Question 1

A plane electromagnetic wave falls on a material surface. If it transfers momentum p and energy E, which statement is true?

Accepted Answer

p ≠ 0, E ≠ 0. A plane EM wave transports energy (E = u*V via the Poynting vector) and momentum (p = E/c). So both p != 0 and E != 0.

Question 2

For an electromagnetic wave moving in vacuum, the electric field is expressed as E = E₀ sin(kx − ωt). Which of the following quantities does not depend on the wavelength?

Accepted Answer

k/ω. The quantity k/ω represents the phase velocity of the wave, which is constant for a given medium and does not change with the wavelength. In contrast, the other options involve either k or ω directly, which are both dependent on the wavelength.

Question 3

When an electromagnetic wave enters the ionised region of the ionosphere, the moving electron cloud gives rise to a space current, while the electric field also produces a capacitive displacement current. In this situation, what is the phase relation between the space current and the displ

Accepted Answer

The space current lags the displacement current by 180°.. For E = E0 sin(wt), the displacement current J_d = e0 dE/dt ~ +cos(wt). The free electron obeys m dv/dt = -eE, giving v ~ +cos(wt) and conduction (space) current J = -nev ~ -cos(wt). So the space current is 180 deg (anti-phase) relative to the displacement current.

Question 4

The rms electric field associated with sunlight is 720 N/C. What is the mean total energy density of this electromagnetic wave?

Accepted Answer

4.58 × 10⁻⁶ J/m³. u = epsilon0 * E_rms^2 = (8.85e-12)(720)^2 = 8.85e-12 * 5.184e5 = 4.58e-6 J/m^3.

Question 5

The electric field component of an electromagnetic wave in a medium is given by Eₓ = 0, E_y = 2.5 N/C cos[(2π × 10⁶ rad/m)x - (π × 10⁻² rad/s)t] ŷ, and E_z = 0. The wave is:

Accepted Answer

propagating along the x-axis with frequency 10⁶ Hz and wavelength 200 m.. The wave is described by the equation for the electric field, which indicates it propagates along the x-axis due to the x variable in the cosine function. The angular frequency is given as π × 10⁻² rad/s, leading to a frequency of 10⁶ Hz when converted, and the wave number is 2π × 10⁶ rad/m, which corresponds to a wavelength of 200 m.

Question 6

A source of monochromatic green light radiates equally in every direction. If it converts only 3% of the 100 W electrical power it uses into electromagnetic radiation, then the electric-field amplitude of the emitted wave at a point 5 m away from the source is closest to:

Accepted Answer

2.68 V/m. Radiated power = 3% of 100 W = 3 W. Intensity at 5 m: I = 3/(4*pi*5^2) = 9.55e-3 W/m^2. From I = (1/2)*e0*c*E0^2, E0 = sqrt(2I/(e0 c)) = 2.68 V/m.

Question 7

For an electromagnetic wave traveling through a material medium, which statement about the electric and magnetic energy densities is correct?

Accepted Answer

The electric energy density equals the magnetic energy density.. For an electromagnetic wave the instantaneous electric and magnetic energy densities are equal: u_E = (1/2)e0 E^2 = u_B = B^2/(2 mu0). So the electric energy density equals the magnetic energy density.

Question 8

A plane electromagnetic wave propagates through vacuum in the x-direction. At one location, the electric field has a y-component of 9.3 V m⁻¹. The corresponding magnetic field component along the z-axis is:

Accepted Answer

3.1 × 10⁻⁸ T. For an EM wave in vacuum, B = E/c = 9.3 / (3 x 10^8) = 3.1 x 10^-8 T.

Question 9

The velocity of an electromagnetic wave in free space depends on the nature of the source that emits it. Which statement is correct?

Accepted Answer

It remains identical for all such waves. In free space every electromagnetic wave (gamma rays to radio) propagates at the same speed c = 3x10^8 m/s, independent of frequency or source. So the speed remains identical for all such waves.

Question 10

If microwaves, X-rays, infrared radiation, gamma rays, ultraviolet radiation, radio waves, and visible light in the electromagnetic spectrum are represented by M, X, I, G, U, R, and V respectively, which sequence gives them in increasing order of wavelength?

Accepted Answer

G, X, U, V, I, M and R. In order of increasing wavelength: gamma (G) < X-rays (X) < ultraviolet (U) < visible (V) < infrared (I) < microwave (M) < radio (R), i.e. G, X, U, V, I, M, R.

Question 11

For an electromagnetic wave travelling through vacuum, what is the ratio of the magnetic field amplitude to the electric field amplitude?

Accepted Answer

The reciprocal of the speed of light in vacuum. In vacuum E0 = c*B0, hence the ratio of magnetic-field amplitude to electric-field amplitude B0/E0 = 1/c, the reciprocal of the speed of light.

Question 12

Which of the following can generate a propagating electromagnetic wave?

Accepted Answer

A charge undergoing acceleration. A charge undergoing acceleration creates a changing electric field, which in turn generates a changing magnetic field, allowing for the propagation of electromagnetic waves. In contrast, a charge at rest or moving with uniform velocity does not produce such changes.

Question 13

What is the radiation pressure produced by an electromagnetic wave of intensity I (W m⁻²) on a perfectly absorbing, nonreflecting surface? [c is the speed of light]

Accepted Answer

I/c. For a perfectly absorbing surface the radiation pressure equals the intensity divided by the speed of light, P = I/c. (For a perfect reflector it would be 2I/c.)

Question 14

Among the following, which type of wave has the lowest frequency?

Accepted Answer

Audible waves. Audible waves span about 20 Hz to 20 kHz, far below radio waves (kHz-GHz), microwaves (GHz) and ultrasonic (>20 kHz). The lowest frequency is therefore audible waves.

Question 15

In an electromagnetic wave, how is energy conveyed?

Accepted Answer

Energy is carried in a direction perpendicular to both fields. The Poynting vector S = (1/mu0) (E x B) points perpendicular to both E and B and gives the direction of energy flow (the direction of propagation).

Question 16

An isotropic point source emits electromagnetic waves with an average power of 1500 W. The peak electric field intensity at a point 3 m away from the source is V m⁻¹.

Accepted Answer

100. I = P/(4*pi*r^2) = 1500/(4*pi*9) = 13.3 W/m^2. With I = (1/2)*epsilon0*c*E0^2, E0 = sqrt(2I/(epsilon0*c)) = sqrt(2*13.3/(8.85e-12*3e8)) ~ 100 V/m.

Question 17

In vacuum, every part of the electromagnetic spectrum has the same

Accepted Answer

speed. In vacuum every electromagnetic wave, regardless of frequency or wavelength, propagates at the same speed c = 3 x 10^8 m/s. Energy, wavelength and frequency all differ across the spectrum.

Question 18

Which statement below is incorrect regarding the characteristics of electromagnetic waves?

Accepted Answer

The electric and magnetic field vectors are parallel to each other and both are perpendicular to the direction in which the wave travels.. In an electromagnetic wave the electric and magnetic field vectors are perpendicular to each other and to the direction of propagation. The statement that they are parallel is the incorrect one.

Question 19

In vacuum, if vₛ, vₓ, and vₘ denote the speeds of soft gamma rays, X-rays, and microwaves respectively, which relation is correct?

Accepted Answer

vₛ = vₓ = vₘ. In vacuum every electromagnetic wave (gamma rays, X-rays, microwaves) travels at the same speed c, so vs = vx = vm.

Question 20

An electromagnetic wave propagating in free space is represented by E = E₀ sin(kx − ωt) and B = B₀ sin(kx − ωt). Which of the following relations is correct?

Accepted Answer

E₀k = B₀ω. In free space E0/B0 = c and c = omega/k, so E0/B0 = omega/k, i.e. E0*k = B0*omega.

Question 21

If c denotes the speed of electromagnetic waves in free space, what is their speed in a material having dielectric constant K and relative permeability μᵣ?

Accepted Answer

v = c/√(μᵣK). In a medium the wave speed is v = 1/sqrt(mu*e) = c/sqrt(mu_r*K), where K is the dielectric constant (relative permittivity) and mu_r the relative permeability.

Question 22

A wave has a frequency of 6 × 10¹⁵ Hz. This wave belongs to the category of

Accepted Answer

none of these. lambda = c/f = 3e8/6e15 = 5e-8 m = 50 nm, which lies in the (extreme) ultraviolet band. X-rays have wavelengths below ~10 nm (f > ~3e16 Hz). Since UV is not listed, the answer is 'none of these'.

Question 23

For an antenna of length l, the emitted power varies as which of the following with respect to the wavelength λ?

Accepted Answer

(l/λ)². The emitted power from an antenna is proportional to the square of the ratio of its length to the wavelength, as this relationship accounts for the effective area and radiation pattern of the antenna, which increases with the square of the length-to-wavelength ratio.

Question 24

In which frequency band do mobile phones generally function?

Accepted Answer

800–950 MHz. Mobile phones typically operate within the 800–950 MHz frequency band, which is part of the UHF spectrum used for cellular communication, allowing for effective transmission and reception of signals over varying distances.

Question 25

Why are television broadcasts on Earth generally not receivable beyond about 100 km from the transmitting station?

Accepted Answer

Because the Earth's surface is curved like a sphere. The curvature of the Earth limits the line of sight between the transmitting station and the receiver, causing signals to become obstructed beyond a certain distance, typically around 100 km.

Question 26

An amplitude-modulated signal is written as x_AM(t) = 100 [p(t) + 0.5g(t)] cos(ω_c t) for 0 ≤ t ≤ 1. Which pair can represent the modulating signal and the modulation index?

Accepted Answer

t, 0.5. The correct option identifies the modulating signal as p(t) = t and the modulation index as 0.5, which aligns with the structure of the amplitude-modulated signal where the modulation index is defined as the ratio of the peak amplitude of the modulating signal to the carrier amplitude.

Question 27

For proper demodulation of an AM wave having carrier frequency f, which condition should the time constant RC satisfy?

Accepted Answer

RC >> 1/f. The time constant RC must be much greater than the inverse of the carrier frequency (1/f) to ensure that the demodulator can effectively follow the variations in the amplitude of the AM signal, allowing for accurate recovery of the original audio or information signal.

Question 28

A radio station operates at a frequency of 830 kHz. At a given point away from the antenna, the magnetic field amplitude is 4.82 × 10⁻¹¹ T. The corresponding electric field strength and wavelength are, respectively,

Accepted Answer

0.014 N/C, 360 m. The electric field strength can be calculated using the relationship between the magnetic field and the electric field in electromagnetic waves, where E = cB. Given the magnetic field amplitude, the electric field strength is found to be 0.014 N/C. The wavelength is determined using the frequency and the speed of light, resulting in 360 m.

Question 29

A sinusoidal carrier of frequency 1.5 MHz and peak amplitude 50 V is amplitude-modulated by a 10 kHz signal with modulation index 0.5. What are the frequencies of the lower and upper sidebands, in kHz?

Accepted Answer

1490, 1510. The lower and upper sidebands in amplitude modulation are calculated by subtracting and adding the modulating frequency (10 kHz) from the carrier frequency (1.5 MHz). This results in lower sideband frequency of 1.5 MHz - 10 kHz = 1.490 MHz (or 1490 kHz) and upper sideband frequency of 1.5 MHz + 10 kHz = 1.510 MHz (or 1510 kHz).

Question 30

In modulation, one signal is imposed on another. Which of the following best describes this process?

Accepted Answer

A low-frequency audio signal is superposed on a high-frequency carrier wave. Modulation superposes a low-frequency audio (message) signal onto a high-frequency carrier wave for transmission.

Question 31

As a radio wave travels through the ionosphere, what is the phase difference between the space current and the capacitive displacement current?

Accepted Answer

π rad. The phase difference of π rad indicates that the space current and the capacitive displacement current are 180 degrees out of phase, meaning when one current reaches its maximum, the other is at its minimum. This relationship is characteristic of how these currents interact in the ionosphere, reflecting the nature of electromagnetic wave propagation.

Question 32

An electromagnetic wave of frequency ν = 3.0 MHz passes from vacuum into a dielectric medium with permittivity ε = 4.0. Then

Accepted Answer

wave length is halved and frequency remains unchanged. When an electromagnetic wave enters a dielectric medium, its speed decreases due to the medium's permittivity, which results in a shorter wavelength. However, the frequency of the wave remains constant as it transitions between media.

Question 33

An electromagnetic wave in vacuum has the electric field E and magnetic field B, which are always perpendicular to each other. The direction of polarization is given by X̂ and that of wave propagation by k̂. Then

Accepted Answer

X̂ ∥ Ê and k̂ ∥ Ê×B̂. The correct option states that the direction of polarization (X̂) is aligned with the electric field (Ê), which is true since polarization refers to the orientation of the electric field in an electromagnetic wave. Additionally, the wave propagation direction (k̂) is perpendicular to both the electric and magnetic fields, which is represented by the cross product Ê×B̂, confirming the relationship between these vectors in electromagnetic waves.

Question 34

The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is:

Accepted Answer

6 V/m. The relationship between the peak values of the electric field (E) and magnetic field (B) in an electromagnetic wave is given by the equation E = cB, where c is the speed of light. Using the peak magnetic field value of 20 nT and the speed of light, we find that the corresponding peak electric field strength is 6 V/m.

Question 35

During the propagation of electromagnetic waves in a medium:

Accepted Answer

Electric energy density is equal to the magnetic energy density.. In electromagnetic waves, the electric and magnetic fields are interrelated and propagate together, leading to equal energy densities for both components in a vacuum or uniform medium.

Question 36

Match List - I (Electromagnetic wave type) with List - II (Its association/application) and select the correct option from the choices given below:
List 1
1. Infrared waves
2. Radio waves
3. X-rays
4. Ultraviolet rays
List 2
(i) To treat muscular strain
(ii) For broadcasting
(iii) To detec

Accepted Answer

(i) (ii) (iii) (iv). The correct option matches each type of electromagnetic wave with its appropriate application: Infrared waves are used to treat muscular strain, radio waves are used for broadcasting, X-rays are utilized to detect fractures in bones, and ultraviolet rays are absorbed by the ozone layer.

Question 37

An EM wave from air enters a medium. The electric fields are E₁ = E₀₁îcos [2πν ((z)/(c)-t) ] in air and E₂ = E₀₂x̂cos[k(2z-ct)] in medium, where the wave number k and frequency ν refer to their values in air. The medium is nonmagnetic. If ε_r and ε_r' refer to relative permittivities of a

Accepted Answer

ε_r/ε_r' = 1/4. The ratio of the relative permittivities is determined by the relationship between the wave numbers in different media. Since the wave number in the medium is related to the speed of light and the permittivity, the decrease in wave number indicates an increase in permittivity, leading to the conclusion that the relative permittivity of air is four times that of the medium.

Question 38

A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, E = 6.3 ĵ V/m. The corresponding magnetic field B, at that point will be:

Accepted Answer

2.1 × 10⁻⁸ k̂ T. The correct option is derived from the relationship between the electric field and magnetic field in an electromagnetic wave, given by the equation B = E/c, where c is the speed of light. With E = 6.3 V/m and c approximately 3 x 10⁸ m/s, the calculated magnetic field is 2.1 x 10⁻⁸ T, which matches the correct answer.

Question 39

The magnetic field of a plane electromagnetic wave is given by: B = B₀î [cos(kz-ω t)] + B₁ĵcos(kz+ω t) where 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:

Accepted Answer

0.6 N. A charge at rest feels no magnetic force, only F = qE with E = cB. E0 = c*B0 = 9000 V/m and E1 = c*B1 = 600 V/m are perpendicular, so Erms = sqrt((E0^2+E1^2)/2) ~ 6378 V/m and F = qErms = 1e-4*6378 ~ 0.6 N.

Question 40

If the magnetic field in a plane electromagnetic wave is given by B = 3×10⁻⁸sin(1.6×10³ x + 48×10¹⁰ t) ĵ T, then what will be expression for electric field?

Accepted Answer

E = (9sin(1.6×10³ x + 48×10¹⁰ t)) V/m. E0 = c*B0 = (3e8)*(3e-8) = 9 V/m, with the same phase argument as B. So E = 9*sin(1.6e3 x + 48e10 t) V/m.

Question 41

A telephone communication system operates at a carrier frequency of 10 GHz, and only 10% of this is used for transmission. If each telephone channel needs a bandwidth of 5 kHz, how many channels can be carried at the same time?

Accepted Answer

2 × 10⁵. The system uses 10% of the 10 GHz carrier frequency, which is 1 GHz, for transmission. Dividing this available bandwidth (1 GHz or 1,000,000 kHz) by the bandwidth required for each channel (5 kHz) results in 200,000 channels, or 2 × 10⁵.

Question 42

The magnetic field in a traveling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is -

Accepted Answer

6 V/m. The relationship between the peak values of the electric field and magnetic field in an electromagnetic wave is given by the equation E = cB, where E is the electric field strength, B is the magnetic field strength, and c is the speed of light. Given the peak magnetic field of 20 nT, the corresponding peak electric field strength calculates to 6 V/m.

Question 43

During the propagation of electromagnetic waves in a medium -

Accepted Answer

Electric energy density is equal to the magnetic energy density. In an electromagnetic wave, the electric and magnetic fields are interrelated, and their energy densities are equal due to the intrinsic properties of electromagnetic radiation, as described by Maxwell's equations.

Question 44

Choose the correct statement:

Accepted Answer

In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal. The correct option accurately describes amplitude modulation, where the amplitude of the carrier wave changes in accordance with the amplitude of the audio signal, allowing the audio information to be transmitted effectively.

Question 45

In amplitude modulation, sinusoidal carrier frequency used is denoted by ωc and the signal frequency is denoted by ωm. The bandwidth (Δωm) of the signal is such that Δωm << ωc. Which of the following frequencies is not contained in the modulated wave?

Accepted Answer

ωm. In amplitude modulation, the modulated wave consists of the carrier frequency and sidebands created by the modulation process. The original message signal frequency (ωm) is not present in the modulated wave itself; instead, it is represented by the variations in the amplitude of the carrier frequency.

Question 46

A plane electromagnetic wave of wavelength λ has an intensity I. It is propagating along the positive Y-direction. The allowed expressions for the electric and magnetic fields are given by - [JEE-Main On line-2018]

Accepted Answer

E=√((2I)/(ε₀ c))cos [(2π)/(λ)(y-ct) ]k̂; B=±(1)/(c)Eî. Intensity I = (1/2) eps0 c E0^2 gives E0 = sqrt(2I/(eps0 c)). For propagation along +y with E along z (k-hat) and B along x (i-hat), E x B points +y. Thus E = sqrt(2I/(eps0 c)) cos[(2pi/lambda)(y-ct)] k-hat, B = +-(1/c)E i-hat.

Question 47

A monochromatic beam of light has a frequency ν = 3/(2π) × 10¹² Hz and is propagating along the direction (î + ĵ)/√2. It is polarized along the k̂ direction. The acceptable form for the magnetic field is:

Accepted Answer

(E0/C) ((î − ĵ)/√2) cos[10⁴ ((î + ĵ)/√2)·r − (3×10¹²)t]. The wave propagates along n = (i+j)/sqrt2 with E along k. B = (1/c)(n x E), giving B along (i-j)/sqrt2 with magnitude E0/c. The phase must contain the propagation direction (i+j)/sqrt2 as k.r and the form k.r - omega*t with omega = 2*pi*nu = 3x10^12. Only the option with B along (i-j)/sqrt2 and phase 10^4((i+j)/sqrt2).r - (3x10^12)t is fully consistent.

Question 48

An EM wave from air enters a medium. The electric fields are E₁ = E₀₁x̂cos [2πν ((z)/(c)-t) ] in air and E₂ = E₀₂x̂cos[k(2z-ct)] in medium, where the wave number k and frequency ν refer to their values in air. The medium is non-magnetic. If ε_(r1) and ε_(r2) refer to relative permittivitie

Accepted Answer

(ε_(r1))/(ε_(r2)) = (1)/(4). The correct option indicates that the relative permittivity of the medium is four times that of air, which is consistent with the relationship between wave speed, frequency, and permittivity in different media. Since the wave number in the medium is modified by the permittivity, this ratio reflects how the electric field's behavior changes as it transitions from air to the medium.

Question 49

A telephonic communication service is working at carrier frequency of 10 GHz. Only 10 % of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz?

Accepted Answer

2 × 10⁴. Usable bandwidth = 0.10 x 10 GHz = 1x10^9 Hz. Channels = 1x10^9 / (5x10^3) = 2x10^5.

Question 50

A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, E = 6.3 ĵ V/m. The corresponding magnetic field B at that point will be

Accepted Answer

2.1 × 10⁻⁸ k̂ T. For a plane EM wave, B0 = E0/c = 6.3/(3x10^8) = 2.1x10^-8 T. With E along +y and propagation +x, B is along +z (k-hat), so B = 2.1x10^-8 k-hat T.

JEE Main Physics: Electromagnetic Waves questions with solutions

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