Electromagnetic Waves – JEE Main PhysicsPractice Questions & PYQs

The intensity of an electromagnetic wave is given by the relation $I = \frac{1}{2} c \varepsilon_0 E_0^2$, where $E_0$ is the amplitude of the electric field. Given the relationship between the electric field and magnetic field amplitudes in a vacuum, $E_0 = c B_0$, the intensity can be expressed in terms of the magnetic field amplitude $B_0$ as $I = \frac{1}{2} c^3 \varepsilon_0 B_0^2$. Rearranging this equation to solve for $B_0$ gives $B_0 = \sqrt{\frac{2I}{\varepsilon_0 c^3}}$. Substituting the provided values $I = 4.0 \times 10^{14} \, \text{W/m}^2$, $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2$, and $c = 3 \times 10^8 \, \text{m/s}$ yields $B_0 = \sqrt{\frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times (3 \times 10^8)^3}}$. Simplifying the expression inside the square root results in $B_0 = \sqrt{\frac{8 \times 10^{14}}{8.85 \times 10^{-12} \times 27 \times 10^{24}}} = \sqrt{\frac{8 \times 10^{14}}{238.95 \times 10^{12}}} = \sqrt{\frac{800}{238.95}} \approx \sqrt{3.348} \approx 1.83 \, \text{T}$.

Radio waves are generated by the rapid acceleration and oscillation of electrons in conducting circuits like antennas, whereas microwaves are typically produced by specialized electronic devices known as Magnetron valves. Infrared waves arise from changes in the internal vibrational and rotational energy states of atoms and molecules due to thermal excitation. High-energy X-rays result from electronic transitions involving inner-shell electrons in heavy atoms, where an electron jumps from a higher energy shell to fill a vacancy in a lower, tightly bound shell.

Maxwell's equations and foundational laws of electromagnetism describe the interaction between electric and magnetic fields. Faraday's law of electromagnetic induction is represented by $\oint \overrightarrow{E} \cdot d\overrightarrow{l} = -\frac{d}{dt} \oint \overrightarrow{B} \cdot d\overrightarrow{a}$, while the Ampere-Maxwell law, which generalizes Ampere's law by including displacement current for time-varying electric fields, is expressed as $\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 (I + \epsilon_0 \frac{d\varphi_E}{dt})$. Gauss's law of electrostatics relates electric flux through a closed surface to enclosed charge via $\oint \overrightarrow{E} \cdot d\overrightarrow{a} = \frac{1}{\epsilon_0} \int_{v} \rho \, dv$, and Ampere's circuital law defines the magnetic field circulation around a steady current as $\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I$, resulting in the correct mapping of A-II, B-III, C-IV, and D-I.

The magnitude of the magnetic field $B_0$ is determined by the relation $B_0 = E_0 / c$. Substituting $E_0 = 69$ V/m and $c = 3 \times 10^8$ m/s, the amplitude is $B_0 = 2.3 \times 10^{-7}$ T. Since the wave propagates in the $+x$ direction and the electric field oscillates along the $y$-axis, the magnetic field must oscillate along the $z$-axis to satisfy $\hat{E} \times \hat{B} = \hat{k}$. With the phase term remaining consistent, the magnetic field is $B_z = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t]$ T.

Electromagnetic waves carry linear momentum, which allows them to exert radiation pressure on any surface they strike. This pressure is fundamentally a result of momentum transfer between the field and the material, governed by the relation $P = \frac{I}{c}$, where $I$ represents intensity and $c$ is the speed of light. Although it is true that electromagnetic waves possess no rest mass, this absence of mass is an independent physical property and does not explain the origin of the pressure, which is caused specifically by the momentum carried by the waves.

The average intensity of an electromagnetic wave, which represents the average power per unit area, is given by the formula $I_{avg} = \frac{E_0^2}{2Z_0}$, where $E_0$ is the peak electric field amplitude and $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$ is the intrinsic impedance of free space. Substituting the provided values $E_0 = \sqrt{377}$ and $Z_0 = 377$ into this expression results in $I_{avg} = \frac{(\sqrt{377})^2}{2 \times 377}$. Simplifying this calculation yields $I_{avg} = \frac{377}{2 \times 377} = \frac{1}{2}$. By comparing this result to the given expression $\frac{1}{\alpha}$, it is determined that $\alpha = 2$.

The standard wave equation for an electromagnetic wave is $E(x, t) = E_0 \sin(\omega t - kx)$, where $\omega$ is the angular frequency and $k$ is the wave number. Comparing the given wave equation to this standard form yields $\omega = 2.0 \times 10^{15}$ rad/s and $k = 10^7$ rad/m.

The velocity of the wave in the medium, $v$, is calculated as:
$v = \frac{\omega}{k} = \frac{2.0 \times 10^{15}}{10^7} = 2.0 \times 10^8 \text{ m/s}$

The refractive index $n$ is defined by the ratio of the speed of light in a vacuum, $c \approx 3.0 \times 10^8$ m/s, to the speed of light in the medium $v$:
$n = \frac{c}{v} = \frac{3.0 \times 10^8}{2.0 \times 10^8} = 1.5$

The displacement current between the plates of a capacitor is directly proportional to the rate at which the potential difference across those plates changes over time. This relationship is defined by $I_d = C \frac{dV}{dt}$, where $I_d$ represents the displacement current, $C$ is the capacitance, and $\frac{dV}{dt}$ is the rate of change of voltage. In this scenario, the displacement current is 4.0 A and the capacitance is 6 μF, which is equivalent to $6 \times 10^{-6}$ F.

Given that the rate of change of potential is specified as $\alpha \times 10^6$ V/s, we substitute these known values into the equation:
$4 = (6 \times 10^{-6}) \times (\alpha \times 10^6)$
Since the product of $10^{-6}$ and $10^6$ is 1, the equation simplifies to $4 = 6\alpha$. Dividing both sides by 6, we find that $\alpha = \frac{4}{6}$, which simplifies to $\frac{2}{3}$ or approximately 0.67.

The speed of an electromagnetic wave in vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, while the speed in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$. By substituting the relative permittivity $\epsilon_r = \frac{\epsilon}{\epsilon_0} = 3$ and relative permeability $\mu_r = \frac{\mu}{\mu_0} = 2$, the speed in the medium is $v = \frac{1}{\sqrt{\mu_r \mu_0 \epsilon_r \epsilon_0}} = \frac{c}{\sqrt{\mu_r \epsilon_r}}$. The ratio of speeds in vacuum to the medium is therefore $\frac{c}{v} = \sqrt{\mu_r \epsilon_r} = \sqrt{2 \times 3} = \sqrt{6}$. This yields a final ratio of $\sqrt{6}:1$.

When a charged particle like an electron moves through an electromagnetic wave, it is subjected to both an electric force and a magnetic force. The magnitude of the electric force is $F_E = qE$, while the magnetic force acting on the moving electron is $F_B = qvB$. Because the electric field and magnetic field amplitudes in an electromagnetic wave are coupled by the speed of light $c$, they satisfy the relation $E = cB$.

Dividing the expression for the electric force by the magnetic force reveals that the ratio depends entirely on the speed of light relative to the velocity of the particle, specifically $\frac{F_E}{F_B} = \frac{qE}{qvB} = \frac{c}{v}$. Substituting the speed of light, 3 × 10^8 m/s, and the electron's speed, 1.5 × 10^6 m/s, into this ratio involves dividing 3 × 10^8 by 1.5 × 10^6, which results in 200.

A point source of electromagnetic waves radiates energy symmetrically in all directions, creating a spherical wavefront where the intensity depends strictly on the radial distance $R$ from the source. This relationship is defined by the formula
$I = \frac{P}{4 \pi R^2}$
where $P$ is the power of the source. Since the detector is moved to a new location on the same spherical surface, its radial distance from the source remains $L$, ensuring that the measured intensity at the new position is equal to the original value $I_0$.

The ratio of maximum conduction current density ($J_c$) to maximum displacement current density ($J_d$) in an electromagnetic wave is given by the formula $\frac{J_c}{J_d} = \frac{\sigma}{\omega \epsilon_0}$ where $\sigma$ is the conductivity, $\omega = 2\pi f$ is the angular frequency, and $\epsilon_0$ is the permittivity of free space. Given $\sigma = 10$, $f = 10^8$, and the relation $\epsilon_0 = \frac{1}{36\pi × 10^9}$ derived from the provided constant, the ratio is calculated as $\frac{J_c}{J_d} = \frac{10}{2\pi \times 10^8 \times \left( \frac{1}{36\pi \times 10^9} \right)}$ Simplifying the denominator yields $\frac{J_c}{J_d} = \frac{10}{\frac{2 \times 10^8}{36 \times 10^9}} = \frac{10}{\frac{1}{180}} = 1800$

The amplitude of the magnetic field relates to the electric field amplitude via the formula $B_0 = E_0 / c$. Substituting the given values results in $B_0 = 54 / (3 \times 10^8) = 1.8 \times 10^{-7}$ T. The wave propagates along the $z$-axis, which aligns with the direction of the vector product $\vec{E} \times \vec{B}$, corresponding to the unit vector $\hat{k}$. Since the electric field oscillates along $\hat{j}$, the relation $\hat{j} \times \hat{B} = \hat{k}$ confirms the unit vector of the magnetic field is $-\hat{i}$. Thus, the magnetic field vector is $\vec{B} = -1.8 \times 10^{-7} \sin(kz - \omega t) \hat{i}$ T.

The speed of an electromagnetic wave $v$ in a medium is determined by the ratio of the angular frequency $\omega$ to the wave number $k$ using the formula $v = \frac{\omega}{k}$, and this speed relates to the speed of light in a vacuum $c$ and the dielectric constant $\epsilon_r$ of a non-magnetic medium by the expression $v = \frac{c}{\sqrt{\epsilon_r}}$. Comparing the given equation $E_y = 20 \sin(3 \times 10^6 x - 4.5 \times 10^{14} t)$ to the standard form $E_y = E_0 \sin(kx - \omega t)$, we identify $\omega = 4.5 \times 10^{14} \text{ rad/s}$ and $k = 3 \times 10^6 \text{ m}^{-1}$. Calculating the velocity yields $v = \frac{4.5 \times 10^{14}}{3 \times 10^6} = 1.5 \times 10^8 \text{ m/s}$. Substituting the speed of light $c = 3 \times 10^8 \text{ m/s}$ into the relationship $\sqrt{\epsilon_r} = \frac{c}{v}$, we obtain $\sqrt{\epsilon_r} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$, which results in $\epsilon_r = 4$.

The intensity $I$ of an electromagnetic wave is related to the electric field amplitude by the expression $I = \frac{1}{2} c E_0 E^2$, which implies that the term $\sqrt{\frac{2I}{E_0 c}}$ represents the electric field. To determine its units, first express the intensity $I$ in fundamental units as $\frac{J}{s \cdot m^2}$, which simplifies to $\frac{N}{s \cdot m}$ because one Joule is equivalent to one Newton-meter. Given that the permittivity $E_0$ has units of $\frac{C^2}{N \cdot m^2}$ and the speed of light $c$ has units of $\frac{m}{s}$, their product $E_0 c$ results in $\frac{C^2}{N \cdot m \cdot s}$. Dividing the intensity by this product yields $\frac{I}{E_0 c} = \frac{N}{s \cdot m} \cdot \frac{N \cdot m \cdot s}{C^2} = \frac{N^2}{C^2}$, and taking the square root of this value produces the final unit of $\frac{N}{C}$, which is $\text{NC}^{-1}$.

The amplitude of the electric field $\vec{E} = 57(4 \hat{i} - 3 \hat{j}) \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3 x+4 y)]$ is calculated by finding the magnitude of the vector component, $E_0 = 57 \times \sqrt{4^2 + (-3)^2} = 57 \times 5$ N/C. Using the relationship $B_0 = \frac{E_0}{c}$ with the speed of light $c = 3 \times 10^8$ m/s, the amplitude of the magnetic field is $B_0 = \frac{57 \times 5}{3 \times 10^8}$ T. Since the magnetic field oscillates in phase with the electric field and must be perpendicular to both the electric field and the propagation direction $3 \hat{i} + 4 \hat{j}$, the resulting vector expression maintains the same cosine term and is aligned with the required direction. Combining these factors yields the magnetic field \vec{B} = \frac{57}{3 \times 10^8} \cos\[7.5 × 10^6 t - 5 × 10^{-3}(3 x+4 y)]$ (5 \hat{k})$.

The total electromagnetic energy stored within a cylinder is the product of its energy density and volume. Since the energy density depends on the square of the electric field amplitude, expressed as $u = \frac{1}{2} \varepsilon E^2$, the total trapped energy remains constant only if the product of the square of the amplitude and the cross-sectional area remains unchanged across the two cylinders.

The total energy can be equated as $U = (\frac{1}{2} \varepsilon E^2) \pi R^2 L$. Because the cylinder lengths are identical, equating the total energy of both scenarios allows the common terms to cancel, resulting in $E_1^2 R_1^2 = E_2^2 R_2^2$. Since the second cylinder possesses half the diameter of the first, its radius $R_2$ is equal to $R_1 / 2$. Substituting this relationship reveals that $E_2^2 (R_1 / 2)^2 = E_1^2 R_1^2$, which simplifies to $E_2^2 / 4 = E_1^2$. Consequently, the new electric field amplitude must be twice the initial value of 100, resulting in an amplitude of 200 NC^-1 and the wave equation $E = 200 \sin (\omega t - k x)$ NC^-1.

Electromagnetic waves are transverse in nature, requiring the electric and magnetic field oscillations to always occur perpendicular to the path of energy propagation. This direction is physically determined by the vector cross product $\vec{E} \times \vec{B}$. For a wave moving along the positive x direction, represented by the unit vector $\hat{i}$, the constituent fields must be oriented such that their cross product yields this specific orientation. Following the vector cross product rule

$\hat{j} \times \hat{k} = \hat{i}$

it becomes clear that an electric field oriented along the y axis and a magnetic field oriented along the z axis correctly satisfy the requirements for propagation along the x axis.

In an electromagnetic wave, the oscillating electric field vector $\vec{E}$ and magnetic field vector $\vec{B}$ are perpendicular to each other and to the direction of wave propagation, adhering to the fundamental relationship $\vec{E} = c(\vec{B} \times \hat{c})$. Given the phase argument $\omega(t - \frac{z}{c})$, the wave propagates along the positive $z$-axis, meaning the direction vector is $\hat{k}$. The amplitude of the electric field is directly proportional to the magnetic field amplitude by the speed of light, yielding a magnitude of $30c$.

To determine the spatial orientation of the electric field, we compute the cross product of the magnetic field's directional component with the propagation unit vector, $(\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) \times \hat{k}$. Applying the vector identities $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$, the resulting direction vector is $\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}$. Combining this orientation with the calculated magnitude and the original phase, the expression for the electric field is $\vec{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 30 c \sin \left[ \omega \left( t - \frac{z}{c} \right) \right]$.

Electromagnetic waves are not merely carriers of energy; they also possess linear momentum, which allows them to exert radiation pressure upon objects they strike. While the rest mass of a photon is zero, it still carries momentum because, in relativistic physics, momentum is defined by the relationship $p = \frac{E}{c}$, where $E$ is the energy and $c$ is the speed of light. Consequently, the assertion that electromagnetic waves lack momentum is false, while the statement regarding the zero rest mass of a photon is scientifically accurate.