Electromagnetic Waves – JEE Main PhysicsPractice Questions & PYQs

The electromagnetic spectrum is organized by wavelength, where high-frequency, high-energy radiation occupies the shortest end of the spectrum and lower-frequency waves occupy the longer end. Within the given set, X-rays exhibit the shortest wavelength $\lambda_4$, while ultraviolet rays $\lambda_2$ follow as the next longest. Moving toward lower frequencies, infrared rays $\lambda_3$ are longer than ultraviolet rays, and microwaves $\lambda_1$ possess the longest wavelength of the four, establishing the ascending order as $\lambda_4 < \lambda_2 < \lambda_3 < \lambda_1$.

The displacement current between the plates of a parallel plate capacitor is directly proportional to the rate at which the potential difference changes. This relationship is defined by the equation $I = C \frac{dV}{dt}$, where $I$ represents the displacement current and $C$ is the capacitance.

To find the rate of change of the potential difference, we rearrange the equation to $\frac{dV}{dt} = \frac{I}{C}$. Using the provided values of 0.25 mA, which is 0.25 x 10^-3 A, and 2.5 μF, which is 2.5 x 10^-6 F, we substitute these into the expression to obtain $\frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}}$. Evaluating this division results in 100 Vs⁻¹.

In any electromagnetic wave propagating through free space, the magnitudes of the electric field and magnetic field vectors are directly related by the constant speed of light, $c$, according to the equation $E = cB$. Since the wave moves along the x-axis and the electric field is oriented along the y-axis, the magnetic field must oscillate along the z-axis to maintain the transverse properties of the wave.

To calculate the magnetic field, rearrange the fundamental relationship to solve for the magnetic component, yielding $B = \frac{E}{c}$. By substituting the given electric field of 9.3 Vm^-1 and the speed of light of 3 x 10^8 ms^-1 into this formula, you simply divide the electric field by the speed of light. Performing this calculation results in a magnetic field magnitude of 3.1 x 10^-8 T.

Displacement current in a parallel plate capacitor with a plate separation of 10 cm is uniform across the plates because it is determined by the rate of change of electric flux density. Since the total displacement current is equal to the external conduction current of 6 A that charges the capacitor, the current is distributed consistently over the entire plate area of 16 cm².

The current passing through a smaller cross-sectional area $A_0 = 3.2 \text{ cm}^2$ is proportional to the ratio of that specific area to the total plate area $A$. This calculation is performed using the relationship $I_{A_0} = \left( \frac{A_0}{A} \right) I$. Substituting the known values yields $I_{A_0} = \left( \frac{3.2}{16} \right) \times 6 = 1.2 \text{ A}$. Converting this result into the requested units provides a final value of $1200 \text{ mA}$.

When electromagnetic radiation strikes a non-reflecting surface, it exerts a radiation pressure equal to the intensity of the light divided by the speed of light. Because pressure is defined as force per unit area, we can relate the intensity $I$, the speed of light $c$, the force $F$, and the area $A$ through the relation $I/c = F/A$. The provided intensity of $360 \mathrm{W/cm^2}$ must be converted into SI units to be consistent, resulting in $360 \times 10^4 \mathrm{W/m^2}$ since $1 \mathrm{cm^2}$ corresponds to $10^{-4} \mathrm{m^2}$.

Using the given average force of $2.4 \times 10^{-4} \mathrm{N}$ and the speed of light $c = 3 \times 10^8 \mathrm{m/s}$, we establish the following equation:

$\frac{360 \times 10^4}{3 \times 10^8} = \frac{2.4 \times 10^{-4}}{A}$

Rearranging these terms to isolate $A$ gives the expression $A = \frac{2.4 \times 10^{-4} \times 3 \times 10^8}{3.6 \times 10^6}$. Performing this arithmetic leads to a surface area of $0.02 \mathrm{m^2}$.

The propagation speed of electromagnetic waves within a material medium is defined by the relationship $v = \frac{c}{\sqrt{\varepsilon_r \mu_r}}$, where $c$ is the speed of light in a vacuum, $\varepsilon_r$ is the relative permittivity, and $\mu_r$ is the relative permeability. By squaring both sides of this equation, we obtain the expression $\varepsilon_r \mu_r = \frac{c^2}{v^2}$, which allows us to rearrange for the relative permittivity as $\varepsilon_r = \frac{c^2}{v^2 \mu_r}$. Substituting the given values of $c = 3 \times 10^8 \text{ ms}^{-1}$, $v = 1.5 \times 10^8 \text{ ms}^{-1}$, and $\mu_r = 2.0$ leads to the equation $\varepsilon_r = \frac{(3 \times 10^8)^2}{(1.5 \times 10^8)^2 \times 2}$. Since the ratio of the squared speeds is 4, the calculation simplifies to $\varepsilon_r = \frac{4}{2}$, resulting in a final value of $\varepsilon_r = 2$.

The electromagnetic spectrum is ordered by photon energy, which is inversely proportional to wavelength according to the relationship $E = \frac{hc}{\lambda}$, meaning higher energy corresponds to shorter wavelengths. Infrared radiation constitutes the longest wavelength range, spanning from 1 mm to 700 nm. Moving towards higher energy, Ultraviolet radiation occupies 400 nm to 1 nm, while X-rays follow in the range of 1 nm to 10^-3 nm. Finally, Gamma rays represent the highest energy with the smallest wavelengths, categorized as less than 10^-3 nm.

The amplitude of the magnetic field in an electromagnetic wave is related to the electric field amplitude by the speed of light, such that the magnetic field magnitude is $B_0 = \frac{E_0}{C}$. Electromagnetic waves are transverse, meaning the direction of propagation is determined by the cross product of the electric and magnetic field vectors while the phase term remains identical for both fields. Since the given electric field oscillates along the $\hat{i}$ direction and the wave propagates in the positive $z$-direction, the magnetic field must oscillate along the $\hat{j}$ axis to satisfy the propagation geometry, resulting in:

$\vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j}$

The relationship between the electric field and the magnetic field in an electromagnetic wave is governed by the speed of light, where the amplitude of the electric field $E_0$ is the product of the magnetic field amplitude $B_0$ and the speed of light $c$. Using the provided amplitude $B_0 = 3.5 \times 10^{-7} \text{ T}$ and the speed of light $c = 3 \times 10^8 \text{ m/s}$, the calculation for the electric field amplitude yields $E_0 = (3.5 \times 10^{-7}) \times (3 \times 10^8) = 105 \text{ V/m}$.

Regarding orientation, electromagnetic waves are transverse, meaning the electric field, magnetic field, and the direction of wave propagation are mutually perpendicular. Since the wave propagates along the $x$-axis and the magnetic field oscillates along the $y$-axis, the electric field must oscillate along the $z$-axis. Combining these factors results in the expression $E_z = 105 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \text{ Vm}^{-1}$.

The intensity of an electromagnetic wave, representing the average energy flux, is determined by the expression $I = \frac{1}{2} \epsilon_0 E_0^2 c$. From the provided wave equation $E_y = 600 \sin(\omega t - k x)$, we extract the peak amplitude of the electric field as $E_0 = 600 \text{ V/m}$. Using the permittivity of free space $\epsilon_0 = 9 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2}$ and the constant speed of light $c = 3 \times 10^8 \text{ m/s}$, the calculation becomes $I = \frac{1}{2} \times 9 \times 10^{-12} \times (600)^2 \times 3 \times 10^8$. Simplifying this product results in an intensity of $486 \text{ W/m}^2$.

The intensity of an electromagnetic wave defines the average energy transported across a unit area perpendicular to the direction of propagation and is proportional to the square of the electric field amplitude. This relationship is determined using the permittivity of free space and the speed of light through the following expression:

$I = \frac{1}{2} \varepsilon_0 E_0^2 c$

The wave equation $E_y = (200 \text{ Vm}^{-1}) \sin[1.5 \times 10^7 t - 0.05 x]$ identifies the peak amplitude of the electric field as $E_0 = 200 \text{ Vm}^{-1}$. By squaring this amplitude, we obtain $E_0^2 = (200)^2 = 4 \times 10^4 \text{ V}^2 \text{ m}^{-2}$. Using the permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$ and the speed of light $c = 3 \times 10^8 \text{ ms}^{-1}$, the intensity calculation becomes:

$I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^4 \times 3 \times 10^8$

Completing this multiplication results in an intensity of $53.1 \text{ Wm}^{-2}$, which represents the energy flux carried by the wave.

Electromagnetic waves are transverse, characterized by mutually perpendicular oscillations of the electric field, magnetic field, and the direction of wave propagation. The magnitude of the magnetic field induction is related to the electric field amplitude by the expression $B = \frac{E}{c}$, where $c$ represents the speed of light in a vacuum. In the provided equation, the electric field oscillates with an amplitude of $40 \, \mathrm{N}\,\mathrm{C}^{-1}$ along the x-axis, denoted by the unit vector $\hat{i}$, while the wave travels in the positive z-direction. Because the direction of energy propagation is determined by the cross product of the electric and magnetic field directions, specifically $\hat{i} \times \hat{B} = \hat{k}$, the magnetic field must oscillate along the y-axis, represented by the unit vector $\hat{j}$. Applying these conditions to the wave function yields the magnetic field induction $\vec{B} = \hat{j} \frac{40}{c} \cos \omega \left(t - \frac{z}{c}\right)$.

The energy carried by an electromagnetic wave is stored in both its oscillating electric and magnetic fields. Because these fields share the energy equally, the total average energy density is determined by the following formula:

$\langle u \rangle = \frac{1}{2} \varepsilon_0 E_0^2$

In this expression, $\varepsilon_0$ is the permittivity of free space and $E_0$ is the peak amplitude of the electric field. Substituting the provided values, we calculate the energy density using $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and an electric field amplitude $E_0 = 50 \text{ Vm}^{-1}$. Squaring the amplitude gives $2500 \text{ V}^2\text{m}^{-2}$, which is then multiplied by $\varepsilon_0$ and a factor of one-half, yielding a total average energy density of $1.106 \times 10^{-8} \text{ Jm}^{-3}$.

The electromagnetic spectrum is ordered by wavelength, with gamma rays ($\lambda_1$), X-rays ($\lambda_2$), infrared waves ($\lambda_3$), and microwaves ($\lambda_4$) occupying distinct positions from the high-frequency, shortest wavelength extreme toward the low-frequency, longest wavelength region. Consequently, arranging these radiations in ascending order of their wavelengths results in the sequence $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$.

Radiation pressure represents the force per unit area exerted by electromagnetic radiation upon a surface that absorbs it completely. When light propagates through a medium with a refractive index $\mu$, its velocity is effectively reduced, which alters the momentum transfer compared to propagation in a vacuum. Under these conditions, the radiation pressure is determined by the relationship $P = \frac{I \cdot \mu}{c}$, where $I$ is the intensity of the incident wave and $c$ is the speed of light in a vacuum.

Substituting the provided values of $I = 6 \times 10^8 \, \text{W/m}^2$, $\mu = 3$, and $c = 3 \times 10^8 \, \text{m/s}$ into the equation, we get $P = \frac{6 \times 10^8 \times 3}{3 \times 10^8}$. By simplifying this expression, the terms involving $10^8$ and the factor of 3 \in the numerator and denominator cancel out, leaving a final result of $6 \, \text{N/m}^2$.

Electromagnetic waves propagate through air at the speed of light, denoted by $c$. This speed relates the wave's frequency $f$ and wavelength $\lambda$ through the fundamental equation $c = f \lambda$ where the speed of light is $3 × 10^8$ m/s.

Isolating the wavelength yields the expression $\lambda = \frac{c}{f}$ Substituting the given frequency of $60 × 10^6$ Hz, the calculation becomes $\lambda = \frac{3 \times 10^8}{60 \times 10^6} = 5\ \mathrm{m}$ which confirms the wave's length is 5 m.

Electromagnetic waves distribute their energy equally between the electric and magnetic field components, which is reflected in the equality of their respective energy densities:
$\frac{1}{2} \varepsilon_0 E^2 = \frac{B^2}{2 \mu_0}$
This balance holds true because the electric field and magnetic field are related by $E = cB$, and the wave speed is tied to the fundamental constants of free space. Furthermore, because electromagnetic waves carry momentum, they exert radiation pressure when they encounter and strike a surface, which is a consistent property of their physical nature.

Electromagnetic waves are transverse, meaning that the electric field, the magnetic field, and the direction of wave propagation are all mutually perpendicular. Given that the wave propagates along the $x$-direction and the electric field oscillates in the $y$-direction, the magnetic field must oscillate in the $z$-direction to satisfy the cross-product relationship between these vectors.

The amplitude of the magnetic field, $B_0$, is determined by the relationship $E_0 = c B_0$, where $E_0$ is the electric field magnitude and $c$ is the speed of light. Substituting the values $60$ Vm$^{-1}$ and $3 \times 10^8$ ms$^{-1}$, we find $B_0 = 60 / (3 \times 10^8)$, which equals $2 \times 10^{-7}$ T. To determine the spatial frequency, or wave number $k$, we use the definition $k = 2\pi / \lambda$. With a wavelength of $4$ mm, or $4 \times 10^{-3}$ m, this becomes $2\pi / (4 \times 10^{-3})$, simplifying to $\frac{\pi}{2} \times 10^3$ rad/m.

Combining these components into the standard wave equation form $B_z = B_0 \sin[k(x - ct)]$, we incorporate the calculated amplitude, the wave number, and the propagation speed. This results in the final equation B_z = 2 \times 10^{-7} \sin \[\frac{\pi}{2} × 10^3 (x - 3 × 10^8 t)]$ \hat{k}$ T.

The energy density of an electromagnetic wave is proportional to the square of its electric field, which follows a sinusoidal variation represented by $E = E_0 \sin(\omega t - kx)$. When calculating the density, this term becomes squared, resulting in a $\sin^2(\omega t - kx)$ dependency that simplifies via the trigonometric identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$. This transformation introduces an oscillating term with an angular frequency of $2\omega$, meaning the energy density oscillates at twice the frequency of the original electromagnetic wave.

Maxwell's equations dictate the propagation of electromagnetic waves in a vacuum, where the electric field, magnetic field, and wave propagation vector are mutually perpendicular. For plane waves, Faraday's law of induction establishes that the cross product of the wave propagation vector and the electric field vector is equal to the product of the angular frequency and the magnetic field vector, expressed as $\vec{K} \times \vec{E} = \omega \vec{B}$. Dividing both sides by the angular frequency to isolate the magnetic field provides the relationship $\vec{B} = \frac{1}{\omega} (\vec{K} \times \vec{E})$.