Electromagnetic Waves – JEE Main PhysicsPractice Questions & PYQs

The electric field amplitude $E_0$ and the magnetic field amplitude $B_0$ in an electromagnetic wave are linked by the speed of the wave $v$ through the expression
$E_0 = v B_0$
Identifying the magnetic field amplitude from the given equation yields $B_0 = 5 \times 10^{-6} \mathrm{T}$. The speed of the wave is determined by the ratio of the time coefficient and the spatial coefficient in the wave equation, which results in
$v = \frac{4 \times 10^8}{5} = 8 \times 10^7 \mathrm{m/s}$
Substituting these values into the initial relationship provides the electric field amplitude of $4 \times 10^2 \mathrm{Vm}^{-1}$.

Radiation pressure is the physical consequence of the momentum carried by light. When a surface completely absorbs incident light normally, the force $F$ exerted by the radiation is related to the intensity $I$, the surface area $A$, and the speed of light $c$ by the equation $F = \frac{I A}{c}$. The time duration of 20 minutes is irrelevant to this relationship because the radiation pressure is defined by the energy flux, or intensity, rather than the total accumulated energy delivered over a period of time.

The energy flux is determined by rearranging the equation to $I = \frac{F c}{A}$. Substituting the provided values of $F = 7.2 \times 10^{-9} \text{ N}$, $c = 3 \times 10^8 \text{ m/s}$, and $A = 36 \times 10^{-4} \text{ m}^2$, the expression becomes:

$I = \frac{7.2 \times 10^{-9} \times 3 \times 10^8}{36 \times 10^{-4}}$

Evaluating this formula results in 600 W/m$^2$. Since the final unit requirement is W/cm$^2$, dividing the result by 10,000 converts the energy flux to 0.06 W/cm$^2$.

Electromagnetic wave propagation in a medium is fundamentally determined by the medium's ability to permit and store electrical and magnetic fields, quantified by relative permittivity, $\epsilon_r$, and relative permeability, $\mu_r$. The refractive index of the medium, $n$, acts as the bridge between these electromagnetic properties and the speed of light, defined by the relationship $n = \frac{c}{v}$, where $c$ is the speed of light in a vacuum and $v$ is the wave speed in the medium. By taking the speed of light in a vacuum as 3 x 10^8 m/s and the given wave speed in the medium as 2.0 x 10^8 m/s, the refractive index is calculated to be 1.5.

This refractive index is also related to the electromagnetic parameters of the material through the equation $n = \sqrt{\epsilon_r \mu_r}$. Substituting the known values into the equation, we have $1.5 = \sqrt{\epsilon_r \times 1.0}$. Squaring both sides of this equation isolates the relative permittivity, resulting in a value of 2.25.

Electromagnetic radiation emitted from a point source spreads out over the surface area of a sphere, implying that the intensity at any distance is determined by the total radiated power divided by $4 \pi r^2$. Given the bulb has a power rating of 200 W and an efficiency of 3.5%, the actual power converted into radiation is 7 W. At a distance of 4 m, the intensity is defined as $I = \frac{7}{4 \pi \times 4^2}$. The peak magnetic field $B_0$ relates to this intensity through the equation $I = \frac{B_0^2 C}{2 \mu_0}$, where the constants are $C = 3 \times 10^8 \text{ m/s}$ and $\mu_0 = 4 \pi \times 10^{-7} \text{ Tm/A}$. Rearranging the formula to solve for the magnetic field gives $B_0 = \sqrt{\frac{2 \mu_0 I}{C}}$, which yields $B_0 = 1.71 \times 10^{-8} \text{ T}$.

Electromagnetic waves are formed by oscillating electric and magnetic fields that exhibit a precise orthogonal geometric relationship with the wave's direction of travel. This spatial alignment is strictly governed by the cross-product relationship $\vec{V} = \vec{E} \times \vec{B}$, which requires the electric field vector, the magnetic field vector, and the velocity vector to all be mutually perpendicular. As these waves propagate, the total energy is balanced precisely, resulting in an equal division of energy between the oscillating electric and magnetic field components.

Intensity describes the energy flow of light, which relates to the amplitudes of the oscillating electric and magnetic fields through the formula:
$I = \frac{1}{2} \epsilon_0 c^3 B_0^2$
To determine the magnetic field amplitude, we rearrange this relationship to $B_0 = \sqrt{\frac{2I}{\epsilon_0 c^3}}$. Substituting the given values of 0.22 W/m^2 for intensity, 8.85 × 10^-12 C^2 N^-1 m^-2 for permittivity, and 3 × 10^8 m/s for the speed of light, we can solve for the field strength. The denominator requires calculating the cube of the speed of light, which is 27 × 10^24, and multiplying it by the permittivity. Dividing the doubled intensity by this product and taking the square root yields a magnetic field amplitude of approximately 43 × 10^-9 T.

When an AC voltage is applied across a capacitor, the opposition to the flow of current, known as capacitive reactance, is inversely proportional to both the angular frequency of the source and the capacitance of the device. This relationship is defined by $X_C = \frac{1}{\omega C}$. According to Ohm's law for AC circuits, this reactance also equals the ratio of the root-mean-square voltage to the root-mean-square current, expressed as $X_C = \frac{V}{I}$.

Equating these two expressions allows us to rearrange the variables to solve for capacitance:
$C = \frac{I}{V\omega}$
Substituting the given values, where the current is 6.9 μA, the voltage is 230 V, and the angular frequency is 600 rad/s, leads to the equation $C = \frac{6.9 \times 10^{-6}}{230 \times 600}$. Performing the division results in a final capacitance of 50 pF.

Electromagnetic waves consist of mutually perpendicular electric and magnetic field oscillations that propagate through space. For any electromagnetic wave traveling in a vacuum or air, the peak electric field strength $E_0$ and the peak magnetic field strength $B_0$ are linked by the speed of light $c$ through the relationship $B_0 = E_0/c$.

Looking at the given wave equation, the amplitude of the electric field is identified as 540 V/m. By substituting this value and the speed of light, 3 x 10^8 m/s, into the relationship, the peak magnetic field is calculated as:
$B_0 = \frac{540}{3 \times 10^8}$
Performing the division results in 180 x 10^-8 T, which simplifies to 18 x 10^-7 T.

The electromagnetic spectrum is organized by wavelength, where high-energy radiation consistently features shorter wavelengths compared to low-energy counterparts. Because gamma rays contain the most energy, they exhibit the smallest wavelength, progressing through X-rays and visible light until reaching microwaves, which possess the longest wavelength in this set. This physical relationship establishes the correct ascending order as $\lambda_{\text{\gamma-ray}} < \lambda_{\text{X-ray}} < \lambda_{\text{visible}} < \lambda_{\text{microwave}}$.

The relationship between electric field intensity $E$ and magnetic field intensity $H$ in a medium is governed by the wave propagation speed $v$, expressed as $E = vB$, where $B$ is the magnetic flux density. To find $B$ from the given magnetic intensity $H$, one uses the constitutive relation $B = \mu_0 \mu_r H$. Simultaneously, the speed of the electromagnetic wave in the dielectric medium is determined by $v = \frac{c}{n}$, where $n = \sqrt{\mu_r \epsilon_r}$ represents the refractive index of the medium.

By combining these principles, the expression for the electric field intensity becomes $E = \frac{c}{\sqrt{\mu_r \epsilon_r}} \times \mu_0 \mu_r H$. Substituting the known parameters, we calculate the magnetic flux density as $B = 4\pi \times 10^{-7} \ \mathrm{NA}^{-2} \times 1.61 \times 4.5 \times 10^{-2} \ \mathrm{Am}^{-1}$ and the refractive index as $n = \sqrt{1.61 \times 6.44} = \sqrt{10.3684} = 3.22$. Dividing the speed of light $c = 3 \times 10^8 \ \mathrm{ms}^{-1}$ by this refractive index gives the speed of the wave in the medium. Completing the calculation $E = \frac{3 \times 10^8 \ \mathrm{ms}^{-1}}{3.22} \times B$ yields an approximate magnitude for the electric field intensity of $8.48 \ \mathrm{Vm}^{-1}$.

Intensity from a point source follows the inverse-square law, where energy emitted by the bulb spreads uniformly over the surface of an expanding sphere, defined by the relationship $I = \frac{P}{4\pi r^2}$. Because intensity decreases as the distance from the source increases, the change in intensity between two points is determined by calculating the difference between the intensity at the closer distance and the intensity at the further distance.

The bulb has a total power of 110 W, but since only 10% is converted to visible radiation, the effective power is $P = 11$ W. To calculate the change in intensity from 1 m to 5 m, subtract the intensity at the further distance from the intensity at the closer distance using $\Delta I = \frac{P}{4\pi(1)^2} - \frac{P}{4\pi(5)^2}$. Substituting the effective power into the expression leads to $\Delta I = \frac{11}{4\pi} (1 - \frac{1}{25})$, which simplifies to $\Delta I = \frac{11}{4\pi} \times \frac{24}{25}$. This product further reduces to $\frac{264}{100\pi}$, equivalent to $\frac{264}{\pi} \times 10^{-2}$ W/m^2. Using the approximation $\pi \approx 3.14$ to divide 264 by 3.14 confirms that the value of 'a' is 84.

In electromagnetic waves, the relationship between the electric field amplitude $E_0$ and the magnetic field amplitude $B_0$ is governed by the speed of light $c$, expressed as $E_0 = c B_0$. The speed of the wave is calculated by taking the ratio of the angular frequency $\omega$ to the wave number $k$, yielding $c = \frac{1.5 \times 10^{11}}{0.5 \times 10^3} = 3 \times 10^8 \mathrm{m/s}$. Applying the given magnetic field amplitude $B_0 = 2 \times 10^{-8} \mathrm{T}$ results in an electric field amplitude of $E_0 = 3 \times 10^8 \times 2 \times 10^{-8} = 6 \mathrm{V/m}$. Since the wave propagates in the negative $x$ direction and the magnetic field oscillates along the $y$ axis, the electric field must oscillate along the $z$ axis to ensure that the cross product of the electric and magnetic fields aligns with the direction of propagation.

The interaction between an electromagnetic wave and an obstacle depends primarily on the ratio of the object's size to the wavelength. Reflection typically occurs when the object is much larger than the wavelength, while diffraction takes place when the dimensions are comparable to the wavelength. Since the object here has a size of $\frac{\lambda}{100}$, which is significantly smaller than the wavelength of the 3 GHz wave, the electromagnetic energy disperses in multiple directions through the phenomenon known as scattering.

The displacement current in a parallel plate capacitor is fundamentally driven by the changing electric field between the plates, which occurs when the voltage across them changes over time. The relationship between displacement current $I_d$, permittivity of free space $\epsilon_0$, plate area $A$, plate separation $d$, and the rate of change of voltage $\frac{dV}{dt}$ is defined by the formula $I_d = \frac{\epsilon_0 A}{d} \frac{dV}{dt}$.

To determine the unknown distance $x$ between the plates, rearrange this equation to solve for $d$, resulting in $d = \frac{\epsilon_0 A}{I_d} \frac{dV}{dt}$. Convert the given values into standard SI units: the area is $40 \times 10^{-4}\,\mathrm{m}^2$, the displacement current is $4.425 \times 10^{-6}\,\mathrm{A}$, the permittivity is $8.85 \times 10^{-12}\,\mathrm{C}^2\,\mathrm{N}^{-1}\,\mathrm{m}^{-2}$, and the rate of change of voltage is $10^6\,\mathrm{V/s}$. Substituting these values into the rearranged formula provides the following calculation:

$d = \frac{8.85 \times 10^{-12} \times 40 \times 10^{-4} \times 10^6}{4.425 \times 10^{-6}}$

Evaluating this expression results in $d = 8 \times 10^{-3}\,\mathrm{m}$. Comparing this result to the given form $x \times 10^{-3}\,\mathrm{m}$ identifies the value of $x$ as 8.

The practical utility of electromagnetic radiation is defined by the unique interaction between wave properties and their environment, which allows for specialized uses across the spectrum. Ultraviolet light is utilized to purify water by effectively neutralizing harmful microorganisms, while X-rays function as essential medical diagnostic tools for viewing internal structures like bones. Microwaves facilitate radar technology and long-range communication due to their specific wavelength characteristics, whereas infrared waves are relied upon to improve visibility in foggy conditions because they experience significantly less scattering than visible light.

Electromagnetic waves follow a fundamental relationship between their electric and magnetic field components. For a wave traveling through a vacuum at the speed of light $c = 3 \times 10^8 \text{ m/s}$, the ratio of the maximum electric field strength to the maximum magnetic field strength is defined by the equation $E_0 = c B_0$. Given a maximum electric field of $60 \text{ V/m}$, the magnetic field amplitude calculates to $B_0 = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \text{ T}$. Because the electric field vibrates in the $y$-direction and the wave propagates along the $x$-axis, the magnetic field must oscillate in the $z$-direction to satisfy the right-hand rule for the direction of energy propagation.

The wave equation depends on the wave number $k$, which relates to the wavelength $\lambda = 8 \text{ mm} = 8 \times 10^{-3} \text{ m}$ through the formula $k = \frac{2\pi}{\lambda}$. This calculation yields $k = \frac{2\pi}{8 \times 10^{-3}} = \frac{\pi}{4} \times 10^3 \text{ m}^{-1}$. Applying these values to the general wave expression $\sin(k(x - ct))$ with $c = 3 \times 10^8 \text{ m/s}$ results in the following field equations:

$E_y = 60 \sin\left[ \frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t) \right]\$ \hat{j} \text{ V m}^{-1}$

$B_z = 2 \times 10^{-7} \sin\left[ \frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t) \right]\$ \hat{k} \text{ T}$

The intrinsic impedance of free space, denoted by $\eta_0$, serves as the ratio between the amplitudes of the electric and magnetic field vectors in an electromagnetic wave. This value is determined by the permeability of vacuum $\mu_0$ and the speed of light $c$ via the relation $\eta_0 = \mu_0 c$, which yields approximately $377 \, \Omega$ given the standard constants.

For a uniform plane wave propagating in the positive $z$-direction, the magnetic field intensity vector $H$ is related to the electric field vector $E$ through the cross product expression $H = \frac{1}{\eta_0} (\hat{a}_z \times E)$. Distributing the propagation vector $\hat{a}_z$ across the electric field components requires applying the cross-product identities $\hat{a}_z \times \hat{a}_x = \hat{a}_y$ and $\hat{a}_z \times \hat{a}_y = -\hat{a}_x$ to transform the vector fields.

Substituting the electric field components $E_x = -301.6 \sin(kz-\omega t)$ and $E_y = 452.4 \sin(kz-\omega t)$ into the equation $H = \frac{1}{\eta_0} (E_x \hat{a}_y - E_y \hat{a}_x)$ allows for the calculation of the magnetic field coefficients. Dividing each component amplitude by $377$ results in $-0.8$ for the $y$-directed term and $-1.2$ for the $x$-directed term, yielding the magnetic intensity $H = -0.8 \sin(kz-\omega t) \hat{a}_y - 1.2 \sin(kz-\omega t) \hat{a}_x \, \text{Am}^{-1}$.

Electromagnetic waves transport energy through space, and the intensity is defined as the average power per unit area carried by the wave. This value is fundamentally related to the amplitude of the oscillating electric field, the permittivity of free space, and the speed at which the wave propagates. The average intensity of a plane electromagnetic wave is given by the expression $I = \frac{1}{2} \varepsilon_0 E_0^2 c$.

From the provided electric field equation, the peak amplitude is identified as $E_0 = 56.5 \text{ NC}^{-1}$. Using the constant speed of light $c = 3 \times 10^8 \text{ ms}^{-1}$ and the given vacuum permittivity $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2}$, the intensity calculation becomes $I = \frac{1}{2} \times 8.85 \times 10^{-12} \times (56.5)^2 \times 3 \times 10^8$. Squaring the field amplitude yields $3192.25$, and performing the multiplication results in an intensity of approximately $4.24 \text{ Wm}^{-2}$.

Electromagnetic waves transport energy through space by coupling time-varying electric and magnetic fields. The instantaneous energy density stored within the electric field is defined by the relation $U_{e} = \frac{1}{2} \varepsilon_{0} E^{2}$, whereas the energy density associated with the magnetic field is given by $U_{m} = \frac{B^{2}}{2 \mu_{0}}$. In free space, the relationship between the amplitudes of these fields is governed by the speed of light $c$, expressed as $E = cB$, where the speed of light is related to the fundamental constants of free space by $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$. Substituting the expression for $E$ \in terms of $B$ and these constants into the electric energy density formula yields $U_{e} = \frac{1}{2} \varepsilon_{0} \left( \frac{B}{\sqrt{\mu_{0} \varepsilon_{0}}} \right)^{2}$. Expanding this results \in $U_{e} = \frac{1}{2} \varepsilon_{0} \left( \frac{B^{2}}{\mu_{0} \varepsilon_{0}} \right)$, which simplifies directly to $U_{e} = \frac{B^{2}}{2 \mu_{0}}$. Because this simplified expression is mathematically identical to the magnetic energy density, it confirms that the average energy density of the electric field is equal to that of the magnetic field.

Electromagnetic waves are transverse in nature, meaning the electric field vector $\vec{E}$, the magnetic field vector $\vec{B}$, and the direction of wave propagation are mutually orthogonal to one another. Since the wave is propagating along the $y$-direction, neither the electric field nor the magnetic field can have a component along that axis, effectively restricting their oscillations to the $xz$-plane.

Because the cross product $\vec{E} \times \vec{B}$ determines the direction of propagation, the fields must be chosen such that they remain perpendicular to the $y$-axis and to each other. This geometry requires that if the electric field oscillates along the $x$-axis, the magnetic field must oscillate along the $z$-axis, or if the electric field oscillates along the $z$-axis, the magnetic field must oscillate along the $x$-axis.