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Electromagnetic Waves

JEE Physics · 127 questions · Page 5 of 13 · Click an option or "Show Solution" to reveal answer

Q41
Intensity of sunlight is observed as 0.092 Wm−-2 at a point in free space. What will be the peak value of magnetic field at the point? (ε0=8.85×10−12C2N−1m−2{\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}})
A 2.77 ×\times 10−-8 T
B 1.96 ×\times 10−-8 T
C 8.31 T
D 5.88 T
Correct Answer
Option A
Solution
IC=12ε0.E02{I \over C} = {1 \over 2}{\varepsilon _0}.E_0^2
⇒E0=2ICε0\Rightarrow {E_0} = \sqrt {{{2I} \over {C{\varepsilon _0}}}}
E0B0=C⇒B0=E0C{{{E_0}} \over {{B_0}}} = C \Rightarrow {B_0} = {{{E_0}} \over C}
⇒B0=2Iε0C3=2×0.0928.85×10−12×27×10+24\Rightarrow {B_0} = \sqrt {{{2I} \over {{\varepsilon _0}{C^3}}}} = \sqrt {{{2 \times 0.092} \over {8.85 \times {{10}^{ - 12}} \times 27 \times {{10}^{ + 24}}}}}
=2.77×10−8= 2.77 \times {10^{ - 8}}

T

Q42
A light beam is described by E=800sin⁡ω(t−xc)E = 800\sin \omega \left( {t - {x \over c}} \right). An electron is allowed to move normal to the propagation of light beam with a speed of 3 ×\times 107 ms−-1. What is the maximum magnetic force exerted on the electron?
A 1.28 ×\times 10−-18 N
B 1.28 ×\times 10−-21 N
C 12.8 ×\times 10−-17 N
D 12.8 ×\times 10−-18 N
Correct Answer
Option D
Solution
E0C=B0{{{E_0}} \over C} = {B_0}
Fmax⁡=eB0V{F_{\max }} = e{B_0}V
=1.6×10−19×8003×108×3×107= 1.6 \times {10^{ - 19}} \times {{800} \over {3 \times {{10}^8}}} \times 3 \times {10^7}
=12.8×10−18= 12.8 \times {10^{ - 18}}

N

Q43
Electric field in a plane electromagnetic wave is given by E = 50 sin(500x −- 10 ×\times 1010 t) V/m The velocity of electromagnetic wave in this medium is : (Given C = speed of light in vacuum)
A 32{3 \over 2}C
B C
C 23{2 \over 3}C
D C2{C \over 2}
Correct Answer
Option C
Solution
V=ωK=10×1010500=2×108V = {\omega \over K} = {{10 \times {{10}^{10}}} \over {500}} = 2 \times {10^8}
V=2C3V = {{2C} \over 3}

.

Q44
A radar sends an electromagnetic signal of electric field (E0) = 2.25 V/m and magnetic field (B0) = 1.5 ×\times 10−-8 T which strikes a target on line of sight at a distance of 3 km in a medium. After that, a part of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time t = 0 from radar, then after how much time echo will reach to the radar?
A 2.0 ×\times 10−-5 s
B 4.0 ×\times 10−-5 s
C 1.0 ×\times 10−-5 s
D 8.0 ×\times 10−-5 s
Correct Answer
Option B
Solution

E0 = 2.25 V/m B0 = 1.5 ×\times 10−-8 T

⇒E0B0=1.5×108\Rightarrow {{{E_0}} \over {{B_0}}} = 1.5 \times {10^8}

m/s ⇒\Rightarrow Refractive index = 2 Distance to be travelled = 6 km Time taken

=6×1031.5×108=4×10−5= {{6 \times {{10}^3}} \over {1.5 \times {{10}^8}}} = 4 \times {10^{ - 5}}

s

Q45
If Electric field intensity of a uniform plane electromagnetic wave is given as E=−301.6sin⁡(kz−ωt)a^x+452.4sin⁡(kz−ωt)a^yVmE = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}{V \over m}. Then magnetic intensity 'H' of this wave in Am−-1 will be : [Given : Speed of light in vacuum c=3×108c = 3 \times {10^8} ms−-1, Permeability of vacuum μ0=4π×10−7{\mu _0} = 4\pi \times {10^{ - 7}} NA−-2]
A +0.8sin⁡(kz−ωt)a^y+0.8sin⁡(kz−ωt)a^x + 0.8\sin (kz - \omega t){\widehat a_y} + 0.8\sin (kz - \omega t){\widehat a_x}
B +1.0×10−6sin⁡(kz−ωt)a^y+1.5×10−6(kz−ωt)a^x + 1.0 \times {10^{ - 6}}\sin (kz - \omega t){\widehat a_y} + 1.5 \times {10^{ - 6}}(kz - \omega t){\widehat a_x}
C −0.8sin⁡(kz−ωt)a^y−1.2sin⁡(kz−ωt)a^x - 0.8\sin (kz - \omega t){\widehat a_y} - 1.2\sin (kz - \omega t){\widehat a_x}
D −1.0×10−6sin⁡(kz−ωt)a^y−1.5×10−6sin⁡(kz−ωt)a^x - 1.0 \times {10^{ - 6}}\sin (kz - \omega t){\widehat a_y} - 1.5 \times {10^{ - 6}}\sin (kz - \omega t){\widehat a_x}
Correct Answer
Option C
Solution
E→=−301.6sin⁡(kz−ωt)a^x+452.4sin⁡(kz−ωt)a^y\overrightarrow E = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}
E0x=301.6{E_{0x}} = 301.6
E0y=+452.4{E_{0y}} = + 452.4
E0=E0x2+E0y2{E_0} = \sqrt {E_{0x}^2 + E_{0y}^2}

Now,

E0B0=C⇒B0=E0C=E0x2+E0y2C{{{E_0}} \over {{B_0}}} = C \Rightarrow {B_0} = {{{E_0}} \over C} = {{\sqrt {E_{0x}^2 + E_{0y}^2} } \over C}

Also,

B^=C^×E^⇒k^×(E0xi^+E0yj^)E0x2+E0y2\widehat B = \widehat C \times \widehat E \Rightarrow \widehat k \times {{({E_{0x}}\widehat i + {E_{0y}}\widehat j)} \over {\sqrt {E_{0x}^2 + E_{0y}^2} }}
B^=−E0xj^−E0yi^E0x2+E0y2\widehat B = {{ - {E_{0x}}\widehat j - {E_{0y}}\widehat i} \over {\sqrt {E_{0x}^2 + E_{0y}^2} }}
B→=−E0xCsin⁡(kz−ωt)a^y−E0yCsin⁡(kz−ωt)a^x\overrightarrow B = - {{{E_{0x}}} \over C}\sin (kz - \omega t){\widehat a_y} - {{{E_{0y}}} \over C}\sin (kz - \omega t){\widehat a_x}
H→=B→μ0\overrightarrow H = {{\overrightarrow B } \over {{\mu _0}}}
H→=−E0xμ0Csin⁡(kz−ωt)a^y−E0yμ0Csin⁡(kz−ωt)a^x\overrightarrow H = - {{{E_{0x}}} \over {{\mu _0}C}}\sin (kz - \omega t){\widehat a_y} - {{{E_{0y}}} \over {{\mu _0}C}}\sin (kz - \omega t){\widehat a_x}
Q46
In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size λ100{\lambda \over {100}}, where λ\lambda is the wavelength of the wave in free space. The phenomenon, which happens there will be :
A Reflection
B Refraction
C Diffraction
D Scattering
Correct Answer
Option D
Solution
aλ=1100\frac{\mathrm{a}}{\lambda}=\frac{1}{100}

For reflection size of obstacle must be much larger than wavelength, for diffraction size should be order of wavelength.

Since the object is of size λ100\dfrac{\lambda}{100}, much smaller than wavelength, so scattering will occur.

Q47
Which is the correct ascending order of wavelengths?
A λ\lambdavisible X-ray gamma-ray microwave
B λ\lambdagamma-ray X-ray visible microwave
C λ\lambdaX-ray gamma-ray visible microwave
D λ\lambdamicrowave visible gamma-ray X-ray
Correct Answer
Option B
Solution

Wavelength of microwave is maximum then visible light then X-rays and then gamma rays so the correct order will be λ\lambdagamma-ray Xray visible microwave

Q48
The electromagnetic waves travel in a medium at a speed of 2.0 ×\times 108 m/s. The relative permeability of the medium is 1.0. The relative permittivity of the medium will be :
A 2.25
B 4.25
C 6.25
D 8.25
Correct Answer
Option A
Solution

The speed of electromagnetic waves in a medium is given by the formula:

v=1μεv = \frac{1}{\sqrt{\mu \varepsilon}}

where μ\mu and ε\varepsilon are the absolute permeability and absolute permittivity of the medium, respectively.

Given that μ=μ0μr\mu = \mu_0 \mu_r and ε=ε0εr\varepsilon = \varepsilon_0 \varepsilon_r, we can rewrite the above equation as :

v=1μ0μrε0εrv = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}}

which simplifies to :

v=1μ0ε0×1μrεrv = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \times \frac{1}{\sqrt{\mu_r \varepsilon_r}}

Substituting cc for 1μ0ε0\dfrac{1}{\sqrt{\mu_0 \varepsilon_0}} (the speed of light in a vacuum), we get:

v=cμrεrv = \frac{c}{\sqrt{\mu_r \varepsilon_r}}

We can rearrange this equation to solve for εr\varepsilon_r :

εr=c2v2μr=(3×108m/s)2(2×108m/s)2×1=2.25\varepsilon_r = \frac{c^2}{v^2 \mu_r} = \frac{(3 \times 10^8 m/s)^2}{(2 \times 10^8 m/s)^2 \times 1} = 2.25

So, the relative permittivity of the medium is 2.25.

Q49
An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.
A 1.19 ×\times 10−-8T
B 1.71 ×\times 10−-8T
C 0.84 ×\times 10−-8T
D 3.36 ×\times 10−-8T
Correct Answer
Option B
Solution

The total power (PT) of the light bulb is given as 200 W, but only 3.5% of this power is actually emitted as radiation, which we will call P.

So, Effective power output of the bulb

P=3⋅5100×200=7 W\mathrm{P}=\frac{3 \cdot 5}{100} \times 200=7 \mathrm{~W}
 Intensity I=P4πr2=74π(4)2 W/m2 Intensity I= Average energy density ×c=12∈0E02c=12B02μ0c\begin{aligned} \text{ Intensity } \mathrm{I} & =\frac{\mathrm{P}}{4 \pi r^2}=\frac{7}{4 \pi(4)^2} \mathrm{~W} / \mathrm{m}^2 \\\\ \text{ Intensity } \mathrm{I} & =\text{ Average energy density } \times c \\\\ & =\frac{1}{2} \in_0 \mathrm{E}_0^2 c \\\\ & =\frac{1}{2} \frac{\mathrm{B}_0^2}{\mu_0} c \end{aligned}
∴B0=2μ0Ic=2×4π×10−7×73×108×4π×16=1.71×10−8 T\begin{aligned} \therefore \quad \mathrm{B}_0 & =\sqrt{\frac{2 \mu_0 \mathrm{I}}{c}} \\\\ & =\sqrt{\frac{2 \times 4 \pi \times 10^{-7} \times 7}{3 \times 10^8 \times 4 \pi \times 16}} \\\\ & =1.71 \times 10^{-8} \mathrm{~T} \end{aligned}
Q50
An expression for oscillating electric field in a plane electromagnetic wave is given as Ez = 300 sin(5π\pi ×\times 103x −- 3π\pi ×\times 1011t) Vm−-1 Then, the value of magnetic field amplitude will be : (Given : speed of light in Vacuum c = 3 ×\times 108 ms−-1)
A 1 ×\times 10−-6 T
B 5 ×\times 10−-6 T
C 18 ×\times 109 T
D 21 ×\times 109 T
Correct Answer
Option B
Solution

Given the electric field expression: Ez=300sin⁡(5π×103x−3π×1011t) Vm−1E_z = 300 \sin(5\pi \times 10^3 x - 3\pi \times 10^{11} t) \, \text{Vm}^{-1} The amplitude of the electric field (E0E_0) is 300 V/m300 \, \text{V/m}.

The velocity (vv) of the wave in the medium is given by the ratio of the coefficients of time and displacement in the wave equation, which can be calculated as: v=Coefficient of tCoefficient of x=3π×10115π×103=35×108 m/sv = \dfrac{\text{Coefficient of } t}{\text{Coefficient of } x} = \dfrac{3\pi \times 10^{11}}{5\pi \times 10^3} = \dfrac{3}{5} \times 10^8 \, \text{m/s} The relationship between the electric field amplitude and the magnetic field amplitude in an electromagnetic wave is given by: B0=E0vB_0 = \dfrac{E_0}{v} Substituting the values for E0E_0 and vv into this formula: B0=30035×108B_0 = \dfrac{300}{\dfrac{3}{5} \times 10^8} B0=300×53×108B_0 = \dfrac{300 \times 5}{3 \times 10^8} B0=15003×108B_0 = \dfrac{1500}{3 \times 10^8} B0=5×10−6 TB_0 = 5 \times 10^{-6} \, \text{T} Therefore, the amplitude of the magnetic field (B0B_0) is 5×10−65 \times 10^{-6} T.

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