Electromagnetic Waves — Page 11 | JEE Physics

Q101

Match the (Phenomenon associated with electromagnetic radiation) with (Part of electromagnetic spectrum) and select the correct code from the choices given the lists : .tg .tg

List - I		List - II
(A)	Visible radiation	(I)	Doublet of sodium
(B)	Microwave (III Wavelength emitted by atomic hydrogen in interstellar space	(II)	Wavelength corresponding to temperature associated with the isotropic radiation filling all space
(C)	Short radiowave	(IV)	Wavelength of radiation arising from two close energy levels in hydrogen
(D)	X - rays	()

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

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

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

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

Correct Answer

Option B

Solution

(I) These have a wavelength of 589 nm – 589.6 nm that corresponds to visible light in the yellow region.

The correct matching is (I)-(A), Visible radiation.

The given wavelength range falls within the visible spectrum.

(II) The temperature is 2.7 K, which corresponds to the emission of microwaves.

The correct matching is (II)-(B), Microwaves.

The given temperature corresponds to the cosmic microwave background radiation.

(III) The wavelength associated is 21 cm, which corresponds to radiowaves of short wavelength.

The correct matching is (III)-(C), Short radiowave.

The given wavelength of 21 cm falls within the range of radiowaves.

(IV) This radiation has a frequency of 1,057 MHz, which corresponds to radiowaves of high frequency or short wavelength.

The correct matching is (IV)-(B), Microwaves.

The given frequency falls within the microwave range.

Therefore, the correct answer is Option B : (I)-(A), (II)-(B), (III)-(C), (IV)-(B).

Q102

The magnetic field vector of an electromagnetic wave is given by

B = {B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos (kz - \omega t)

; where

\widehat i,\widehat j

represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q1 of 4

\pi

coulomb and q2 of 2

\pi

coulomb located at

\left( {0,0,{\pi \over k}} \right)

and

\left( {0,0,{{3\pi } \over k}} \right)

, respectively, have the same velocity of 0.5 c

\widehat i

, (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is :-

A

2\sqrt 2 :1

B

1:\sqrt 2

C 2 : 1

D

\sqrt 2 :1

Correct Answer

Option C

Solution

\overrightarrow F = q\left( {\overrightarrow V \times \overrightarrow B } \right)

{\overrightarrow F _1} = 4\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{\pi \over K} - 0} \right)} \right]

{\overrightarrow F _2} = 2\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{{3\pi } \over K} - 0} \right)} \right]

\cos \pi = - 1

,

\cos 3\pi = - 1

$\therefore$

{{{F_1}} \over {{F_2}}} = 2

Q103

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A : EM waves used for optical communication have longer wavelengths than that of microwave, employed in Radar technology. Reason R : Infrared EM waves are more energetic than microwaves, (used in Radar) In the light of given statements, choose the correct answer from the options given below.

A Both

\mathrm{A}

and

\mathrm{R}

are true but

\mathrm{R}

is NOT the correct explanation of

\mathrm{A}

B

\mathrm{A}

is true but

\mathrm{R}

is false

C Both

\mathrm{A}

and

\mathrm{R}

are true and

\mathrm{r}

is the correct explanation of

\mathrm{A}

D

\mathrm{A}

is false but

\mathrm{R}

is true

Correct Answer

Option D

Solution

Optical communication is performed in the frequency range of $1 \mathrm{THz}$ to $1000 \mathrm{THz}$ .

(Microwave to UV) So, EM waves used for optical communication have shorter wavelength than that of microwaves used in RADAR.

Also, $U_{\text{INFRARED }}>U_{\text{MICROWAVE }}$ $\therefore$ Infrared EM waves are more energetic than microwave

Q104

If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6 sin

\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]

then the maximum electric field associated with it is -

A 4.5

\times

104 N/C

B 4

\times

104 N/C

C 6

\times

104 N/C

D 3

\times

104 N/C

Correct Answer

Option D

Solution

E0 = B0 $\times$ C = 100 $\times$ 10 $-$ 6 $\times$ 3 $\times$ 108 = 3 $\times$ 104 N/C $\therefore$ correct answer is 3 $\times$ 104 N/C

Q105

In an electromagnetic wave, at an instant and at particular position, the electric field is along the negative

z

-axis and magnetic field is along the positive

x

-axis. Then the direction of propagation of electromagnetic wave is:

A at

45^{\circ}

angle from positive y-axis

B positive

y

-axis

C negative

\mathrm{y}

-axis

D positive z-axis

Correct Answer

Option C

Solution

As the electric field is along the negative $z$ -axis and the magnetic field is along the positive $x$ -axis, the direction of propagation of the wave is given by the vector product $\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$ , which is in the direction of the negative $y$ -axis.

Using the right-hand rule, if you point your right thumb in the direction of the vector product $\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$ , your fingers will curl in the direction of the propagation of the wave.

In this case, the fingers would curl in the direction of the negative $y$ -axis.

Therefore, the direction of propagation of the electromagnetic wave in this scenario is along the negative $y$ -axis.

Q106

In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by

\widehat k

and

2\widehat i - 2\widehat j

, respectively. What is the unit vector along direction of propagation of the wave?

A

{1 \over {\sqrt 5 }}\left( {\widehat i + 2\widehat j} \right)

B

{1 \over {\sqrt 5 }}\left( {2\widehat i + \widehat j} \right)

C

{1 \over {\sqrt 2 }}\left( {\widehat i + \widehat j} \right)

D

{1 \over {\sqrt 2 }}\left( {\widehat j + \widehat k} \right)

Correct Answer

Option C

Solution

\overrightarrow E \times \overrightarrow B = \widehat k \times \left( {2\widehat i - 2\widehat j} \right)

=

2\widehat k \times \widehat i - 2\widehat k \times \widehat j

=

\left( {2\widehat j + 2\widehat i} \right)

Unit vector along

\overrightarrow E \times \overrightarrow B

=

{{\left( {2\widehat j + 2\widehat i} \right)} \over {2\sqrt 2 }}

=

{1 \over {\sqrt 2 }}\left( {\widehat i + \widehat j} \right)

Q107

A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by : [Given permittivity of space

\in

0 = 9

\times

10–12 SI units, Speed of light c = 3

\times

108 m/s]

A 2 kV/m

B 1 kV/m

C 1.4 kV/m

D 0.7 kV/m

Correct Answer

Option C

Solution

Intensity of EM wave is given by

{\rm I} = {{Power} \over {Area}} = {1 \over 2}{\varepsilon _0}E_0^2C

= {{27 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}

= {1 \over 2} \times 9 \times {10^{ - 12}} \times {E^2} \times 3 \times {10^8}

E = \sqrt 2 \times {10^3}kv/m

= 1.4

kv/m

Q108

Arrange the following in the ascending order of wavelength: A. Gamma rays

\left(\lambda_1\right)

B.

x

- rays

\left(\lambda_2\right)

C. Infrared waves

\left(\lambda_3\right)

D. Microwaves

\left(\lambda_4\right)

Choose the most appropriate answer from the options given below

A

\lambda_1<\lambda_2<\lambda_3<\lambda_4

B

\lambda_2<\lambda_1<\lambda_4<\lambda_3

C

\lambda_4<\lambda_3<\lambda_2<\lambda_1

D

\lambda_4<\lambda_3<\lambda_1<\lambda_2

Correct Answer

Option A

Solution

Wavelengths are as Gamma rays

x-ray Infrared

Microwave $$

Q109

A radiation of energy

E

falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

A

Ec

B

2E/c

C

E/c

D

E/{c^2}

Correct Answer

Option B

Solution

Momentum of photon

= {E \over c}

Change in momentum

= {{2E} \over c}

$=$ momentum transferred to the surface (the photon will reflect with same magnitude of momentum in opposite direction)

Q110

The magnetic field of a plane electromagnetic wave is

\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i

T where c = 3

\times

108 ms–1 is the speed of light. The corresponding electric field is :

A

\overrightarrow E = - {10^{ - 6}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k

V/m

B

\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k

V/m

C

\overrightarrow E = 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k

V/m

D

\overrightarrow E = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k

Correct Answer

Option B

Solution

Given,

\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i

T $\therefore$ E0 = CB0 = 3 × 108 × 3 × 10–8 = 9 V/m We know,

\left( {\overrightarrow E \times \overrightarrow B } \right)||\overrightarrow C

And here

\widehat B = \widehat i\& \widehat C = - \widehat j

$\therefore$

\widehat E = - \widehat k

So,

\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k

V/m