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Electromagnetic Waves

JEE Physics · 127 questions · Page 3 of 13 · Click an option or "Show Solution" to reveal answer

Q21
Microwave oven acts on the principle of :
A transferring electrons from lower to higher energy levels in water molecule
B giving rotational energy to water molecules
C giving vibrational energy to water molecules
D giving translational energy to water molecules
Correct Answer
Option C
Solution

Microwave over use the principle of giving vibrational energy to water molecule.

Q22
Arrange the following electromagnetic radiations per quantum in the order of increasing energy : A : Blue light B : Yellow light C : X-ray D : Radiowave.
A C, A, B, D
B B, A, D, C
C D, B, A, C
D A, B, D, C
Correct Answer
Option C
Solution

Energy is in terms of increasing wavelength, we have λ\lambdaX-rays Blue light Yellow light Radio waves Since

E=hcλE = {{hc} \over \lambda }

Therefore, order of increasing energy is ERadio waves Yellow light Blue light X-rays

Q23
In E→\overrightarrow E and K→\overrightarrow K represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by : (ω\omega - angular frequency) :
A 1ω(K‾×E‾){1 \over \omega }\left( {\overline K \times \overline E } \right)
B K‾×E‾\overline K \times \overline E
C ω(K‾×E‾)\omega \left( {\overline K \times \overline E } \right)
D ω(E‾×K‾)\omega \left( {\overline E \times \overline K } \right)
Correct Answer
Option A
Solution

Magnetic field vector will be in the direction of K^×E^\hat{\mathrm{K}} \times \hat{\mathrm{E}} magnitude of B=EC=KωEB=\dfrac{E}{C}=\dfrac{K}{\omega} E Or B→=1ω(K→×E→)\overrightarrow{\mathrm{B}}=\dfrac{1}{\omega}(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})

Q24
Magnetic field in a plane electromagnetic wave is given by B→\overrightarrow B = B0 sin (k x + ω\omega t) j^ T\widehat j\,T Expression for corresponding electric field will be : Where c is speed of light.
A E→\overrightarrow E = B0 c sin (k x + ω\omega t) k^\widehat k V/m
B E→\overrightarrow E = B0c{{{B_0}} \over c} sin (k x + ω\omega t) k^\widehat k V/m
C E→\overrightarrow E = −- B0 c sin (kx +ω\omega t) k^\widehat k V/m
D E→\overrightarrow E = B0 c sin (kx −-ω\omega t) k^\widehat k V/m
Correct Answer
Option A
Solution

The relation between electric and magnetic field is , C =

E→B→{{\overrightarrow E } \over {\overrightarrow B }}

⇒\Rightarrow

   \,\,\,
E→\overrightarrow E

= C

B→\overrightarrow B

Electric field component is perpendicular to the direction of magnetic field.

Given magnetic field is along y −- axis, So, electric field along z −- axis will be

E→\overrightarrow E

= B0 C sin (kx + ω\omegat)

 k^\,\widehat k

v/m

Q25
A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electroagnetic wave is :
A 3×108∣z2−z1∣{{3 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}
B 1.5×108∣z2−z1∣{{1.5 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}
C 6×108∣z2−z1∣{{6 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}
D 1t1+∣z2−z1∣3×108{1 \over {{t_1} + {{\left| {{z_2} - {z_1}} \right|} \over {3 \times {{10}^8}}}}}
Correct Answer
Option B
Solution

Since

c=fλ⇒f=cλc = f\lambda \Rightarrow f = {c \over \lambda }

. Here,

λ=2∣z2−z1∣\lambda = 2|{z_2} - {z_1}|

and

c=3×108c = 3 \times {10^8}

. Therefore,

f=3×1082∣z2−z1∣=1.5×108∣z2−z1∣f = {{3 \times {{10}^8}} \over {2|{z_2} - {z_1}|}} = {{1.5 \times {{10}^8}} \over {|{z_2} - {z_1}|}}
Q26
A plane electromagnetic wave having a frequency v = 23.9 GHz propagates along the positive z-direction in free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave ?
A B→\overrightarrow B = 2 × 10–7 sin(1.5 × 102 x + 0.5 × 1011t) j^\widehat j
B B→\overrightarrow B = 60 sin(0.5 × 103x + 0.5 × 1011t) k^\widehat k
C B→\overrightarrow B = 2 × 10–7 sin(0.5 × 103 z + 1.5 × 1011t) i^\widehat i
D B→\overrightarrow B = 2 × 10–7 sin(0.5 × 103 z - 1.5 × 1011t) i^\widehat i
Correct Answer
Option D
Solution

Since the wave is propagating in positive z-direction So acceptable magnetic field component will be

B→=B0sin⁡(kz−ωt)\overrightarrow B = {B_0}\sin \left( {kz - \omega t} \right)
i^\widehat i

∵\because C =

EoBo{{{E_o}} \over {{B_o}}}

⇒\Rightarrow

Bo=EoC{B_o} = {{{E_o}} \over C}

=

603×108{{60} \over {3 \times {{10}^8}}}

= 2 ×\times 10-7 and

ω=2πf\omega = 2\pi f

= 2π\pi×\times23.9×\times109 = 1.5 × 1011 Hz and k =

ωc{\omega \over c}

=

1.5×10113×108{{1.5 \times {{10}^{11}}} \over {3 \times {{10}^8}}}

= 0.5 × 103 Note : When wave is propagating in positive z-direction then sign of kz and

ωt{\omega t}

should be opposite. From option you can see only option D can be correct.

Q27
An electromagnetic wave is represented by the electric field E→=E0n^sin⁡[ωt+(6y−8z)]\overrightarrow E = {E_0}\widehat n\sin \left[ {\omega t + \left( {6y - 8z} \right)} \right] . Taking unit vectors in x, y and z directions to be i^,j^,k^\widehat i,\widehat j,\widehat k , the direction of propagation s^\widehat s, is :
A s^=3i^−4j^5\widehat s = {{3\widehat i - 4\widehat j} \over 5}
B s^=−4k^+3j^5\widehat s = {{ - 4\widehat k + 3\widehat j} \over 5}
C s^=(−3j^+4k^5)\widehat s = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)
D s^=4j^−3k^5\widehat s = {{4\widehat j - 3\widehat k} \over 5}
Correct Answer
Option C
Solution
E→=E0n^sin⁡(ωt+(6y−8z))\overrightarrow E = {E_0}\widehat n\sin \left( {\omega t + \left( {6y - 8z} \right)} \right)
=E0n^sin⁡(ωt+k→.r→)= {E_0}\widehat n\sin \left( {\omega t + \overrightarrow k .\overrightarrow r } \right)

where

r→=xi^+yj^+zk^\overrightarrow r = x\widehat i + y\widehat j + z\widehat k

and

k→.r→=6y−8z\overrightarrow k .\overrightarrow r = 6y - 8z
⇒k→=6j^−8k^\Rightarrow \overrightarrow k = 6\widehat j - 8\widehat k

direction of propagation

s^=−k^=(−3j^+4k^5)\widehat s = - \widehat k = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)
Q28
The electric field of a plane electromagnetic wave is given by E→=E0i^cos⁡(kz)cos(ωt)\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t) The corresponding magnetic field B→\overrightarrow B is then given by
A B→=E0Cj^sin⁡(kz)sin⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\sin (\omega t)
B B→=E0Cj^sin⁡(kz)cos⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\cos (\omega t)
C B→=E0Cj^cos⁡(kz)sin⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\cos (kz)\sin (\omega t)
D B→=E0Ck^sin⁡(kz)cos⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat k\sin (kz)\cos (\omega t)
Correct Answer
Option A
Solution
∴E→×B→∥v→\therefore \overrightarrow E \times \overrightarrow B \parallel \overrightarrow v

Given that wave is propagating along positive z-axis and

E→\overrightarrow E

along positive x-axis. Hence

B→\overrightarrow B

along y-axis. From Maxwell equation

V→×E→=−∂B∂t\overrightarrow V \times \overrightarrow E = - {{\partial B} \over {\partial t}}

i.e.

∂E∂Z=−∂Bdt{{\partial E} \over {\partial Z}} = - {{\partial B} \over {dt}}

and

B0=E0C{B_0} = {{{E_0}} \over C}
Q29
50 W/m2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1m2 surface area will be close to (c = 3 × 108 m/s) :-
A 20 × 10–8 N
B 35 × 10–8 N
C 10 × 10–8 N
D 15 × 10–8 N
Correct Answer
Option A
Solution

Radiation pressure for 100% reflection =

2IC{{2I} \over C}

Radiation pressure for 0% reflection =

IC{I \over C}

Hence, in given case, radiation pressure =

(0.25)(2IC)+(0.75)(IC)\left( {0.25} \right)\left( {{{2I} \over C}} \right) + \left( {0.75} \right)\left( {{I \over C}} \right)
(1.25)(IC)\left( {1.25} \right)\left( {{I \over C}} \right)

∴\therefore Force = P × (Area) = 20.83 × 10–8 N

Q30
The magnetic field of a plane electromagnetic wave is given by : B‾=B0i^[cos⁡(kz−ωt)]+Bij^cos⁡(kz+ωt)\overline B = {B_0}\widehat i\left[ {\cos (kz - \omega t)} \right] + {B_i}\widehat j\cos (kz + \omega t)$ B0 = 3 × 10–5 T and B1 = 2 × 10–6 T. The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :
A 0.6 N
B 0.9 N
C 3 × 10–2 N
D 0.1 N
Correct Answer
Option A
Solution

Maximum electric field E = (B) (C)

E0→=(3×10−5)c(−j^)\overrightarrow {{E_0}} = \left( {3 \times {{10}^{ - 5}}} \right)c\left( { - \widehat j} \right)
E1→=(2×10−6)c(−i^)\overrightarrow {{E_1}} = \left( {2 \times {{10}^{ - 6}}} \right)c\left( { - \widehat i} \right)

Maximum force

F→net=10−4×3×108(3×10−5)2+(2×10−6)2=0.9N{\overrightarrow F _{net}} = {10^{ - 4}} \times 3 \times {10^8}\sqrt {{{\left( {3 \times {{10}^{ - 5}}} \right)}^2} + {{\left( {2 \times {{10}^{ - 6}}} \right)}^2}} = 0.9N
Frms=F02=0.6 N{F_{rms}} = {{{F_0}} \over {\sqrt 2 }} = 0.6\,N

(approx)

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