Electromagnetic Waves – JEE Main PhysicsPractice Questions & PYQs

In an electromagnetic wave propagating through a vacuum, energy is distributed equally between the oscillating electric and magnetic fields. This symmetry arises because the two fields are intrinsically linked by the fundamental properties of space, ensuring that the energy contained in the electric field remains equivalent to the energy contained in the magnetic field as the wave travels.

To demonstrate this relationship mathematically, we examine the average energy density for each field, which is expressed as $u_E = \frac{1}{2}\epsilon_0 E_0^2$ for the electric field and $u_B = \frac{B_0^2}{2\mu_0}$ for the magnetic field. Using the relationship between the peak electric field amplitude $E_0$ and the peak magnetic field amplitude $B_0$, defined as $E_0 = c B_0$, we can rewrite the electric energy density in terms of the magnetic field. Substituting $E_0$ into the expression yields $u_E = \frac{1}{2}\epsilon_0 (c B_0)^2 = \frac{1}{2}\epsilon_0 c^2 B_0^2$. Given that the speed of light $c$ relates to the vacuum permeability $\mu_0$ and permittivity $\epsilon_0$ by $c^2 = \frac{1}{\mu_0 \epsilon_0}$, we substitute this into our equation to get $u_E = \frac{1}{2}\epsilon_0 \left( \frac{1}{\mu_0 \epsilon_0} \right) B_0^2$, which simplifies directly to $u_E = \frac{B_0^2}{2\mu_0}$. Since this derivation results in an expression identical to that of $u_B$, the average energy densities of the two fields are confirmed to be equal.

The relationship between the electric field and the magnetic field in an electromagnetic wave is governed by the direction of wave propagation, which is defined by the wave vector $\vec{K}$. From the argument of the cosine term $6x + 8z$, the wave vector components are clearly $6\hat{i}$ and $8\hat{k}$. Normalizing this vector to find the direction of propagation yields the unit vector $\hat{n} = \frac{6\hat{i} + 8\hat{k}}{\sqrt{6^2 + 8^2}} = 0.6\hat{i} + 0.8\hat{k}$.

The magnetic field is determined by the cross product $\vec{B} = \frac{1}{c} (\hat{n} \times \vec{E})$. With the electric field oscillating as $10\hat{j}$, performing this cross product $\vec{B} = \frac{1}{c} ((0.6\hat{i} + 0.8\hat{k}) \times 10\hat{j})$ requires distributing the cross product over the vector components. Given the standard unit vector identities where $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{k} \times \hat{j} = -\hat{i}$, the calculation results in $0.6\hat{k} - 0.8\hat{i}$. Multiplying by the amplitude factor $\frac{10}{c}$ simplifies the vector expression to $\frac{1}{c}(6\hat{k} - 8\hat{i})$.

Finally, the time-dependent phase of the wave is given by the expression $\vec{K} \cdot \vec{r} - \omega t$. Because the magnitude of the wave vector is $\sqrt{6^2 + 8^2} = 10$, the angular frequency $\omega$ is equal to $10c$. Incorporating this frequency into the phase term yields $6x + 8z - 10ct$, which completes the magnetic field expression as $\vec{B} = \frac{1}{c}(6\hat{k} - 8\hat{i}) \cos(6x + 8z - 10ct)$.

The force exerted by electromagnetic radiation on a surface is given by $F = \frac{IA}{c}(1+R)$, where $I$ is the intensity of the incident light, $A$ is the surface area, $c$ is the speed of light, and $R$ is the fraction of incident light reflected. Given that $25\%$ of the energy is reflected, the coefficient of reflection is $R = 0.25$, while the remaining $75\%$ is absorbed.

Substituting the provided values $I = 50 \, W/m^2$, $A = 1 \, m^2$, $c = 3 \times 10^8 \, m/s$, and $R = 0.25$ into the equation:

$F = \frac{50 \times 1}{3 \times 10^8} (1 + 0.25) = \frac{50 \times 1.25}{3 \times 10^8} = \frac{62.5}{3} \times 10^{-8} \, N \approx 20.83 \times 10^{-8} \, N$

This calculated value is closest to $20 \times 10^{-8} \, N$.

Electromagnetic waves consist of oscillating electric and magnetic fields that are mutually perpendicular and travel perpendicular to the direction of wave propagation. The amplitude of the magnetic field $B_0$ is linked to the electric field amplitude $E_0$ by the speed of light $c$, expressed as $B_0 = \frac{E_0}{c}$. By substituting the given values, $B_0 = \frac{60}{3 \times 10^8}$, which results in $2 \times 10^{-7} \text{ T}$.

Since the wave propagates along the positive $z$-direction, the phase factor in the wave equation must be of the form $(kz - \omega t)$. The wave number $k$ is calculated as $k = \frac{2\pi f}{c} = \frac{2\pi \times 23.9 \times 10^9}{3 \times 10^8}$, which simplifies to $500 \text{ m}^{-1}$, or $0.5 \times 10^3 \text{ m}^{-1}$. Similarly, the angular frequency $\omega$ is calculated as $\omega = 2\pi f = 2\pi \times 23.9 \times 10^9$, giving $1.5 \times 10^{11} \text{ rad/s}$. Integrating these values into the wave equation provides the correct mathematical description for the magnetic field.

A stationary charge in an electromagnetic wave is unaffected by the magnetic field, so the force exerted on it arises entirely from the oscillating electric field associated with the wave. In such waves, the electric field amplitude relates to the magnetic field via $E = cB$, and because the provided magnetic components are mutually perpendicular, their combined electric field amplitude is $E_{\text{net}} = c\sqrt{B_0^2 + B_1^2}$. Calculating the root-mean-square force requires using the relation $F_{\text{rms}} = Q E_{\text{rms}}$, where $E_{\text{rms}} = \frac{E_{\text{net}}}{\sqrt{2}}$. Using the values $Q = 10^{-4} \text{ C}$, $c = 3 \times 10^8 \text{ m/s}$, $B_0 = 3 \times 10^{-5} \text{ T}$, and $B_1 = 2 \times 10^{-6} \text{ T}$, this expression yields a force of approximately 0.6 N.

The relationship between the electric field and the magnetic field in an electromagnetic wave is governed by the speed of light, where the electric field amplitude $E_0$ is calculated from the magnetic field magnitude $B_0$. As the wave phase is expressed by $(2 \times 10^7 z + 6 \times 10^{15} t)$, the electromagnetic disturbance propagates in the negative $z$-direction, represented by the unit vector $-\hat{k}$.

The vector product $\vec{E} \times \vec{B}$ must point along this propagation direction. With the magnetic field defined by $(2 \hat{i} + \hat{j})$, evaluating the candidate electric field vectors reveals that only $(-\hat{i} + 2 \hat{j})$ results in a propagation vector of $-\hat{k}$. Combining the derived amplitude with this directional information yields the field expression $\vec{E} = 4.8 \times 10^2 \cos(2 \times 10^7 z + 6 \times 10^{15} t)(-\hat{i} + 2 \hat{j}) \, \text{V/m}$.

Electromagnetic waves are mathematically described by the phase factor $\omega t - \vec{k} \cdot \vec{r}$, where the wave vector $\vec{k}$ determines the direction of propagation. By rewriting the given phase expression $\sin[\omega t + (6y - 8z)]$ as $\sin[\omega t - (-6y + 8z)]$, we identify the effective propagation vector as $-6\hat{j} + 8\hat{k}$. We find the unit direction vector $\hat{s}$ by dividing this propagation vector by its magnitude, $\sqrt{(-6)^2 + 8^2} = 10$. Performing this division, $\frac{-6\hat{j} + 8\hat{k}}{10}$ simplifies to $\frac{-3\hat{j} + 4\hat{k}}{5}$.

The relationship between the electric field and the magnetic field in an electromagnetic wave is governed by Faraday's law of induction, which dictates that the spatial variation of the electric field is tied to the time variation of the magnetic field through the equation $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$. Starting with the given electric field $\vec{E} = E_{0}\hat{i}\cos(kz)\cos(\omega t)$, we find its curl by calculating the partial derivative with respect to position $z$, which yields $-\hat{j} k E_{0}\sin(kz)\cos(\omega t)$. Setting the negative of this curl equal to the time derivative of the magnetic field, we then integrate with respect to time $t$ to obtain $\vec{B} = \hat{j} \frac{k E_{0}}{\omega} \sin(kz) \sin(\omega t)$. Finally, since the relationship between the angular frequency $\omega$ and the wave number $k$ is defined by the speed of light $C = \frac{\omega}{k}$, the ratio $\frac{k}{\omega}$ simplifies to $\frac{1}{C}$, resulting in the magnetic field vector $\vec{B} = \frac{E_{0}}{C}\hat{j}\sin(kz)\sin(\omega t)$.

Electromagnetic radiation exerts force on a surface because photons possess momentum. In the case of a perfectly absorbing surface, each photon deposits its entire momentum upon impact, making the total transferred momentum equivalent to the ratio of the total incident energy to the speed of light.

We calculate the power incident on the surface by multiplying the intensity by the area. Converting the intensity of $25 \, \text{W/cm}^2$ to SI units results in $25 \times 10^4 \, \text{W/m}^2$, while the area of $25 \, \text{cm}^2$ converts to $25 \times 10^{-4} \, \text{m}^2$. Multiplying these figures gives a power of $P = 625 \, \text{W}$.

The total energy incident over the specified duration is found by multiplying this power by the time in seconds. With a time of 40 minutes, the conversion is $40 \times 60 = 2400 \, \text{s}$, leading to a total energy of $E = 625 \times 2400 \, \text{J}$.

Finally, the momentum $p$ is calculated using the relation
$p = \frac{E}{c}$
Using the speed of light $c = 3 \times 10^8 \, \text{m/s}$, the expression becomes
$p = \frac{625 \times 2400}{3 \times 10^8}$
This calculation results in a momentum transfer of $5.0 \times 10^{-3} \, \text{N s}$.

In electromagnetic waves propagating through free space, the electric and magnetic fields are always perpendicular to each other and to the direction of wave motion. These fields are related by the constant speed of light $c$, such that the magnitude of the magnetic field is determined by the relationship $B = \frac{E}{c}$, while their directional orientation follows the cross product rule $\hat{E} \times \hat{B} = \hat{\text{propagation}}$.

Given the electric field is 6.3 V/m along the $\hat{j}$ axis and the wave propagates along the $\hat{i}$ axis, we calculate the magnitude of the magnetic field using the speed of light, which is approximately 3 × 10^8 m/s, yielding $B = \frac{6.3}{3 \times 10^8} = 2.1 \times 10^{-8}$ T. To satisfy the propagation condition, since $\hat{j} \times \hat{k} = \hat{i}$, the magnetic field must be directed along the $\hat{k}$ axis. Combining these findings, the magnetic field vector is $2.1 \times 10^{-8} \hat{k}$ T.

Because electromagnetic waves transmit energy, the intensity of the wave remains constant as it propagates from a vacuum into a dielectric medium. In a vacuum, the average intensity is given by $\frac{1}{2} \epsilon_0 E_0^2 c$, whereas inside the medium with refractive index $n$, the intensity is $\frac{1}{2} \epsilon_0 \kappa E^2 v$. By substituting the refractive index relationship $n^2 = \kappa$ and the wave velocity $v = c/n$ into the equality of these two intensities, the terms rearrange to show that the ratio of the electric field magnitude in vacuum to that in the medium is $\sqrt{n}$.

A parallel derivation applies to the magnetic component of the wave. The magnetic energy density is related to the electric field through $B = E/v$. Comparing the intensity expressions formulated for the magnetic field, $\frac{1}{2} \frac{B_0^2 c}{\mu_0}$ and $\frac{1}{2} \frac{B^2 v}{\mu_0}$, indicates that the square of the ratio of the magnetic fields $B_0/B$ is equal to the ratio of the velocities $v/c$. Since the velocity ratio is $1/n$, taking the square root reveals that the ratio of the magnetic field magnitude in vacuum to that in the medium is $1/\sqrt{n}$.

Electromagnetic waves transport energy, and the intensity of this radiation relates directly to the amplitudes of its electric and magnetic fields. The intensity is defined by the energy flux, which can be expressed in terms of the root-mean-square electric field as follows:

$I = \varepsilon_0 c E_{\text{rms}}^2$

Because the electric and magnetic fields in an electromagnetic wave are coupled by the speed of light through the relation $E_{\text{rms}} = c B_{\text{rms}}$, we can substitute this expression into the intensity formula to link the intensity directly to the magnetic field:

$I = \varepsilon_0 c^3 B_{\text{rms}}^2$

Rearranging this relationship to solve for the magnetic field gives the formula:

$B_{\text{rms}} = \sqrt{\frac{I}{\varepsilon_0 c^3}}$

Using the provided solar intensity value of 10^8 W/m^2, along with the physical constants for the permittivity of free space and the speed of light, the calculation yields a root-mean-square magnetic field strength that is approximately 10^-4 T.

In any electromagnetic wave propagating through a vacuum, the amplitude of the electric field oscillation, denoted as $E_0$, relates directly to the amplitude of the magnetic field oscillation, $B_0$, through the speed of light, $c$. This fundamental relationship is defined by the equation $E_0 = B_0 c$, reflecting how the two fields fluctuate in phase and are proportional to one another based on the constant velocity of the wave.

From the provided wave equation, the amplitude of the magnetic field is identifiable as $B_0 = 100 \times 10^{-6}$ Tesla. Substituting this value along with the speed of light $c = 3 \times 10^8$ m/s into the relationship produces the calculation $E_0 = (100 \times 10^{-6}) \times (3 \times 10^8)$ which results \in a maximum electric field of $3 × 10^4$ N/C.

The propagation speed of an electromagnetic wave is fundamentally governed by the dielectric properties of the medium. By comparing the phase arguments of the given electric field expressions, one can identify the wave speed in each region. The phase in air, represented by $2\pi v(\frac{z}{c} - t)$, implies a wave speed $c_1 = c$. The phase in the medium, $k(2z - ct)$, can be expanded to $2kz - kct$, where the angular frequency is $\omega = kc$ and the wave number is $2k$, resulting in a wave speed $c_2 = \frac{\omega}{k_{medium}} = \frac{kc}{2k} = \frac{c}{2}$.

Since the speed of light in a non-magnetic medium is inversely proportional to the square root of its relative permittivity, the relationship between speed and permittivity is defined as $c \propto \frac{1}{\sqrt{\varepsilon_r}}$. Taking the ratio of these speeds, $\frac{c_1}{c_2} = \sqrt{\frac{\varepsilon_{r2}}{\varepsilon_{r1}}}$, allows for the calculation of the permittivity ratio by squaring both sides. Given that $\frac{c_1}{c_2} = 2$, it follows that $\frac{\varepsilon_{r2}}{\varepsilon_{r1}} = 4$, which implies that $\frac{\varepsilon_{r1}}{\varepsilon_{r2}} = \frac{1}{4}$.

The intensity $I$ of an electromagnetic wave is related to the electric field amplitude $E_0$ by the formula $I = \frac{1}{2} \epsilon_0 c E_0^2$, which yields an amplitude of $E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$. For a wave propagating in the positive Y-direction, the spatial and temporal variation is described by the phase factor $(y - ct)$, and the direction of propagation $\hat{j}$ must satisfy the vector relationship $\hat{j} = \hat{E} \times \hat{B}$. Given that the electric field $\vec{E}$ is oriented along the $\hat{k}$ direction, the magnetic field $\vec{B}$ must be oriented along the $\hat{i}$ direction with magnitude $B_0 = \frac{E_0}{c}$ to satisfy the cross product and the wave properties, resulting in $\vec{B} = \frac{1}{c} E \hat{i}$.

The distance between two consecutive zeros of a sinusoidal electromagnetic wave in space corresponds to half of its wavelength, such that $\left|z_2 - z_1\right| = \frac{\lambda}{2}$, which implies $\lambda = 2 \left|z_2 - z_1\right|$. Using the fundamental wave equation $c = f\lambda$, where $c = 3 \times 10^8$ m/s is the speed of light in a vacuum and $f$ is the frequency, the expression for frequency becomes $f = \frac{c}{\lambda}$. Substituting the expression for wavelength yields $f = \frac{3 \times 10^8}{2 \left|z_2 - z_1\right|}$, which simplifies to $\frac{1.5 \times 10^8}{\left|z_2 - z_1\right|}$.

Faraday's law of induction defines the relationship between the electric field $\vec{E}$ and the magnetic field $\vec{B}$ as $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$. Given the electric field $\vec{E} = 2E_0 \cos(kz) \cos(\omega t) \hat{i}$, the curl is calculated as $\nabla \times \vec{E} = \frac{\partial}{\partial z} (2E_0 \cos(kz) \cos(\omega t)) \hat{j} = -2E_0 k \sin(kz) \cos(\omega t) \hat{j}$. Substituting this result into Faraday's law yields $-\frac{\partial \vec{B}}{\partial t} = -2E_0 k \sin(kz) \cos(\omega t) \hat{j}$, which simplifies to $\frac{\partial \vec{B}}{\partial t} = 2E_0 k \sin(kz) \cos(\omega t) \hat{j}$. Integrating this equation with respect to time $t$ gives $\vec{B} = \int 2E_0 k \sin(kz) \cos(\omega t) \hat{j} \, dt = \frac{2E_0 k}{\omega} \sin(kz) \sin(\omega t) \hat{j}$. Replacing the term $\frac{k}{\omega}$ with $\frac{1}{c}$ provides the final magnetic field expression $\vec{B} = \frac{2E_0}{c} \sin(kz) \sin(\omega t) \hat{j}$.

The energy associated with a quantum of electromagnetic radiation is calculated using the relationship $E = h\nu$, indicating that energy is directly proportional to the radiation's frequency. By arranging these radiations according to their position on the electromagnetic spectrum from the lowest to the highest frequency, we find that radio waves possess the least energy, followed by the visible light spectrum where yellow light has a lower frequency than blue light. Since X-rays have a significantly higher frequency than both visible light and radio waves, they contain the most energy per quantum, resulting in an ascending order of radio wave, yellow light, blue light, and X-ray.

The operational principle of a microwave oven relies on the phenomenon of resonance, where the electromagnetic energy of the waves matches the vibrational energy levels of water molecules. Because the energy of microwaves ($E$) falls directly within the range corresponding to these molecular vibrations, the incident radiation is efficiently absorbed by the substance. This absorption forces the water molecules into states of higher vibrational motion, which manifests as the internal thermal energy required to cook food.

For an electromagnetic wave, the electric field $\vec{E}$ and magnetic field $\vec{B}$ are always perpendicular to the direction of propagation and must lie in the plane transverse to that direction. Because the wave propagates in the $+x$ direction, the fields must be functions of the coordinate $x$ and time $t$, and the cross product $\vec{E} \times \vec{B}$ must point in the $+x$ direction. Calculating the cross product for the fields in the first option, where $\vec{E}$ is proportional to $(\hat{y} - \hat{z})$ and $\vec{B}$ is proportional to $(\hat{y} + \hat{z})$, yields:
$(\hat{y} - \hat{z}) \times (\hat{y} + \hat{z}) = (\hat{y} \times \hat{y}) + (\hat{y} \times \hat{z}) - (\hat{z} \times \hat{y}) - (\hat{z} \times \hat{z}) = 2\hat{x}$
This confirms the resulting direction of wave propagation is indeed $+x$, satisfying the necessary physical constraints.