Electromagnetic Waves – JEE Main PhysicsPractice Questions & PYQs

When a light bulb emits radiation, it functions as a point source, directing electromagnetic waves outward into space. Only a fraction of the total electrical power consumed is converted into this radiant energy, which subsequently distributes itself uniformly across the surface of an expanding sphere. The relationship between the resulting intensity of the wave and the electric field generated is dictated by the energy density of the electromagnetic field at any given point.

The effective power radiated by the bulb is determined by its efficiency and total power consumption, calculated as $P = 0.0125 \times 1000 = 12.5$ W. At a distance of 2 m, this energy spreads over a spherical surface area $A = 4\pi (2)^2 = 16\pi$ m$^2$. The intensity $I$ of the radiation at this location is the power per unit area:
$I = \frac{P}{A} = \frac{12.5}{16\pi} \text{ W/m}^2$

To relate this intensity to the peak electric field $E$, we apply the standard energy density expression involving the speed of light $c = 3 \times 10^8$ m/s and the permittivity of free space $\varepsilon_0 = 8.854 \times 10^{-12}$ C$^2$ N$^{-1}$ m$^{-2}$:
$I = \frac{1}{2} \varepsilon_0 E^2 c$

Equating the intensity derived from the power output to the intensity formula allows us to isolate the electric field:
$E = \sqrt{\frac{2I}{\varepsilon_0 c}} = \sqrt{\frac{2 \times 12.5}{16\pi \times 8.854 \times 10^{-12} \times 3 \times 10^8}}$

Evaluating this calculation yields a peak electric field strength of approximately 13.7 V/m. Since the problem requires the value in the form $x \times 10^{-1}$ V/m, we express the result as $137 \times 10^{-1}$ V/m, confirming the value of $x$ is 137.

The conduction current density in a material relates to the electric field through Ohm's law, expressed as $J_c = \sigma E$, where $\sigma = 1/\rho$ is the conductivity. Given an electric field $E = \frac{V(t)}{d} = \frac{V_0 \sin(2 \pi f t)}{d}$, the conduction current density is $J_c = \frac{V_0 \sin(2 \pi f t)}{\rho d}$. Simultaneously, the displacement current density arises from the time-varying electric field and is given by $J_d = \varepsilon \frac{\partial E}{\partial t}$, where the permittivity is $\varepsilon = 80 \varepsilon_0$. Substituting the expression for the electric field, we find $J_d = \frac{80 \varepsilon_0}{d} \frac{\partial}{\partial t} [V_0 \sin(2 \pi f t)]$, which simplifies to $J_d = \frac{80 \varepsilon_0 V_0 (2 \pi f) \cos(2 \pi f t)}{d}$.

To determine how many times greater the conduction current density is compared to the displacement current density, we evaluate the ratio $\frac{J_c}{J_d} = \frac{\sin(2 \pi f t)}{\rho (80 \varepsilon_0) (2 \pi f) \cos(2 \pi f t)} = \frac{\tan(2 \pi f t)}{160 \pi \rho \varepsilon_0 f}$. For the given frequency $f = 900$ Hz and time $t = 1/800$ s, the phase is $2 \pi (900) (\frac{1}{800}) = 2.25 \pi$ radians. Since $\tan(2.25 \pi) = 1$, the ratio simplifies to $\frac{1}{160 \pi \rho \varepsilon_0 f}$. Using the provided constant $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{N} \cdot \text{m}^2 \text{C}^{-2}$, we can substitute $\varepsilon_0 = \frac{1}{36 \pi \times 10^9} \, \text{F/m}$.

Plugging these values into the ratio yields $\frac{1}{160 \pi \times 0.25 \times (\frac{1}{36 \pi \times 10^9}) \times 900}$. Simplifying the denominator, we have $40 \pi \times \frac{900}{36 \pi \times 10^9} = \frac{36000 \pi}{36 \pi \times 10^9} = \frac{1000}{10^9} = 10^{-6}$. Because the ratio $\frac{J_c}{J_d} = \frac{1}{10^{-6}} = 10^6$, the conduction current density is $10^6$ times the displacement current density, meaning $x = 6$.

Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation, with their amplitudes related by the speed of light $c$. The electric field amplitude is determined using the relation $E_0 = c B_0$, where the speed of light is 3 × 10^8 m/s and the magnetic field amplitude is 3 × 10^-8 T. Substituting these values into the calculation, $E_0 = (3 \times 10^8)(3 \times 10^{-8}) = 9$ V/m.

Regarding the direction, the argument of the sine function, $(1.6 \times 10^3 x + 48 \times 10^{10} t)$, indicates that the wave is traveling in the negative x-direction, which is represented by the unit vector $-\hat{i}$. Because the direction of energy propagation is given by the cross product $\vec{E} \times \vec{B}$, we must satisfy $\vec{E} \times \vec{B} = -\hat{i}$. Given that the magnetic field points along $\hat{j}$, we identify the required direction for the electric field using the cross product identity $\hat{k} \times \hat{j} = -\hat{i}$. Consequently, the electric field must oscillate along the $\hat{k}$ axis, resulting in the full expression $\vec{E} = 9 \sin(1.6 \times 10^3 x + 48 \times 10^{10} t) \hat{k}$ V/m.

The total energy density of an electromagnetic wave in a vacuum is shared equally between its electric and magnetic fields, with the magnetic contribution defined by the equation $u = \frac{B_0^2}{2\mu_0}$ where $B_0$ is the magnetic field amplitude and $\mu_0 = 4\pi × 10^{-7} \text{ T m/A}$. Rearranging this for the amplitude gives the expression $B_0 = \sqrt{2\mu_0 u}$, and substituting the given energy density $u = 1.02 × 10^{-8} \text{ J/m}^3$ yields the calculation $B_0 = \sqrt{2 \times 4\pi \times 10^{-7} \times 1.02 \times 10^{-8}}$. This simplifies to the equation $B_0 = \sqrt{256 \times 10^{-16}} \text{ T}$, which results in a magnetic field amplitude of 160 nT.

The relationship between the electric field and magnetic field amplitudes in an electromagnetic wave is governed by the speed of light, expressed by the equation $E_0 = cB_0$. Substituting the given values, the electric field amplitude is calculated as $E_0 = (3 \times 10^8)(3 \times 10^{-8}) = 9 \text{ V/m}$.

Determining the direction of the electric field requires identifying the direction of wave propagation relative to the magnetic field vector. The phase term $y+ct$ indicates that the wave travels along the negative y-axis, represented by the unit vector $-\hat{j}$. Since the direction of propagation is given by the cross product of the electric and magnetic fields, we must solve for $\hat{E}$ in the relation $\hat{E} \times \hat{i} = -\hat{j}$, where $\hat{i}$ represents the direction of the magnetic field. Using the properties of cross products, specifically that $\hat{k} \times \hat{i} = \hat{j}$, we conclude that $-\hat{k} \times \hat{i} = -\hat{j}$, meaning the electric field must oscillate along the negative z-axis. Combining the magnitude and direction, the complete expression for the electric field is $\vec{E} = -9 \sin[200 \pi (y+ct)] \hat{k} \text{ V/m}$.

In an electromagnetic wave, the electric field vector, magnetic field vector, and direction of propagation form a right-handed orthogonal system defined by the relationship that the propagation direction equals the cross product of the electric and magnetic field directions. Because the wave propagates in the direction $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$ and the electric field is polarized along $\hat{k}$, the magnetic field must satisfy the cross product $\hat{k} \times \hat{B} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$. Evaluating the cross product of $\hat{k}$ with the vector $\frac{\hat{i}-\hat{j}}{\sqrt{2}}$ yields $\frac{1}{\sqrt{2}}(\hat{k} \times \hat{i} - \hat{k} \times \hat{j})$, which simplifies to $\frac{\hat{j} + \hat{i}}{\sqrt{2}}$, confirming that this is the correct orientation for the magnetic field. Applying this directional vector to the wave's phase expression, which depends on the propagation direction, results in the magnetic field $B_0 \frac{\hat{i}-\hat{j}}{\sqrt{2}} \cos\left(\omega t - k \frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$.

In an electromagnetic wave, the amplitudes of the electric field and magnetic field are related by the equation $E_0 = c B_0$. Using the provided speed of light $c = 3 \times 10^8 \text{ m/s}$ and the magnetic field amplitude $B_0 = 1.2 \times 10^{-7} \text{ T}$, we calculate the electric field amplitude as $E_0 = (3 \times 10^8)(1.2 \times 10^{-7}) = 36 \text{ V/m}$.

The direction of wave propagation corresponds to the direction of the cross product $\vec{E} \times \vec{B}$. The argument of the sine function, $(0.5 \times 10^{3}x + 1.5 \times 10^{11}t)$, indicates that the wave propagates along the negative x-direction, denoted by the unit vector $-\hat{i}$. Because the magnetic field vector oscillates along the z-axis, represented by $\hat{k}$, the electric field vector $\vec{E}$ must be oriented such that its cross product with $\hat{k}$ yields the negative x-direction. Recalling that the cross product of unit vectors $\hat{j} \times \hat{k}$ results in $\hat{i}$, it follows that $(-\hat{j}) \times \hat{k}$ equals $-\hat{i}$, confirming that the electric field must lie along the negative y-axis. Since the electric and magnetic fields remain in phase, the resulting expression is $36 \sin(0.5 \times 10^{3}x + 1.5 \times 10^{11}t)$ directed along $-\hat{j}$.

The behavior of a charged particle moving through an electromagnetic field is determined by the Lorentz force law, where the magnetic component is given by $F = qvB$. To find the force, we must identify the magnetic field amplitude associated with the wave, which is inherently linked to the electric field amplitude via the phase velocity. Specifically, the ratio of the peak electric field to the peak magnetic field equals the phase velocity of the wave, $v_{wave} = \omega / k$, where $\omega$ is the angular frequency and $k$ is the wave number.

From the provided wave equation, $E = 30 \sin(1.5 \times 10^7 t - 5 \times 10^{-2} x)$, the phase velocity is calculated as $v_{wave} = (1.5 \times 10^7) / (5 \times 10^{-2}) = 3 \times 10^8 \text{ m/s}$. The corresponding magnetic field amplitude is then $B = 30 / (3 \times 10^8) = 10^{-7} \text{ T}$. Given the electron moves at a speed of $0.1c$, which is $3 \times 10^7 \text{ m/s}$, the maximum magnetic force is determined by substituting these values into the force expression: $F = (1.6 \times 10^{-19} \text{ C}) \times (3 \times 10^7 \text{ m/s}) \times (10^{-7} \text{ T}) = 4.8 \times 10^{-19} \text{ N}$.

The propagation of an electromagnetic wave is governed by the relationship $\hat{k} = \hat{E} \times \hat{B}$, where the direction of travel, the electric field vector, and the magnetic field vector form a mutually orthogonal set. In this specific wave, the electric field oscillates in the direction of $(\hat{x} + \hat{y})$ while the wave propagates along the positive $z$-axis, which is denoted by the unit vector $\hat{k}$. To determine the magnetic field, we must identify a vector that, when crossed with the electric field vector, results in a direction parallel to the propagation vector $\hat{k}$.

Evaluating the vector cross product, we find that $(\hat{x} + \hat{y}) \times (-\hat{x} + \hat{y})$ yields $2\hat{z}$, which aligns with the wave's direction of propagation. By applying the fundamental relation for electromagnetic waves where the magnitude of the magnetic field is related to the electric field by the speed of light $c$ as $B = \frac{E}{c}$, we incorporate the amplitude scaling factor. Consequently, the magnetic field is represented by $\vec{B} = \frac{E_0}{c}(-\hat{x} + \hat{y})\sin(kz - \omega t)$.

The electromagnetic spectrum is organized by wavelength, where energy and wavelength are related by the equation $E = \frac{hc}{\lambda}$, meaning higher energy radiation corresponds to shorter wavelengths. AM radio waves possess the longest wavelength in this group at 100 m, whereas microwaves have a shorter wavelength of 10^-3 m. Proceeding toward higher energies, X-rays are characterized by a wavelength of 10^-10 m, while gamma rays exhibit the shortest wavelength at 10^-15 m. Aligning each radiation type with its corresponding scale confirms the relationships established between the two columns.

The intensity of an electromagnetic wave is related to the root-mean-square electric field $E_{rms}$ through the equation:
$I = c \varepsilon_0 E_{rms}^2$
To solve for the field, we rearrange this to isolate $E_{rms}$:
$E_{rms} = \sqrt{\frac{I}{c \varepsilon_0}}$

Substituting the provided values $I = 315/\pi$, $\varepsilon_0 = 8.86 \times 10^{-12}$, and $c = 3 \times 10^8$ into the expression gives $E_{rms} = \sqrt{315 / (\pi \varepsilon_0 c)}$. By multiplying the numerator and denominator by 4, we introduce the constant $k = 1 / (4 \pi \varepsilon_0) = 9 \times 10^9$, which allows us to rewrite the expression as:
$E_{rms} = \sqrt{\frac{4 \times 315 \times 9 \times 10^9}{3 \times 10^8}}$

Simplifying the terms under the square root yields $E_{rms} = \sqrt{4 \times 315 \times 30}$, which is equal to $\sqrt{37800}$. Evaluating this value results in approximately 194.42, which rounds to the nearest integer of 194 V/m.

In an electromagnetic wave, the electric field $\vec{E}$ and magnetic field $\vec{B}$ are mutually perpendicular to each other and to the direction of propagation, with their magnitudes linked by the relationship $E = cB$. Given a magnetic field of $\vec{B} = 5 \times 10^{-8} \hat{j} \text{ T}$ and a speed of light of $c = 3 \times 10^8 \text{ m/s}$, the magnitude of the electric field is calculated as:
$E = (5 \times 10^{-8}) \times (3 \times 10^8) = 15 \text{ V/m}$
Because the wave propagates along the z-axis, designated by the unit vector $\hat{k}$, the cross product of the electric and magnetic fields must point in this direction. Since the magnetic field is oriented along the y-axis, the electric field must be oriented along the x-axis, defined by the unit vector $\hat{i}$, because the cross product $\hat{i} \times \hat{j} = \hat{k}$ aligns with the direction of propagation. Therefore, the corresponding electric field vector is $\vec{E} = 15 \hat{i} \text{ V/m}$.

The electromagnetic spectrum is organized by wavelength, where waves with lower frequencies possess longer wavelengths, while those with higher frequencies correspond to significantly shorter wavelengths. Arranging these parts from the longest to the shortest yields a clear sequence where radio waves have the greatest wavelength, followed by microwaves, then the visible light range, and finally, the high-energy, short-wavelength region occupied by X-rays. This hierarchy establishes the relationship $\lambda_{\text{radio waves}} > \lambda_{\text{micro waves}} > \lambda_{\text{visible}} > \lambda_{\text{x-rays}}$.

The relationship between the electric field and magnetic field amplitudes in a vacuum is governed by the speed of light $c$, where the magnetic field magnitude is given by $B_0 = E_0 / c$. Since the speed of light in free space is defined as $c = 1 / \sqrt{\mu_0 \varepsilon_0}$, substituting this value into the relationship results in an amplitude of $B_0 = E_0 \sqrt{\mu_0 \varepsilon_0}$. Because electromagnetic waves are transverse, the direction of propagation must align with the cross product of the electric and magnetic field vectors, following the relation $\hat{E} \times \hat{B} = \hat{Propagation}$. With the wave traveling along the $\hat{i}$ direction and the electric field oscillating along the $\hat{j}$ direction, the magnetic field must oscillate along the $\hat{k}$ direction to satisfy this cross-product requirement. Setting the time variable to zero simplifies the phase term $\cos(\omega t - kx)$ to $\cos(-kx)$, which is equivalent to $\cos(kx)$ due to the even nature of the cosine function. Combining these components, the magnetic field at the requested instant is $\vec{B} = E_0 \sqrt{\mu_0 \varepsilon_0} \cos(kx) \hat{k}$.

In an electromagnetic wave, the direction of propagation is defined by the cross product of the electric field vector and the magnetic field vector, represented by the unit vector relation $\hat{C} = \hat{E} \times \hat{B}$. We are provided with the electric field direction as $\hat{E} = \hat{k}$ and the magnetic field vector as $\vec{B} = 2\hat{i} - 2\hat{j}$. To determine the unit vector $\hat{B}$, we must divide $\vec{B}$ by its magnitude, which is calculated as $\sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$, resulting in $\hat{B} = \frac{1}{\sqrt{2}}(\hat{i} - \hat{j})$.

Calculating the propagation direction involves the cross product $\hat{C} = \hat{k} \times [\frac{1}{\sqrt{2}}(\hat{i} - \hat{j})]$. Distributing this cross product across the terms inside the brackets gives $\frac{1}{\sqrt{2}}(\hat{k} \times \hat{i} - \hat{k} \times \hat{j})$. Using the standard cyclic properties of unit vectors, where $\hat{k} \times \hat{i} = \hat{j}$ and $\hat{k} \times \hat{j} = -\hat{i}$, the expression simplifies to $\frac{1}{\sqrt{2}}(\hat{j} - (-\hat{i}))$, which yields $\frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$.

The instantaneous force experienced by a charged particle moving through electromagnetic fields is determined by the Lorentz force law, which sums the contributions from both the electric and magnetic components: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$. At the origin at time $t = 0$, the spatial and temporal arguments within the cosine functions evaluate to zero, simplifying $\cos(0)$ to $1$. Consequently, the electric field vectors at this point are $\vec{E}_1 = E_0 \hat{j}$ and $\vec{E}_2 = E_0 \hat{k}$.

To determine the magnetic fields associated with these plane waves, we use the relationship $\vec{B} = \frac{1}{c}(\hat{n} \times \vec{E})$, where $\hat{n}$ represents the direction of wave propagation. For the first wave propagating along the $x$-axis, the magnetic field is $\vec{B}_1 = \frac{1}{c}(\hat{i} \times E_0 \hat{j}) = \frac{E_0}{c}\hat{k}$. For the second wave propagating along the $y$-axis, the magnetic field is $\vec{B}_2 = \frac{1}{c}(\hat{j} \times E_0 \hat{k}) = \frac{E_0}{c}\hat{i}$.

With the particle velocity given as $\vec{v} = 0.8c \hat{j}$, we substitute these electric and magnetic components into the total force equation:
$\vec{F} = q(\vec{E}_1 + \vec{E}_2) + q(\vec{v} \times \vec{B}_1) + q(\vec{v} \times \vec{B}_2)$
Expanding this expression with the known vectors:
$\vec{F} = q(E_0 \hat{j} + E_0 \hat{k}) + q(0.8c \hat{j} \times \frac{E_0}{c} \hat{k}) + q(0.8c \hat{j} \times \frac{E_0}{c} \hat{i})$
Calculating the cross products, where $\hat{j} \times \hat{k} = \hat{i}$ and $\hat{j} \times \hat{i} = -\hat{k}$, the expression becomes:
$\vec{F} = qE_0 \hat{j} + qE_0 \hat{k} + 0.8qE_0 \hat{i} - 0.8qE_0 \hat{k}$
Grouping the terms by their unit vectors results in the final force:
$\vec{F} = qE_0(0.8 \hat{i} + \hat{j} + 0.2 \hat{k})$

The motion and force on a charged particle within an electromagnetic field are described by the Lorentz force expression $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$. Evaluating the electric field at the specified coordinate $z = \frac{\pi}{k}$ and time $t = 0$ yields a cosine term of $\cos(\pi) = -1$, which results in an electric field vector directed in the opposite sense of the wave's polarization, specifically $\vec{E} = -E_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.

Because the particle's instantaneous velocity is non-relativistic compared to the speed of light, the magnetic force contribution $q(\vec{v} \times \vec{B})$ remains negligible relative to the electric force term $q\vec{E}$. Consequently, the total force acting on the particle is dominated by the electric component. Since this electric field vector points in the direction exactly opposite to the polarization vector $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$, the resulting force vector is antiparallel to that direction.

Intensity of an electromagnetic wave characterizes the energy transfer rate per unit area, defined as the ratio of power to cross-sectional area. This intensity is also intrinsically linked to the maximum electric field amplitude through the relationship $I = \frac{1}{2} \varepsilon_0 c E_0^2$, where $\varepsilon_0$ denotes the permittivity of space and $c$ represents the speed of light.

Dividing the laser power of $27 \times 10^{-3}$ W by the cross-sectional area of $10 \times 10^{-6}$ m$^2$ results in an intensity of $2.7 \times 10^3$ W/m$^2$. Equating this to the electromagnetic wave formula gives $2.7 \times 10^3 = \frac{1}{2} \times (9 \times 10^{-12}) \times (3 \times 10^8) \times E_0^2$, which simplifies to $2.7 \times 10^3 = 13.5 \times 10^{-4} \times E_0^2$. Solving for $E_0^2$ yields $2 \times 10^6$, leading to $E_0 = \sqrt{2} \times 10^3$ V/m, which evaluates to approximately $1.4$ kV/m.

When a light wave hits a glass slab, the conservation of energy implies that the intensity transmitted into the glass is 96% of the incident intensity. Because the intensity of an electromagnetic wave is proportional to the product of the medium's refractive index and the square of the electric field amplitude, we relate the transmitted and incident fields using the equation:
$n_2 A_t^2 = 0.96 n_1 A_i^2$
Plugging in the refractive index of glass $n_2 = 1.5$ and air $n_1 = 1$ with an incident amplitude of $A_i = 30$ V/m, we get $1.5 A_t^2 = 0.96(1)(30)^2$. Solving this gives $A_t^2 = 0.64 \times 900 = 576$, which confirms the amplitude in the glass is 24 V/m.

Electromagnetic waves propagate such that the electric field, magnetic field, and the direction of wave travel form an orthogonal set, with the magnitudes related by the speed of light in vacuum, $c = 3 \times 10^8 \text{ m/s}$. Dividing the electric field magnitude by this speed determines the magnetic field strength, yielding $B = \frac{6}{3 \times 10^8} = 2 \times 10^{-8} \text{ T}$. Given the wave propagates along the $x$-axis and the electric field oscillates along the $y$-axis, the magnetic field must be oriented along the $z$-axis because the direction of propagation corresponds to the cross product $\vec{E} \times \vec{B}$, which satisfies $\hat{j} \times \hat{k} = \hat{i}$.