Electromagnetic Waves – JEE Main PhysicsPractice Questions & PYQs

Although the wave has a frequency of 5 GHz, its speed in this dielectric medium is determined by the refractive index, which is calculated from the medium's relative permeability and relative permittivity through the relation $n = \sqrt{\mu_r \varepsilon_r}$. With both relative values equal to 2, the refractive index simplifies to 2. Using the relationship $v = \frac{c}{n}$, where $c$ is 300,000,000 m/s, the resulting velocity is $v = \frac{3 \times 10^8}{2} = 15 \times 10^7$ m/s, which indicates that the missing factor is 15.

When light hits a non-reflecting surface at normal incidence, it exerts a radiation pressure derived from the momentum transfer of the photons, defined by the equation $F = \frac{IA}{c}$, where $F$ is the average force, $I$ is the energy flux, $A$ is the area, and $c$ is the speed of light. Rearranging this to solve for the flux gives $I = \frac{Fc}{A}$, where the time span of 40 minutes provided in the problem is extraneous information. By substituting the given values $F = 2.5 \times 10^{-6} \text{ N}$, $c = 3 \times 10^8 \text{ m/s}$, and $A = 30 \text{ cm}^2$ into the expression $I = \frac{2.5 \times 10^{-6} \times 3 \times 10^8}{30}$, we arrive at an energy flux of $25 \text{ W/cm}^2$.

The magnitude of the electric field $E$ in an electromagnetic wave is related to the magnitude of the magnetic field $B$ by the speed of light $c$, governed by the equation $E = cB$. Substituting the speed of 3 × 10^8 m/s and the magnetic field strength of 8.0 × 10^-8 T, we calculate the electric field magnitude to be 24 V/m.

Determining the direction requires applying the cross-product rule for electromagnetic waves, where the direction of propagation is given by the orientation of the vector product $\hat{E} \times \hat{B}$. Given the wave travels along the y-direction and the magnetic field vector points in the z-direction, we must satisfy the condition that $\hat{E} \times \hat{z}$ points in the y-direction. Because the cross product of the unit vector $\hat{x}$ with $\hat{z}$ results in the negative y-direction, the electric field must be oriented in the negative x-direction to satisfy this relation. Applying this directional component to our calculated magnitude results in an electric field vector of -24 $\hat{x}$ V/m.

In any electromagnetic wave traveling through free space, the electric and magnetic fields are linked by the equation $B = \frac{E}{c}$, where the speed of light is 3 x 10^8 m/s. Although the frequency is 30 MHz, this value is unnecessary for finding the magnetic field strength when the electric field is 6 V/m. Substituting the known values into the equation $B = \frac{6}{3 \times 10^8}$ gives a magnetic field magnitude of $B = 2 \times 10^{-8}$ T. Aligning this result with the expression x x 10^-8 T, we conclude that $x = 2$.

Radiation pressure on a perfectly reflecting surface is equal to twice the wave intensity divided by the speed of light. Since the intensity of the electromagnetic wave is defined as $I = \frac{1}{2} \epsilon_0 E_0^2 c$, substituting this into the pressure equation simplifies the relationship to $P = \epsilon_0 E_0^2$. This value represents an instantaneous pressure exerted by the wave, meaning it remains independent of the duration of exposure or the surface area of the reflector.

From the wave equation, the electric field amplitude $E_0$ is 200 V/m. Substituting the permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}$ F/m and $E_0 = 200$ V/m into the simplified pressure formula gives $P = 8.85 \times 10^{-12} \times (200)^2$. Squaring 200 results in 40,000, and multiplying this by $8.85 \times 10^{-12}$ produces $354 \times 10^{-9}$ Pa. Matching this result to the expression $\frac{x}{10^9}$ confirms that the value of $x$ is 354.

Any traveling wave described by an equation of the form $E = E_0 \sin(kx - \omega t)$ moves with a phase velocity determined by its angular frequency $\omega$ and wave number $k$. The relationship defining this speed is given by the formula $v = \frac{\omega}{k}$.

In the expression $E = 50 \sin(500x - 10 \times 10^{10} t)$, the wave number is $500$ rad/m and the angular frequency is $10 \times 10^{10}$ rad/s. Dividing the angular frequency by the wave number yields $v = \frac{10 \times 10^{10}}{500}$, which simplifies to $2 \times 10^8$ m/s. Given that the speed of light in a vacuum is $c = 3 \times 10^8$ m/s, this calculated velocity can be written as $\frac{2}{3} \times 3 \times 10^8$ m/s, resulting in a wave speed of $\frac{2}{3} c$.

The force exerted on a charged particle moving through a magnetic field is determined by the magnetic component of the Lorentz force, expressed as $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$. Because the velocity $\mathbf{v} = 0.5c \hat{i}$ is identical for both charges, the specific force experienced by each particle depends solely on its charge magnitude and the local value of the magnetic field.

At time $t=0$, we evaluate the magnetic field at the specified coordinates. For charge $q_1$ located at $z = \frac{\pi}{k}$, the spatial component is $\cos(k \frac{\pi}{k} - 0) = \cos(\pi) = -1$. For charge $q_2$ located at $z = \frac{3\pi}{k}$, the spatial component is $\cos(k \frac{3\pi}{k} - 0) = \cos(3\pi) = -1$. Since both charges are subject to the same instantaneous magnetic field $\mathbf{B} = -\frac{B_0}{\sqrt{2}}(\hat{i}+\hat{j})$ and move with the same velocity, the magnetic cross product $\mathbf{v} \times \mathbf{B}$ is identical for both cases.

Consequently, the ratio of the force on $q_1$ to the force on $q_2$ simplifies to the ratio of the charge values:
$\frac{F_1}{F_2} = \frac{q_1 |\mathbf{v} \times \mathbf{B}|}{q_2 |\mathbf{v} \times \mathbf{B}|} = \frac{q_1}{q_2}$
Substituting the given values $q_1 = 4\pi \text{ C}$ and $q_2 = 2\pi \text{ C}$ into this expression gives a ratio of $\frac{4\pi}{2\pi} = \frac{2}{1}$. Thus, the final ratio of the forces acting on the two charges is $2:1$.

The energy density of an electromagnetic wave represents the amount of energy stored per unit volume and is determined by the amplitude of the electric field. This energy density is given by the expression $u = \frac{1}{2} \epsilon_0 E_0^2$, where $E_0$ is the peak electric field strength and $\epsilon_0$ is the permittivity of free space. Since the total energy $U$ contained in a volume $V$ is simply the product of energy density and volume, we have $U = V \times \frac{1}{2} \epsilon_0 E_0^2$. Rearranging this relationship to solve for the volume gives:

$V = \frac{2U}{\epsilon_0 E_0^2}$

Substituting the provided values of $U = 5.5 \times 10^{-12} \, \mathrm{J}$, $E_0 = 50 \, \mathrm{NC}^{-1}$, and $\epsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-2}$ into the formula yields:

$V = \frac{2 \times 5.5 \times 10^{-12}}{8.85 \times 10^{-12} \times 50^2} \, \mathrm{m}^3$

The powers of ten cancel out, allowing the expression to be simplified as $V = \frac{11}{8.85 \times 2500} \, \mathrm{m}^3$, which results in approximately $4.97 \times 10^{-4} \, \mathrm{m}^3$. Converting this value into cubic centimeters by multiplying by $10^6$ gives roughly $497 \, \mathrm{cm}^3$, which corresponds to the final value of 500 cm$^3$.

The magnetic force experienced by a charged particle moving through a magnetic field is governed by the relationship $F = qvB\sin\theta$. When an electron moves at a right angle to the propagation of an electromagnetic wave, the angle between its velocity and the magnetic field is $90^\circ$, meaning the force reaches its maximum value as $\sin\theta = 1$. The required formula for this maximum force is $F_{max} = evB_0$, where $e$ is the elementary charge, $v$ is the electron speed, and $B_0$ is the peak magnetic field strength associated with the light beam.

From the provided electric field equation $E = 800 \sin\left(\omega t - \frac{x}{c}\right)$, the peak electric field strength is identified as $E_0 = 800 \, \text{Vm}^{-1}$. The magnetic field amplitude relates to the electric field via the speed of light $c$, yielding $B_0 = \frac{E_0}{c} = \frac{800}{3 \times 10^8} \, \text{T}$. Substituting the electron speed of $3 \times 10^7 \, \text{ms}^{-1}$ and the elementary charge of $1.6 \times 10^{-19} \, \text{C}$ into the force equation results in $F_{max} = (1.6 \times 10^{-19}) \times (3 \times 10^7) \times \frac{800}{3 \times 10^8}$. Simplifying this calculation leads to a maximum magnetic force of $12.8 \times 10^{-18} \, \text{N}$.

When an electromagnetic wave strikes a perfectly reflecting boundary at normal incidence, it undergoes reflection, which forces the direction of propagation to invert. While the incident wave is described by a spatial coordinate and time variable with opposite signs within the cosine function, indicating motion in the positive $z$-direction, the reflected wave travels in the negative $z$-direction. Mathematically, this change in direction is represented by switching the sign between the spatial and temporal terms in the wave equation, resulting in an expression where these components are added. Consequently, the reflected wave is defined by the function 3.1 \cos\[(1.8)z + (5.4 × 10^6)t]$ \hat{i}$ N/C.

The intensity of electromagnetic radiation spreading from a source is directly proportional to the square of the electric field intensity, while intensity itself is defined as power divided by surface area. Because the distance from the source remains 3 meters in both cases, the area through which the radiation passes is identical, making the power directly proportional to the square of the electric field intensity. Consequently, the electric field strength relates to the power by the relationship $E \propto \sqrt{P}$.

Comparing the radiation from the 60 W source to the 100 W source requires calculating the ratio of their electric field intensities based on the square root of their respective power outputs. The ratio of the new field intensity to the original field $E$ is $\sqrt{\frac{60}{100}}$. Simplifying the fraction under the square root yields $\sqrt{\frac{6}{10}}$, which is equivalent to $\sqrt{\frac{3}{5}}$. By equating the derived field intensity $\sqrt{\frac{3}{5}}E$ with the given form $\sqrt{\frac{x}{5}}E$, it is clear that $x$ is 3.

The direction of propagation for an electromagnetic wave is determined by the cross product of the electric field vector and the magnetic field vector, expressed as $\vec{E} \times \vec{B}$. Given the electric field vector $\hat{i}$ and the magnetic field vector $\hat{k}$, computing the cross product $\hat{i} \times \hat{k}$ results in $-\hat{j}$, which defines the path along which the wave travels.

The propagation of an electromagnetic wave with a frequency of 100 MHz is defined by the interaction between the electric field, the magnetic field, and the direction of wave travel. The magnitude of the electric field relates to the magnetic field through the constant speed of light $c = 3.0 \times 10^8$ m/s. Given the magnetic field magnitude $B = 2.0 \times 10^{-8}$ T, the electric field is calculated using the equation:

$E = cB = (3.0 \times 10^8) \times (2.0 \times 10^{-8}) = 6.0 \text{ V/m}$

The direction of the electric field is determined by the cross product relationship $\hat{E} \times \hat{B} = \hat{v}$, where $\hat{v}$ is the direction of propagation. Substituting the given directions, $\hat{E} \times \hat{k} = \hat{i}$, which indicates that the electric field must point along the $\hat{j}$ axis. Therefore, the electric field vector is 6.0 $\hat{j}$ V/m.

Electromagnetic waves originate from distinct physical mechanisms tied to their specific energy regimes. Microwaves are generated by electron oscillations within specialized devices known as magnetrons, whereas infrared radiation arises from the vibrations of atoms and molecules. While the radioactive decay of atomic nuclei acts as the source for gamma rays, X-rays are typically emitted when electrons undergo transitions within the inner energy shells of an atom.

The speed at which electromagnetic waves travel through a medium depends on the material's specific electrical and magnetic properties, defined by its relative permittivity and relative permeability. This velocity is calculated using the equation $v = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$, where $c$ represents the speed of light \in a vacuum, approximately $3 × 10^8$ m/s. By substituting the given values of $\mu_r = 1$ and $\varepsilon_r = 81$ into this expression, the calculation becomes $v = \frac{3 \times 10^8}{\sqrt{1 \times 81}} = \frac{3 \times 10^8}{9}$. Dividing the speed of light by the square root of 81 results \in a velocity of $3.33 × 10^7$ m/s.

The wavelength of an electromagnetic wave changes upon entering a dielectric medium because the wave's phase velocity is reduced relative to its speed in vacuum. For a non-magnetic material, this reduction is governed by the refractive index $n = \sqrt{\varepsilon_r}$. With the relative permittivity $\varepsilon_r = 2.25$, the refractive index is $n = \sqrt{2.25} = 1.5$. First, determine the vacuum wavelength using the relationship $\lambda = \frac{c}{f}$, where the speed of light $c = 3 \times 10^8$ m/s and the frequency $f = 3 \times 10^9$ Hz. This calculation yields a vacuum wavelength of $\lambda = 0.1$ m. Since the wavelength in the medium is related to the vacuum wavelength by the refractive index via $\lambda_m = \frac{\lambda}{n}$, substituting the known values results in $\lambda_m = \frac{0.1}{1.5} = \frac{1}{15}$ m, which is approximately $0.0667$ m. Converting this measurement into centimeters gives $6.67$ cm, which corresponds to $667 \times 10^{-2}$ cm.

The energy of an X-ray photon is defined by the relationship $E = \frac{hc}{\lambda}$, while the mass-energy equivalence principle relates this energy to the mass of a fictitious particle via $E = mc^2$. By setting these two expressions equal to each other, we determine that the mass $m$ is given by $m = \frac{h}{c\lambda}$. Substituting the known values of $\lambda = 10 \times 10^{-10}\,\text{m}$ and $c = 3 \times 10^8\,\text{m/s}$ into this formula results in $m = \frac{h}{(3 \times 10^8)(10 \times 10^{-10})}$. Evaluating the denominator yields $(3 \times 10^8)(10^{-9}) = 3 \times 10^{-1} = 0.3$, which simplifies the expression for mass to $m = \frac{h}{0.3} = \frac{10h}{3}\,\text{kg}$. Since the problem states that this mass is equal to $\frac{x}{3}h\,\text{kg}$, comparing the two values confirms that $x = 10$.

The dielectric constant of a medium is determined by characterizing the electromagnetic wave's propagation speed. Comparing the given field equation $E = 20 \cos(2 \times 10^{10}t - 200x)$ to the standard wave form $E = E_0 \cos(\omega t - kx)$ reveals an angular frequency of $\omega = 2 \times 10^{10} \text{ rad/s}$ and a wave vector of $k = 200 \text{ rad/m}$. The phase velocity $v$ within the medium is the ratio of these quantities, calculated as $v = \frac{\omega}{k} = \frac{2 \times 10^{10}}{200} = 10^8 \text{ m/s}$.

Calculating the refractive index involves comparing this velocity to the speed of light in a vacuum, $c = 3 \times 10^8 \text{ m/s}$, yielding $n = \frac{c}{v} = \frac{3 \times 10^8}{10^8} = 3$. This refractive index is connected to the medium's electromagnetic properties by the relation $n = \sqrt{\epsilon_r \mu_r}$. Inserting the non-magnetic condition $\mu_r = 1$ leads to $n = \sqrt{\epsilon_r}$, which implies that $\epsilon_r = n^2$. Substituting the value for $n$ results in $\epsilon_r = 3^2 = 9$.

Electromagnetic waves transport energy through space, and the average intensity at any given point is directly related to the amplitudes of the oscillating electric and magnetic fields. To find the relationship between intensity $I$ and the peak magnetic field $B_0$, we consider the permittivity of free space $\varepsilon_0$ and the speed of light $c$. The expression relating these quantities is given by $I = \frac{1}{2} \varepsilon_0 c^3 B_0^2$.

By rearranging this formula to solve for the peak magnetic field, we obtain the expression $B_0 = \sqrt{\frac{2I}{\varepsilon_0 c^3}}$. Using the provided values where the intensity $I = 0.092 \, \text{Wm}^{-2}$, the permittivity of free space $\varepsilon_0 = 8.85 × 10^{-12} \, \text{C}^{2} \text{N}^{-1} \text{m}^{-2}$, and taking the speed of light as $c = 3 × 10^8 \, \text{ms}^{-1}$, we substitute these values into the derived formula. Performing the calculation, which involves dividing the doubled intensity by the product of the permittivity and the cube of the speed of light, yields a peak magnetic field value of approximately $2.77 × 10^{-8} \, \text{T}$.

The energy flux of electromagnetic radiation emitted by a point source is distributed uniformly across the surface of an expanding sphere, represented by the intensity $I = \frac{P}{4\pi d^2}$. This intensity is also related to the peak electric field $E$ by the expression $I = \frac{1}{2} \epsilon_0 c E^2$, where $c$ is the speed of light in vacuum and $\epsilon_0$ is the permittivity of free space. Because the fundamental constants are related by $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, we can express the permittivity as $\epsilon_0 = \frac{1}{\mu_0 c^2}$, allowing us to substitute this value into the intensity equation to link the electric field directly to the magnetic permeability $\mu_0$ and the speed of light $c$.

Equating the two expressions for intensity results in $\frac{1}{2} \left( \frac{1}{\mu_0 c^2} \right) c E^2 = \frac{P}{4\pi d^2}$, which simplifies to $\frac{E^2}{2 \mu_0 c} = \frac{P}{4\pi d^2}$. Solving for the electric field, we find $E = \sqrt{\frac{2 P \mu_0 c}{4 \pi d^2}}$, which can be rearranged as $E = \sqrt{\frac{2P}{4d^2}} \sqrt{\frac{\mu_0 c}{\pi}}$. Substituting the given values of $P = 8$ W and $d = 10$ m into this equation, we get $E = \sqrt{\frac{16}{4 \times 100}} \sqrt{\frac{\mu_0 c}{\pi}} = \sqrt{\frac{16}{400}} \sqrt{\frac{\mu_0 c}{\pi}} = \frac{4}{20} \sqrt{\frac{\mu_0 c}{\pi}}$, which reduces to $\frac{2}{10} \sqrt{\frac{\mu_0 c}{\pi}}$ V/m. By comparing this result to the provided form $\frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}}$, it is evident that $x = 2$.