i

(+ x direction) So,

\widehat E \times \widehat B = \widehat i

i.e.

\widehat j \times \widehat k = \widehat i

So,

\overrightarrow E = {E_y}

\overrightarrow B = {B_2}

As,

\widehat i,\,\widehat j

and

\widehat k

are unit vectors in the directions of x, y and z axes respectively. Hence, option 3 is correct.

Q53

The unit of

\sqrt{\dfrac{2I}{\varepsilon_0 c}}

is :(I = intensity of an electromagnetic wave, c = speed of light)

A Vm

B NC-1

C NC

D Nm

Correct Answer

Option B

Solution

The unit of the expression $\sqrt{\dfrac{2I}{\varepsilon_0 c}}$ can be determined by exploring the relationship between intensity ( $I$ ) and the electric field ( $E_0$ ) in an electromagnetic wave.

The intensity of an electromagnetic wave is given by: $I = \dfrac{1}{2} \varepsilon_0 E_0^2 \times c$ From this equation, we can solve for the electric field $E_0$ : $E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}}$ The symbol $E_0$ represents the electric field, which is measured in units of Newtons per Coulomb (N/C).

Therefore, the unit of the expression $\sqrt{\dfrac{2I}{\varepsilon_0 c}}$ is also $N/C$ .

Thus, the expression $\sqrt{\dfrac{2I}{\varepsilon_0 c}}$ has units of $N/C$ .

Q54

Match List I with List II .tg .tg LIST I LIST II A. Microwaves I. Physiotherapy B. UV rays II. Treatment of cancer C. Infra-red light III. Lasik eye surgery D. X-ray IV. Aircraft navigation Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option A

Solution

A. Microwave → IV B. UV rays → III C. Infra-red → I D. X-ray → II

Q55

Match with : .tg .tg

List - I		List - II
(a)	UV rays	(i)	Diagnostic tool in medicine
(b)	X-rays	(ii)	Water purification
(c)	Microwave	(iii)	Communication, Radar
(d)	Infrared wave	(iv)	Improving visibility in foggy days

A (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)

B (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)

C (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)

D (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)

Correct Answer

Option B

Solution

UV - Water purification X-rays - Diagnostic tool in medicine Microwave - Communication, Radar Infrared wave - Improving visibility in foggy days.

Q56

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) }

Q57

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).

Q58

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.

Q59

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 $=\dfrac{E_{0}}{B_{0}}$ $\dfrac{\omega}{k}=\dfrac{E_{0}}{B_{0}}$ $\omega B_{0}=k E_{0}$

Q60

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.