Electromagnetic Waves – JEE Main PhysicsPractice Questions & PYQs

The speed of light, often denoted as $c$ in a vacuum, serves as a universal constant that remains entirely unaffected by the motion of the source emitting it, which is a fundamental postulate of special relativity. When considering light traveling through a medium, its velocity is strictly determined by the refractive index of that material, meaning the speed is also independent of the intensity or brightness of the light. Consequently, these physical principles confirm that the rate at which light propagates is not influenced by how the source moves or the power density of the wave itself.

In an electromagnetic wave traveling through a vacuum, the electric field amplitude $E_0$ and the magnetic field amplitude $B_0$ are directly related by the speed of light $c$ via the equation $E_0 = B_0 c$ where the speed of light is 3 x 10^8 m/s. To determine the peak electric field, substitute the given magnetic field amplitude of 6.0 x 10^-7 T into the expression. This leads to the equation $E_0 = 6.0 \times 10^{-7} \times 3 \times 10^8$ which results in a value of 18 x 10^1. Performing this multiplication provides a final electric field amplitude of 180 V/m.

The production of electromagnetic waves relies on distinct physical processes tailored to specific frequency ranges. Radio waves are generated by oscillating electrons in antennas, essentially through rapid acceleration and deceleration within those aerials. Microwaves are produced using specialized electronic vacuum tubes known as klystron valves. Furthermore, X-rays originate from high-energy transitions involving the inner-shell electrons of an atom, whereas gamma rays are emitted during the radioactive decay of the atomic nucleus.

Electromagnetic waves propagate in the direction defined by the cross product of the electric field and magnetic field vectors, expressed as $\vec{E} \times \vec{B}$. Given that the wave travels in the negative $z$ direction ($-\hat{k}$) and the electric field points along the positive $y$ direction ($\hat{j}$), the mathematical relationship is defined as $\hat{j} \times \vec{B} = -\hat{k}$. Because the cross product of the positive $y$ unit vector and the positive $x$ unit vector results in the negative $z$ unit vector, the magnetic field must be oriented along the positive $x$ direction.

Faraday's law of induction fundamentally distinguishes dynamic behavior from static electrostatics by linking electric fields directly to changing magnetic flux. In static scenarios, the magnetic field is constant, meaning the time derivative of flux is zero and the line integral of the electric field around a closed loop vanishes. However, under time-varying conditions, the existence of a changing magnetic flux induces an electric field, which is mathematically expressed by the equation $\oint \vec{E} \cdot d\vec{l} = -\frac{\partial \phi_B}{\partial t}$ that captures this specific dynamic phenomenon.

In an electromagnetic wave propagating through a vacuum, the total energy is shared equally between the oscillating electric and magnetic fields over time. Because the average electric energy density, $\langle u_E \rangle$, is mathematically equivalent to the average magnetic energy density, $\langle u_B \rangle$, the total average energy density is defined by the sum $\langle u_{\text{total}} \rangle = \langle u_E \rangle + \langle u_B \rangle = 2\langle u_E \rangle$. When calculating the ratio of the average electric energy density to this total value, the common terms simplify to leave a final result of $\frac{1}{2}$.

Electromagnetic waves propagating through a vacuum have a specific speed that links the oscillation frequency to the wave number, given by the ratio $\frac{\omega}{k}$. This same propagation speed is also determined by the ratio of the maximum electric field amplitude to the maximum magnetic field amplitude, expressed as $\frac{E_0}{B_0}$. Equating these two expressions leads to the following relation:

$\frac{\omega}{k} = \frac{E_0}{B_0}$

Cross-multiplying this identity directly yields the relationship $k E_0 = \omega B_0$.

The direction of propagation for an electromagnetic wave is determined by the cross product of the electric field vector and the magnetic field vector, specifically in the direction of $\vec{E} \times \vec{B}$. Given that the electric field is oriented along the negative $z$-axis, represented as $-\hat{k}$, and the magnetic field is directed along the positive $x$-axis, represented as $\hat{i}$, we calculate the propagation direction by performing the cross product $(-\hat{k}) \times (\hat{i})$. Because the cross product of the unit vectors $\hat{k}$ and $\hat{i}$ yields $\hat{j}$, the final result is $-\hat{j}$, indicating that the wave propagates along the negative $y$-axis.

Electromagnetic waves store energy in space through the interaction of their oscillating electric and magnetic fields. The electric energy density, which quantifies the energy per unit volume stored by the electric field $E$, is expressed as $U_E = \frac{\varepsilon_0 E^2}{2}$ in terms of the permittivity of free space $\varepsilon_0$. Similarly, the magnetic energy density associated with the magnetic field $B$ is determined by the permeability of free space $\mu_0$ and is given by the formula $U_B = \frac{B^2}{2\mu_0}$.

Electromagnetic waves propagate by alternating between electric and magnetic fields that share energy equally, a principle known as the equipartition of energy. The average energy density stored in the electric field is defined as $\langle u_E \rangle = \frac{1}{4}\epsilon_0 E_0^2$, while the average energy density stored in the magnetic field is defined as $\langle u_B \rangle = \frac{1}{4\mu_0} B_0^2$. Because these average energy densities are identical for any plane electromagnetic wave, the ratio of the average electric energy density to the average magnetic energy density is exactly 1.

The energy of an electromagnetic wave is inversely proportional to its wavelength, a relationship expressed by the equation $E = \frac{hc}{\lambda}$. Since the infrared waves used in optical communication have much shorter wavelengths compared to the longer-wavelength microwaves utilized in Radar technology, Assertion A is false. Conversely, because infrared radiation has a shorter wavelength and consequently a higher frequency than microwaves, it carries more energy, confirming that Reason R is true.

The electromagnetic spectrum is categorized by wavelength ranges, where microwaves occupy the longest region among these choices, extending from $0.1\,\text{m}$ to $1\,\text{mm}$. Infrared radiation follows in the spectrum, spanning from $1\,\text{mm}$ down to $700\,\text{nm}$, while ultraviolet radiation covers the range from $400\,\text{nm}$ to $1\,\text{nm}$. Finally, X-rays possess the shortest wavelengths in this set, specifically ranging from $1\,\text{nm}$ to $10^{-3}\,\text{nm}$, which completes the correct associations.

When a point source is located at the center of curvature of a hemispherical surface, all emitted light rays strike the inner reflecting surface at a normal incidence. For a perfectly reflecting surface, radiation pressure exerts a force that is double the momentum flux of the incident light, which is given by $F = \frac{2I}{c}$ per unit area, where $I$ is the intensity and $c$ is the speed of light. Because the hemispherical shape is symmetrical around the axis passing through the source, the lateral force components cancel out, and the net force is determined by the projection of the surface area onto the plane perpendicular to the axis, which equals $\pi R^2$.

The total force can therefore be expressed as $F = \frac{2I}{c} \cdot \pi R^2$. Substituting the intensity $I = \frac{P_0}{4\pi R^2}$ into this expression allows the geometry terms to cancel, resulting \in the simplified relation $F = \frac{2}{c} \left( \frac{P_0}{4\pi R^2} \right) \cdot \pi R^2 = \frac{P_0}{2c}$. By applying the given values of power $P_0 = 24 \text{ W}$ and speed of light $c = 3 × 10^8 \text{ m/s}$, the calculation becomes $F = \frac{24}{2 \times 3 \times 10^8} = 4 \times 10^{-8} \text{ N}$.

The intensity of electromagnetic radiation spreading spherically from a point source is determined by the total power radiated divided by the surface area of a sphere at that distance. Because electromagnetic waves carry energy through both oscillating electric and magnetic fields, the total intensity is evenly split between them. To find the intensity contribution of just the electric field component, one must calculate the total intensity and divide it by two.

The actual radiated power is 5% of 100 W, which equals 5 W. At a distance of 5 m, this energy spreads over a spherical surface area $A = 4 \pi r^2$. Substituting the distance, the area is $A = 4 \pi (5)^2 = 100\pi$ m$^2$. The total intensity at this location is $I = \frac{5}{100\pi} = \frac{1}{20\pi}$ W/m$^2$. Since the electric field accounts for exactly half of this energy transport, its specific intensity component is $I_{EF} = \frac{1}{2} \times \frac{1}{20\pi} = \frac{1}{40\pi}$ W/m$^2$.

The refractive index $\mu$ of a medium is defined as the ratio of the speed of light in a vacuum to its speed in the medium, expressed as $\mu = \frac{c}{v}$. With the speed reduced to $v = 0.2c$, we calculate the refractive index as $\mu = \frac{c}{0.2c} = 5$. Since the refractive index is also related to the relative permittivity $\epsilon_r$ and relative permeability $\mu_r$ by the equation $\mu = \sqrt{\epsilon_r \mu_r}$, we can square both sides to obtain $\mu^2 = \epsilon_r \mu_r$, which implies $\epsilon_r = \frac{\mu^2}{\mu_r}$. The requested ratio of relative permittivity to the refractive index is $\frac{\epsilon_r}{\mu} = \frac{\mu^2 / \mu_r}{\mu} = \frac{\mu}{\mu_r}$. Substituting the known values $\mu = 5$ and $\mu_r = 1$ into this expression, the ratio becomes $\frac{5}{1}$, which confirms that the value of $x$ is $5$.

Electromagnetic waves are applied in technology and medicine based on their specific frequency ranges and interactions with matter. Microwaves function as the foundation for radar systems used in aircraft navigation, while ultraviolet rays are harnessed for the precision required to reshape corneal tissue in Lasik eye surgery. Infrared radiation provides superficial heating effective in physiotherapy for pain relief, and high-energy X-rays are employed in radiotherapy to target and destroy malignant cancer cells.

An electromagnetic signal propagating through a medium travels at a speed determined by the ratio of its electric field amplitude to its magnetic field amplitude. Once this velocity is established, the scenario becomes a simple kinematic calculation, provided we account for the fact that the signal must cover the distance to the target and then traverse that same path back to the radar. Because the radar detects an echo, the total distance covered by the wave is exactly twice the separation distance between the radar and the target.

The velocity of the signal is found by dividing the electric field $E_0 = 2.25\ \text{V/m}$ by the magnetic field $B_0 = 1.5 \times 10^{-8}\ \text{T}$, which results in $v = 1.5 \times 10^8\ \text{m/s}$. Given the distance to the target is $3\ \text{km}$, the total round-trip distance equals $6\ \text{km}$, or $6 \times 10^3\ \text{m}$. Dividing this total distance by the velocity of the signal yields the time required for the echo to reach the radar:
$t = \frac{6 \times 10^3\ \text{m}}{1.5 \times 10^8\ \text{m/s}} = 4 \times 10^{-5}\ \text{s}$

The relationship between the electric field amplitude and the magnetic field amplitude in an electromagnetic wave is governed by the phase velocity, represented as $E_0 = v B_0$. By comparing the expression $E_z = 300 \sin(5\pi \times 10^3 x - 3\pi \times 10^{11} t)$ to the standard wave equation, the angular frequency is identified as $\omega = 3\pi \times 10^{11}$ rad/s and the wave number as $k = 5\pi \times 10^3$ rad/m. The phase velocity is determined using the ratio $v = \frac{\omega}{k}$.

Substituting these parameters results in a phase velocity of $v = 0.6 \times 10^8$ m/s. Given the electric field amplitude $E_0 = 300$ V/m, the magnetic field amplitude $B_0$ is calculated from the rearranged relationship:
$B_0 = \frac{E_0}{v} = \frac{300}{0.6 \times 10^8}$
This calculation yields a final magnetic field amplitude of $B_0 = 5 \times 10^{-6}$ T.

A charge moving through an electromagnetic wave experiences an electric force $F_E = qE$ and a magnetic force $F_B = qvB$. Because the electric and magnetic fields in an electromagnetic wave are inherently linked by the speed of light through the relationship $E = cB$, the magnetic field can be expressed as $B = E/c$. Substituting this into the magnetic force equation yields $F_B = qv(E/c)$.

Dividing the electric force by the magnetic force allows both the charge and the electric field magnitude to cancel out, resulting in the simplified ratio $F_E/F_B = c/v$. Using the provided values of $3 \times 10^8 \, \mathrm{ms}^{-1}$ for the speed of light and $3 \times 10^7 \, \mathrm{ms}^{-1}$ for the velocity of the charge, the ratio becomes $(3 \times 10^8) / (3 \times 10^7)$, which calculates to $10/1$. Consequently, the ratio of the electric force to the magnetic force is 10:1.

The electromagnetic spectrum encompasses a wide range of waves, each uniquely suited for specific practical applications based on its frequency, energy, and interaction with matter. Ultraviolet rays contain enough energy to disrupt biological processes, making them highly effective for sterilizing surgical instruments, while microwaves are well-suited for radar systems due to their ability to interact with distant objects. Infrared rays are crucial to the greenhouse effect by effectively retaining thermal energy within the atmosphere, and the high penetration power of X-rays allows them to diffract through atomic lattices, making them the standard tool for studying crystal structures. Aligning these established properties confirms that ultraviolet rays correspond to sterilization, microwaves to radar technology, infrared radiation to the greenhouse effect, and X-rays to the analysis of crystal arrangements.