Practice Start Test

Electromagnetic Waves

NEET Physics · 97 questions · Page 7 of 10 · Click an option or "Show Solution" to reveal answer

Q61
The electric field of a plane electromagnetic wave is given by E→=E0(x^+y^)sin⁡(kz−ωt)\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right) Its magnetic field will be given by :
A E0c(x^+y^)sin⁡(kz−ωt){{{E_0}} \over c}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)
B E0c(x^−y^)sin⁡(kz−ωt){{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\sin \left( {kz - \omega t} \right)
C E0c(x^−y^)cos⁡(kz−ωt){{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\cos \left( {kz - \omega t} \right)
D E0c(−x^+y^)sin⁡(kz−ωt){{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)
Correct Answer
Option D
Solution

Given,

E→=E0(x^+y^)sin⁡(kz−ωt)\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)

We know, direction of propagation,

C→=E→×B→\overrightarrow C = \overrightarrow E \times \overrightarrow B

Here direction of propagation =

k^\widehat k

∴\therefore

k^\widehat k

=

E→×B→\overrightarrow E \times \overrightarrow B

and

E^=i^+j^2\widehat E = {{\widehat i + \widehat j} \over {\sqrt 2 }}

∴\therefore

k^=(i^+j^2)×B→\widehat k = \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right) \times \overrightarrow B

⇒\Rightarrow

B^=−i^+j^2\widehat B = {{ - \widehat i + \widehat j} \over {\sqrt 2 }}

∴\therefore

B^\widehat B

=

E0c(−x^+y^)sin⁡(kz−ωt){{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)
Q62
A plane electromagnetic wave of frequency 35 MHz35 \mathrm{~MHz} travels in free space along the XX-direction. At a particular point (in space and time) E⃗=9.6j^ V/m\vec{E}=9.6 \hat{j} \mathrm{~V} / \mathrm{m}. The value of magnetic field at this point is :
A 9.6j^T9.6 \hat{j} T
B 3.2×10−8i^T3.2 \times 10^{-8} \hat{i} T
C 9.6×10−8k^T9.6 \times 10^{-8} \hat{k} T
D 3.2×10−8k^T3.2 \times 10^{-8} \hat{k} T
Correct Answer
Option D
Solution
EB=CEB=3×108B=E3×108=9.63×108B=3.2×10−8TB^=v^×E^=i^×j^=k^\begin{aligned} \frac{E}{B} & =C \\ \frac{E}{B} & =3 \times 10^8 \\ B & =\frac{E}{3 \times 10^8}=\frac{9.6}{3 \times 10^8} \\ B & =3.2 \times 10^{-8} T \\ \hat{B} & =\hat{v} \times \hat{E} \\ & =\hat{i} \times \hat{j}=\hat{k} \end{aligned}

So,

B→=3.2×10−8k^T\overrightarrow{\mathrm{B}}=3.2 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}
Q63
An electromagnetic wave of frequency v=3.0v=3.0 MHzMHz passes from vacuum into a dielectric medium with permittivity ∈=4.0. \in = 4.0. Then
A wave length is halved and frequency remains unchanged
B wave length is doubled and the frequency becomes half
C wave length is doubled and the frequency remains unchanged
D wave length and frequency both remain unchanged.
Correct Answer
Option A
Solution

Frequency remains constant during refraction

vmed=1μ0∈0×4=c2{v_{med}} = {1 \over {\sqrt {{\mu _0}{ \in _0} \times 4} }} = {c \over 2}
λmedλair=vmedvair=c/2c=12{{{\lambda _{med}}} \over {{\lambda _{air}}}} = {{{v_{med}}} \over {{v_{air}}}} = {{c/2} \over c} = {1 \over 2}

∴\therefore wavelength is halved and frequency remains unchanged

Q64
Match List I with List II : .tg .tg LIST I EM-Wave LIST II Wavelength Range A. Infra-red I. <10−3<10^{-3} nm B. Ultraviolet II. 400 nm to 1 nm C. X-rays III. 1 mm to 700 nm D. Gamma rays IV. 1 nm to 10−310^{-3} nm Choose the correct answer from the options given below :
A (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
B (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
C (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
D (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Correct Answer
Option B
Solution

To match the given List I (EM-Wave) with List II (Wavelength Range), we need to understand the typical wavelength ranges associated with each type of electromagnetic (EM) wave mentioned in List I.

Here's the matching based on known wavelength ranges: Infra-red: Infrared radiation lies between the visible and microwave portions of the electromagnetic spectrum.

The wavelength range for infrared is typically from about 700 nm (nanometers) to 1 mm (millimeter), which means it matches with option (III) - 1 mm to 700 nm.

Ultraviolet: Ultraviolet (UV) light sits between visible light and X-rays along the electromagnetic spectrum, with wavelengths shorter than visible light but longer than X-rays.

The typical wavelength range for UV light is from 400 nm down to 10 nm, so it matches with option (II) - 400 nm to 1 nm (even though the strict lower bound should be 10 nm, this is the closest option provided).

X-rays: X-rays have a wavelength range shorter than UV rays and longer than gamma rays, typically ranging from 10 nm down to 0.01 nm (or

10−210^{-2}

nm). Thus, X-rays match with option (IV) - 1 nm to

10−310^{-3}

nm, as this is the closest range provided that fits the typical scope of X-ray wavelengths.

Gamma rays: Gamma rays have the shortest wavelength on the electromagnetic spectrum, with wavelength less than 10 picometers (pm), which equates to less than

10−310^{-3}

nm.

This corresponds to option (I) - $$ So, based on the above analysis, the correct option for the matching would be: (A)-(III) (B)-(II) (C)-(IV) (D)-(I) Therefore, the correct answer is Option B: (A)-(III), (B)-(II), (C)-(IV), (D)-(I).

Q65
Consider an electromagnetic wave propagating in vacuum. Choose the correct statement :
A For an electromagnetic wave propagating in +x direction the electric field is E⃗=12Eyz (x,t)(y^−z^)\vec E = {1 \over {\sqrt 2 }}{E_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y - \hat z} \right) and the magnetic field is B⃗=12Byz (x,t)(y^+z^)\vec B = {1 \over {\sqrt 2 }}{B_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y + \hat z} \right)
B For an electromagnetic wave propagating in +x direction the electric field is E⃗=12Eyz (y,z,t)(y^+z^)\vec E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right) and the magnetic field is B⃗=12Byz (y,z,t)(y^+z^)\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right)
C For an electromagnetic wave propagating in + y direction the electric field is E→=12Eyz (x,t)y^\overrightarrow E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat y and the magnetic field is B⃗=12Byz (x,t)z^\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat z
D For an electromagnetic wave propagating in + y direction the electric field is E→=12Eyz (x,t)z^\overrightarrow E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat z and the magnetic field is B→=12Bz (x,t)y^\overrightarrow B = {1 \over {\sqrt 2 }}{B_{z{\mkern 1mu} }}\left( {x,t} \right)\widehat y
Correct Answer
Option A
Solution

As wave is propagating in + x direction, then

E→\overrightarrow E

and

B→\overrightarrow B

should be function of

(x,t)\left( {x,t} \right)

and must be in y −- z plane.

Q66
If frequency of electromagnetic wave is 60 MHz60 \mathrm{~MHz} and it travels in air along zz direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in m\mathrm{m} ) is :
A 2.5
B 5
C 10
D 2
Correct Answer
Option B
Solution

The speed of electromagnetic waves in air (and in a vacuum) is approximately the speed of light, which we denote as

cc

. The speed of light

cc

is

3×1083 \times 10^8

meters per second. The relationship between the speed of light

cc

, the frequency

ff

, and the wavelength λ\lambda of an electromagnetic wave is given by the equation:

c=fλc = f \lambda

Here, we are given that the frequency

ff

of the electromagnetic wave is 60 MHz (megahertz), which can be converted to hertz (Hz) by multiplying by

10610^6

(because 1 MHz =

10610^6

Hz).

f=60×106 Hzf = 60 \times 10^6 \text{ Hz}

To find the wavelength λ\lambda in meters, we can rearrange the wave equation to solve for λ\lambda:

λ=cf\lambda = \frac{c}{f}

Substitute the given values of

cc

and

ff

into the equation to find the wavelength:

λ=3×108 m/s60×106 Hz\lambda = \frac{3 \times 10^8 \text{ m/s}}{60 \times 10^6 \text{ Hz}}

Now, simplify the equation by dividing the numbers:

λ=360×108−6 m\lambda = \frac{3}{60} \times 10^{8-6} \text{ m}
λ=120×102 m\lambda = \frac{1}{20} \times 10^2 \text{ m}
λ=5 m\lambda = 5 \text{ m}

Therefore, the wavelength of the wave is 5 meters, corresponding to option B.

Additionally, in an electromagnetic wave, the electric and magnetic field vectors are indeed mutually perpendicular to each other and to the direction of propagation, which in this case is the

zz

direction.

Q67
Which of the following are true? A. Speed of light in vacuum is dependent on the direction of propagation. B. Speed of light in a medium is independent of the wavelength of light. C. The speed of light is independent of the motion of the source. D. The speed of light in a medium is independent of intensity. Choose the correct answer from the options given below:
A B and D only
B B and C only
C A and C only
D C and D only
Correct Answer
Option D
Solution

The speed of light does not depend on the motion of the source as well as intensity.

Q68
An object is placed in a medium of refractive index 3 . An electromagnetic wave of intensity 6×108 W/m26 \times 10^8 \mathrm{~W} / \mathrm{m}^2 falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space =3×108 m/s=3 \times 10^8 \mathrm{~m} / \mathrm{s} ) :
A 6 Nm−26 \mathrm{~Nm}^{-2}
B 36 Nm−236 \mathrm{~Nm}^{-2}
C 18 Nm−218 \mathrm{~Nm}^{-2}
D 2 Nm−22 \mathrm{~Nm}^{-2}
Correct Answer
Option A
Solution
 Radiation pressure =I V=I⋅μc=6×108×33×108=6 N/m2\begin{aligned} & \text{ Radiation pressure }=\frac{I}{\mathrm{~V}} \\ & =\frac{\mathrm{I} \cdot \mu}{\mathrm{c}} \\ & =\frac{6 \times 10^8 \times 3}{3 \times 10^8} \\ & =6 \mathrm{~N} / \mathrm{m}^2 \end{aligned}
Q69
An electron is constrained to move along the y-axis with a speed of 0.1 c (c is the speed of light) in the presence of electromagnetic wave, whose electric field is E→=30j^sin⁡(1.5×107t−5×10−2x)\overrightarrow E = 30\widehat j\sin \left( {1.5 \times {{10}^7}t - 5 \times {{10}^{ - 2}}x} \right) V/m. The maximum magnetic force experienced by the electron will be : (given c = 3 × \times 108 ms–1 and electron charge = 1.6 × \times 10–19 C)
A 4.8 × \times 10–19 N
B 2.4 × \times 10–18 N
C 3.2 × \times 10–18 N
D 1.6 × \times 10–18 N
Correct Answer
Option A
Solution
E→=30j^sin⁡(1.5×107t−5×10−2x)V/m\overrightarrow E = 30\widehat j\sin (1.5 \times {10^7}t - 5 \times {10^{ - 2}}x)V/m

V =

1.5×1025×10−2{{1.5 \times {{10}^2}} \over {5 \times {{10}^{ - 2}}}}

= 3 ×\times 108 = C

⇒B=E/C=301.5×107×5×10−2\Rightarrow B = E/C = {{30} \over {1.5 \times {{10}^7}}} \times 5 \times {10^{ - 2}}
=10−7Tesla= {10^{ - 7}}Tesla
⇒Fmax=q(V→×B→)=∣qVB∣\Rightarrow {F_{max}} = q\left( {\overrightarrow V \times \overrightarrow B } \right) = \left| {qVB} \right|
=1.6×10−19×0.1×3×108×10−7= 1.6 \times {10^{ - 19}} \times 0.1 \times 3 \times {10^8} \times {10^{ - 7}}
=4.8×10−19N= 4.8 \times {10^{ - 19}}N
Q70
A plane electromagnetic wave is propagating along the direction i^+j^2{{\widehat i + \widehat j} \over {\sqrt 2 }} , with its polarization along the direction k^\widehat k . The correct form of the magnetic field of the wave would be (here B0 is an appropriate constant) :
A B0i^−j^2cos⁡(ωt−ki^+j^2){B_0}{{\widehat i - \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)
B B0i^+j^2cos⁡(ωt−ki^+j^2){B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)
C B0j^−i^2cos⁡(ωt+ki^+j^2){B_0}{{\widehat j - \widehat i} \over {\sqrt 2 }}\cos \left( {\omega t + k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)
D B0k^cos⁡(ωt−ki^+j^2){B_0}\widehat k\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)
Correct Answer
Option A
Solution

Direction of propagation =

i^+j^2{{\widehat i + \widehat j} \over {\sqrt 2 }}

Electric field is in direction =

k^\widehat k

As

E→×B→\overrightarrow E \times \overrightarrow B

=

i^+j^2{{\widehat i + \widehat j} \over {\sqrt 2 }}

Propagation direction of

B→=i^−j^2\overrightarrow B = {{\widehat i - \widehat j} \over {\sqrt 2 }}
Ready for a full NEET mock test? Timed · full syllabus · instant results
Take a Mock Test →