Practice Start Test

Electromagnetic Waves

JEE Physics · 127 questions · Page 2 of 13 · Click an option or "Show Solution" to reveal answer

Q11
The electric field in an electromagnetic wave is given by E = 56.5 sin ω\omega(t −- x/c) NC−-1. Find the intensity of the wave if it is propagating along x-axis in the free space. (Given : ε\varepsilon 0 = 8.85 ×\times 10−-12C2N−-1m−-2)
A 5.65 Wm−-2
B 4.24 Wm−-2
C 1.9 ×\times 10−-7 Wm−-2
D 56.5 Wm−-2
Correct Answer
Option B
Solution
I=12ε0E02cI = {1 \over 2}{\varepsilon _0}E_0^2c
=128.5×10−12×(56.5)2×3×108= {1 \over 2} 8.5 \times {10^{ - 12}} \times {(56.5)^2} \times 3 \times {10^8}
=4.24= 4.24

W/m2

Q12
A beam of light travelling along XX-axis is described by the electric field Ey=900sin⁡ω(t−x/c)E_{y}=900 \sin \omega(\mathrm{t}-x / c). The ratio of electric force to magnetic force on a charge q\mathrm{q} moving along YY-axis with a speed of 3×107 ms−13 \times 10^{7} \mathrm{~ms}^{-1} will be : (Given speed of light =3×108 ms−1=3 \times 10^{8} \mathrm{~ms}^{-1})
A 1 : 1
B 1 : 10
C 10 : 1
D 1 : 2
Correct Answer
Option C
Solution

Ratio

=∣qE→∣∣qv→×B→∣= {{|q\overrightarrow E |} \over {|q\overrightarrow v \times \overrightarrow B |}}
=EvB=vwavev= {E \over {vB}} = {{{v_{wave}}} \over v}

⇒\Rightarrow Ratio

=3×1083×107=10= {{3 \times {{10}^8}} \over {3 \times {{10}^7}}} = 10
Q13
The electric field component of a monochromatic radiation is given by E→\overrightarrow E = 2 E0 i^\widehat i cos kz cos ω\omega t Its magnetic field B→\overrightarrow B is then given by :
A 2E0c{{2{E_0}} \over c} j^\widehat j sin kz cos ω\omega t
B −- 2E0c{{2{E_0}} \over c} j^\widehat j sin kz sin ω\omega t
C 2E0c{{2{E_0}} \over c} j^\widehat j sin kz sin ω\omega t
D 2E0c{{2{E_0}} \over c} j^\widehat j cos kz cos ω\omega t
Correct Answer
Option C
Solution

We have

dEdz=−dEdt{{dE} \over {dz}} = {{ - dE} \over {dt}}
dEdz=−2E0ksin⁡kzcos⁡ωt=−dBdt{{dE} \over {dz}} = - 2{E_0}k\sin kz\cos \omega t = {{ - dB} \over {dt}}

Therefore,

dB=+2E0ksin⁡kzcos⁡ωt dtdB = + 2{E_0}k\sin kz\cos \omega t\,dt

That is,

B=+20E0ksin⁡kzcos⁡ωt dtB = + 20{E_0}k\sin {k_z}\cos \omega t\,dt
B=+2E0ksin⁡k2∫cos⁡ωt dt=+2E0kωsin⁡kzsin⁡ωtB = + 2{E_0}k\sin {k_2}\int {\cos \omega t\,dt = + 2{E_0}{k \over \omega }\sin kz\sin \omega t}

Now,

E0B0=ωk=c{{{E_0}} \over {{B_0}}} = {\omega \over k} = c

Therefore, its magnetic field

B→\overrightarrow B

is given by

B=2E0cj^sin⁡kzsin⁡ωtB = {{2{E_0}} \over c}\widehat j\sin kz\sin \omega t
Q14
The magnetic field in a travelling electromagnetic wave has a peak value of 2020 nnTT. The peak value of electric field strength is :
A 3V/m3V/m
B 6V/m6V/m
C 9V/m9V/m
D 12V/m12V/m
Correct Answer
Option B
Solution

From question,

B0=20nT=20×10−9T{B_0} = 20nT = 20 \times {10^{ - 9}}T

( as velocity of light in vacuum

C=3×108  ms−1C = 3 \times {10^8}\,\,m{s^{ - 1}}

)

E→0=B→0×C→{\overrightarrow E _0} = {\overrightarrow B _0} \times \overrightarrow C
∣E→0∣=∣B→∣.∣C→∣=20×10−9×3×108\left| {{{\overrightarrow E }_0}} \right| = \left| {\overrightarrow B } \right|.\left| {\overrightarrow C } \right| = 20 \times {10^{ - 9}} \times 3 \times {10^8}
=6  V/m.= 6\,\,V/m.
Q15
During the propagation of electromagnetic waves in a medium :
A Electric energy density is double of the magnetic energy density.
B Electric energy density is half of the magnetic energy density.
C Electric energy density is equal to the magnetic energy density.
D Both electric and magnetic energy of densities are zero.
Correct Answer
Option C
Solution
E0=CB0{E_0} = C{B_0}

and

C=1μ0ε0C = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}

Electric energy density

=12ε0E02=μE= {1 \over 2}{\varepsilon _0}{E_0}^2 = {\mu _E}

Magnetic energy density

=12B02μ0=μB= {1 \over 2}{{{B_0}^2} \over {{\mu _0}}} = {\mu _B}

Thus,

μE=μB{\mu _E} = {\mu _B}

Energy is equally divided between electric and magnetic field

Q16
If microwaves, X rays, infracted, gamma rays, ultra-violet, radio waves and visible parts of the electromagnetic spectrum by M, X, I, G, U, R and V, the following is the arrangement in ascending order of wavelength :
A R, M, I, V, U, X and G
B M, R, V, X, U, G and I
C G, X, U, V, I, M and R
D I, M, R, U, V, X and G
Correct Answer
Option C
Solution

The arrangement in ascending order of wavelength will be

λG<λX<λU<λv<λI<λM<λR\lambda_{\mathrm{G}}<\lambda_{\mathrm{X}}<\lambda_{\mathrm{U}}<\lambda_{\mathrm{v}}<\lambda_{\mathrm{I}}<\lambda_{\mathrm{M}}<\lambda_{\mathrm{R}}
Q17
Which of the following are not electromagnetic waves?
A cosmic rays
B gamma rays
C β\beta -rays
D XX-rays
Correct Answer
Option C
Solution

β\beta -rays are fast moving beam of electrons.

Q18
A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be nearly :
A 1.34 V/m
B 2.68 V/m
C 4.02 V/m
D 5.36 V/m
Correct Answer
Option B
Solution

Since UE×c=I=\quad U_E \times c=I= Intensity

12ε0E2×c=I=P4πR2=3/100×P4πR2E=6Pε0×c×100×4πR2=6×1008.85×10−12×3×108×100×4π×(5)2=2.68 V/m\begin{aligned} & \frac{1}{2} \varepsilon_0 E^2 \times c=I=\frac{P}{4 \pi R^2}=\frac{3 / 100 \times P}{4 \pi R^2} \\\\ E & =\sqrt{\frac{6 P}{\varepsilon_0 \times c \times 100 \times 4 \pi R^2}} \\\\ & =\sqrt{\frac{6 \times 100}{8.85 \times 10^{-12} \times 3 \times 10^8 \times 100 \times 4 \pi \times(5)^2}} \\\\ & =2.68 \mathrm{~V} / \mathrm{m} \end{aligned}
Q19
An electromagnetic wave of frequency 1 × \times 1014 hertz is propagating along z - axis. The amplitude of electric field is 4 V/m. If ∈ \in 0 = 8.8 × \times 10−-12 C2/N-m2, then average energy density of electric field will be :
A 35.2 × \times 10−-10 J/m3
B 35.2 × \times 10−-11 J/m3
C 35.2 × \times 10−-12 J/m3
D 35.2 × \times 10−-13 J/m3
Correct Answer
Option C
Solution

We have

E=E0sin⁡(ωt−kx)E=E_0 \sin (\omega t-k x)

Since, energy density is μE=12ε0E2\mu_{\mathrm{E}}=\dfrac{1}{2} \varepsilon_0 E^2

⇒μˉE=12ε0Eˉ2=12ε0Erms2=12ε0E022=14ε0E02=14×8.8×10−12×42=35.2×10−12 J/m3\begin{aligned} \Rightarrow \bar{\mu}_{\mathrm{E}} & =\frac{1}{2} \varepsilon_0 \bar{E}^2=\frac{1}{2} \varepsilon_0 E_{\mathrm{rms}}^2 \\\\ & =\frac{1}{2} \varepsilon_0 \frac{E_0^2}{2}=\frac{1}{4} \varepsilon_0 E_0^2 \\\\ & =\frac{1}{4} \times 8.8 \times 10^{-12} \times 4^2=35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3 \end{aligned}
Q20
For plane electromagnetic waves propagating in the z direction, which one of the following combination gives the correct possible direction for E→\overrightarrow E and B→\overrightarrow B field respectively ?
A (i^+2j^)  \left( {\widehat i + 2\widehat j} \right)\,\, and (2i^−j^)\left( {2\widehat i - \widehat j} \right)
B (− 2i^−3j^)\left(-\, {2\widehat i - 3\widehat j} \right) and (3i^−2j^)\left( {3\widehat i - 2\widehat j} \right)
C (2i^+3j^)\left( {2\widehat i + 3\widehat j} \right) and (i^+2j^)\left( {\widehat i + 2\widehat j} \right)
D (3i^+4j^)\left( {3\widehat i + 4\widehat j} \right) and (4i^−3j^)\left( {4\widehat i - 3\widehat j} \right)
Correct Answer
Option B
Solution

Since E⃗\vec{E} and B⃗\vec{B} are mutually perpendicular. Therefore,

E⃗×B⃗=c⃗=ck^⇒(−2i^−3j^)⋅(3i^−2j^)=−6+6=0⇒(−2i^−3j^)×(3i^−2j^)=(6+9)k^=15k^\begin{gathered} \vec{E} \times \vec{B}=\vec{c}=c \hat{k} \\\\ \Rightarrow(-2 \hat{i}-3 \hat{j}) \cdot(3 \hat{i}-2 \hat{j})=-6+6=0 \\\\ \Rightarrow(-2 \hat{i}-3 \hat{j}) \times(3 \hat{i}-2 \hat{j})=(6+9) \hat{k}=15 \hat{k} \end{gathered}
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