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Electromagnetic Waves

NEET Physics · 97 questions · Page 10 of 10 · Click an option or "Show Solution" to reveal answer

Q91

Match with .

List - IList - II
(a) Source of microwave frequency (i) Radioactive decay of nucleus
(b) Source of infrared frequency (ii) Magnetron
(c) Source of Gamma Rays (iii) Inner shell electrons
(d) Source of X-rays (iv) Vibration of atoms and molecules
() (v) LASER
() (vi) RC circuit
A (a)-(vi), (b)-(v), (c)-(i), (d)-(iv)
B (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
C (a)-(ii), (b)-(iv), (c)-(vi), (d)-(iii)
D (a)-(vi), (b)-(iv), (c)-(i), (d)-(v)
Correct Answer
Option B
Solution

(a) Source of microwave frequency - (ii) Magnetron (b) Source of infra red frequency - (iv) Vibration of atom and molecules (c) Source of gamma ray - (i) Radio active decay of nucleus (d) Source of X-ray - (iii) inner shell electron

Q92
A plane electromagnetic wave, has frequency of 2.0 × \times 1010 Hz and its energy density is 1.02 × \times 10–8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to ( 14πε0=9×109Nm2C2{1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}{{N{m^2}} \over {{C^2}}} and speed of light = 3 × \times 108 ms–1)
A 190 nT
B 150 nT
C 160 nT
D 180 nT
Correct Answer
Option C
Solution

Energy density,

dUdV=B022μ0{{dU} \over {dV}} = {{B_0^2} \over {2{\mu _0}}}

⇒\Rightarrow 1.02 ×\times 10–8 =

B022μ0{{B_0^2} \over {2{\mu _0}}}

Also, c =

1μ0ε0{1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}

⇒\Rightarrow

μ0=1c2ε0{\mu _0} = {1 \over {{c^2}{\varepsilon _0}}}

∴\therefore

B02{B_0^2}

= 1.02 ×\times 10–8 ×\times 2 ×\times

1c2ε0{1 \over {{c^2}{\varepsilon _0}}}

= 1.02 ×\times 10–8 ×\times 2 ×\times

4π×9×1099×1016{{4\pi \times 9 \times {{10}^9}} \over {9 \times {{10}^{16}}}}

⇒\Rightarrow B0 = 16 ×\times 10-8 T = 160 nT

Q93
A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be :
A 6 V/m
B 10 V/m
C 30 V/m
D 24 V/m
Correct Answer
Option D
Solution

Prefracted =

96100Pi{{96} \over {100}}Pi

⇒\Rightarrow K2A

Pit2Pi_t^2

=

96100{{96} \over {100}}

K1A

i2_i^2

⇒\Rightarrow r2A

i2_i^2

=

96100{{96} \over {100}}

r1A

i2_i^2

⇒\Rightarrow A

t2_t^2

=

96100×132×(30)2{{96} \over {100}} \times {1 \over {{3 \over 2}}} \times {\left( {30} \right)^2}

A1

64100×(30)2=24\sqrt {{{64} \over {100}} \times {{\left( {30} \right)}^2}} = 24
Q94
A linearly polarized electromagnetic wave in vacuum is E=3.1cos⁡[(1.8)z−(5.4×106)t]i^N/CE = 3.1\cos \left[ {(1.8)z - (5.4 \times {{10}^6})t} \right]\widehat iN/C is incident normally on a perfectly reflecting wall at z = a. Choose the correct option
A The wavelength is 5.4 m
B The frequency of electromagnetic wave is 54 ×\times 104 Hz.
C The transmitted wave will be 3.1cos⁡[(1.8)z−(5.4×106)t]i^N/C3.1\cos \left[ {(1.8)z - (5.4 \times {{10}^6})t} \right]\widehat iN/C
D The reflected wave will be 3.1cos⁡[(1.8)z+(5.4×106)t]i^N/C3.1\cos \left[ {(1.8)z + (5.4 \times {{10}^6})t} \right]\widehat iN/C
Correct Answer
Option D
Solution

Reflected wave will have direction opposite to incident wave.

Q95
An electromagnetic wave travelling in the x-direction has frequency of 23 × \times 1014 Hz and electric field amplitude of 27 Vm−-1. From the options given below, which one describes the magnetic field for this wave ?
A B→\overrightarrow B (x, t) = (3 × \times 10−-8T) j^\widehat j sin [ 2π\pi (1.5 × \times 10−-8x −- 2 × \times 1014t)]
B B→\overrightarrow B (x, t) = (9 × \times 10−-8T) k^\widehat k sin [ 2π\pi (1.5 × \times 10−-6x −- 2 × \times 1014t)]
C B→\overrightarrow B (x, t) = (9 × \times 10−-8T) i^\widehat i sin [ 2π\pi (1.5 × \times 10−-8x −- 2 × \times 1014t)]
D B→\overrightarrow B (x, t) = (9 × \times 10−-8T) j^\widehat j sin [(1.5 × \times 10−-6 x −- 2 × \times 1014t)]
Correct Answer
Option B
Solution

We know that

Eo=cBo⇒Bo=Eoc=273×108=9×10−8( T)ω=2πf=2π×2×1014rad/s;K=2πλ=2πfC=2π×2×10143×108=2π×(0.67×10−6)/m\begin{gathered} E_o=c B_o \Rightarrow B_o=\frac{E_o}{c}=\frac{27}{3 \times 10^8}=9 \times 10^{-8}(\mathrm{~T}) \\\\ \omega=2 \pi f=2 \pi \times 2 \times 10^{14} \mathrm{rad} / \mathrm{s} ; \\\\ K=\frac{2 \pi}{\lambda}=\frac{2 \pi f}{C}=2 \pi \times \frac{2 \times 10^{14}}{3 \times 10^8}=2 \pi \times\left(0.67 \times 10^{-6}\right) / \mathrm{m} \end{gathered}

Since B⃗⊥\vec{B} \perp wave direction (x-axis)

B=Bosin⁡(Kx−ωt)=Bosin⁡2π(xλ−ft)=9×10−8sin⁡2π(x1.5×106−2×1014t)\begin{aligned} B & =B_o \sin (K x-\omega t)=B_o \sin 2 \pi\left(\frac{x}{\lambda}-f t\right) \\\\ & =9 \times 10^{-8} \sin 2 \pi\left(\frac{x}{1.5 \times 10^6}-2 \times 10^{14} t\right) \end{aligned}
Q96
The magnetic field of an E.M. wave is given by B⃗=(32i^+12j^)30sin⁡[ω(t−zc)]\vec{B} = \left( \dfrac{\sqrt{3}}{2} \hat{i} + \dfrac{1}{2} \hat{j} \right) 30 \sin \left[ \omega \left( t - \dfrac{z}{c} \right) \right] (S.I. Units).The corresponding electric field in S.I. units is:
A E→=(12i^+32j^)30csin⁡[ω(t+zc)]\overrightarrow{\mathrm{E}}=\left(\dfrac{1}{2} \hat{i}+\dfrac{\sqrt{3}}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}+\dfrac{z}{\mathrm{c}}\right)\right]
B E→=(12i^−32j^)30csin⁡[ω(t−zc)]\overrightarrow{\mathrm{E}}=\left(\dfrac{1}{2} \hat{i}-\dfrac{\sqrt{3}}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\dfrac{z}{\mathrm{c}}\right)\right]
C E→=(32i^−12j^)30csin⁡[ω(t+zc)]\overrightarrow{\mathrm{E}}=\left(\dfrac{\sqrt{3}}{2} \hat{i}-\dfrac{1}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}+\dfrac{z}{\mathrm{c}}\right)\right]
D E→=(34i^+14j^)30ccos⁡[ω(t−zc)]\overrightarrow{\mathrm{E}}=\left(\dfrac{3}{4} \hat{i}+\dfrac{1}{4} \hat{j}\right) 30 \mathrm{c} \cos \left[\omega\left(\mathrm{t}-\dfrac{z}{\mathrm{c}}\right)\right]
Correct Answer
Option B
Solution
B⃗=(32i^+12j^)30sin⁡[ω(t−zc)]E⃗=B⃗×c⃗ and E=B0c Here E⃗(32(−j^)+12i^)E0=30cE⃗=(12i^−32j^)30csin⁡[ω(t−zc)]\begin{aligned} & \vec{B}=\left(\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}\right) 30 \sin \left[\omega\left(t-\frac{z}{c}\right)\right] \\ & \vec{E}=\vec{B} \times \vec{c} \text{ and } E=B_0 c \\ & \text{ Here } \vec{E}\left(\frac{\sqrt{3}}{2}(-\hat{j})+\frac{1}{2} \hat{i}\right) \\ & E_0=30 c \\ & \vec{E}=\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right) 30 c \sin \left[\omega\left(t-\frac{z}{c}\right)\right] \end{aligned}
Q97
An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm−-1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :
A Ey=60sin⁡[π4×103(x−3×108t)]j^  Vm−1{E_y} = 60\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}} Bz=2sin⁡[π4×103(x−3×108t)]k^  T{B_z} = 2\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat k\,\,T
B Ey=60sin⁡[π4×103(x−3×108t)]j^  Vm−1{E_y} = 60\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}} Bz=2×10−7sin⁡[π4×103(x−3×108t)]k^  T{B_z} = 2 \times {10^{ - 7}}\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat k\,\,T
C Ey=2×10−7sin⁡[π4×103(x−3×108t)]j^  Vm−1{E_y} = 2 \times {10^{ - 7}}\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}} Bz=60sin⁡[π4×103(x−3×108t)]k^  T{B_z} = 60\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat k\,\,T
D Ey=2×10−7sin⁡[π4×104(x−4×108t)]j^  Vm−1{E_y} = 2 \times {10^{ - 7}}\sin \left[ {{\pi \over 4} \times {{10}^4}(x - 4 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}} Bz=60sin⁡[π4×104(x−4×108t)]k^  T{B_z} = 60\sin \left[ {{\pi \over 4} \times {{10}^4}(x - 4 \times {{10}^8}t)} \right]\widehat k\,\,T
Correct Answer
Option B
Solution

In first 3 options speed of light is 3 ×\times 108 m/sec and in the fourth option it is 4 ×\times 108 m/sec.

Using E = CB We can check the option is B.

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