Electromagnetic Waves – JEE Main PhysicsPractice Questions & PYQs

The energy densities associated with the electric and magnetic fields of an electromagnetic wave are defined as $u_E = \frac{1}{2} \epsilon E^2$ and $u_B = \frac{1}{2\mu} B^2$, respectively. By substituting the relationship $B = \frac{E}{v}$ and the velocity of light in a medium $v = \frac{1}{\sqrt{\mu\epsilon}}$ into the expression for magnetic energy density, we obtain $u_B = \frac{1}{2\mu} \left(\frac{E}{v}\right)^2 = \frac{\epsilon\mu E^2}{2\mu} = \frac{1}{2} \epsilon E^2$. This derivation demonstrates that the energy is shared equally between the electric and magnetic field components throughout the propagation of the wave.

The electromagnetic spectrum consists of various types of radiation, each characterized by distinct frequencies and unique interactions with matter that determine their practical applications. Infrared waves transmit thermal energy effectively, which is why they are utilized to treat muscular strain, while radio waves are employed for broadcasting because of their ability to propagate over long distances. High-energy X-rays penetrate soft tissue but are blocked by dense structures, allowing them to detect fractures of bones, and ultraviolet rays are absorbed by the ozone layer of the atmosphere to shield life from harmful solar radiation.

The amplitude of the electric field in an electromagnetic wave is directly related to the magnetic field amplitude through the speed of light, as described by the formula $E_0 = cB_0$. With the speed of light equal to 3 × 10⁸ m/s and the peak magnetic field given as 20 × 10⁻⁹ T, we can determine the electric field intensity by finding the product of these two values. Solving the equation $E_0 = (3 \times 10^8) \times (20 \times 10^{-9}) = 6$ reveals that the peak electric field strength is 6 V/m.

The polarization of an electromagnetic wave is fundamentally determined by the orientation of its oscillating electric field vector $\vec{E}$, which means the direction of polarization $\vec{X}$ is parallel to $\vec{E}$. Because electromagnetic waves are transverse and the direction of wave propagation $\vec{k}$ aligns with the flow of electromagnetic energy, this direction corresponds to the right-hand rule for the cross product of the oscillating field components, specifically $\vec{k} \parallel \vec{E} \times \vec{B}$.

The total average energy density of an electromagnetic wave arises from the combined contributions of the electric and magnetic fields, which store energy equally. By summing the average energy density of the electric field and the average energy density of the magnetic field, we arrive at a total energy density formula expressed as $u_{avg} = \varepsilon_0 E_{rms}^2$, where $\varepsilon_0$ is the permittivity of free space and $E_{rms}$ is the root-mean-square value of the electric field.

Using the provided value of $E_{rms} = 720\ \mathrm{N}/\mathrm{C}$ and the constant $\varepsilon_0 \approx 8.85 \times 10^{-12}\ \mathrm{F}/\mathrm{m}$, the calculation involves squaring the electric field intensity and multiplying by the permittivity. Squaring $720$ results in $518400$, and when this figure is multiplied by $8.85 \times 10^{-12}$, the final result is $4.58 \times 10^{-6}\ \mathrm{J} / \mathrm{m}^3$.

Photons carrying energy $E$ possess an inherent momentum defined by the expression $p = \frac{E}{c}$. When such a photon strikes a perfectly reflecting surface normally, it rebounds with the same magnitude of momentum but in the opposite direction. The total momentum transferred to the surface is the change in momentum, which is the difference between the final momentum and the initial momentum, resulting in a magnitude of $\frac{2E}{c}$.

An electromagnetic wave maintains its frequency when entering a dielectric medium because frequency is an intrinsic property determined solely by the source. Because the medium has a permittivity of 4.0, the wave speed is reduced by a factor of the refractive index, which is $\sqrt{4} = 2$, resulting in a new speed of $v_{\text{med}} = \frac{c}{2}$. Since wavelength is directly proportional to speed for a constant frequency, $\lambda = \frac{v}{f}$, this drop in velocity causes the wavelength to be halved.

Polarization is a phenomenon unique to transverse waves, occurring because these waves vibrate in directions perpendicular to their path of travel. Unlike longitudinal waves, which oscillate parallel to the direction of propagation and thus cannot be restricted to a single plane, transverse waves can be filtered to block oscillations in all but one specific orientation. Since electromagnetic waves demonstrate this ability to be polarized, it provides clear evidence that they possess the transverse oscillations necessary for this physical behavior.

Electromagnetic waves are defined by the oscillation of electric and magnetic fields that propagate through space without the transfer of matter. In contrast, $\beta$-rays are physically distinct because they consist of a fast-moving beam of electrons, which are subatomic particles possessing both mass and charge. Since these rays are streams of physical matter rather than wave-like field fluctuations, they cannot be classified as electromagnetic radiation.

Electromagnetic waves consist of oscillating electric and magnetic fields but carry no net charge, which renders them immune to the influence of external electric or magnetic fields. Because they are electrically neutral, these waves pass through such fields without experiencing any deflection. The relationship between the amplitude of the electric field $E_0$ and the magnetic field $B_0$ in free space is determined by the speed of light $c$, which is defined by the permittivity $\varepsilon_0$ and permeability $\mu_0$ as follows:

$E_0 = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} B_0$

The expression provided in the statement incorrectly relates the amplitudes using the intrinsic impedance of free space, rather than the velocity of the wave, making that assertion false.

In an electromagnetic wave propagating through a vacuum, the electric field, magnetic field, and propagation vector are mutually perpendicular and constrained by Maxwell's equations. The magnetic field vector is determined by the cross product of the propagation vector and the electric field, scaled by the reciprocal of the angular frequency, which is expressed by the equation $\vec{B} = \frac{1}{\omega} (\vec{k} \times \vec{E})$.

The amplitude of an electric field in an electromagnetic wave is related to the magnetic field amplitude by the speed of light through the relation $E_0 = cB_0$. Because electromagnetic waves propagate in the direction of the vector cross product $\vec{E} \times \vec{B}$, we identify the required unit vector by noting the phase $kx + \omega t$ implies propagation in the negative $x$-direction ($-\hat{i}$) while the magnetic field oscillates along the $y$-axis ($\hat{j}$). Solving the cross product $\hat{E} \times \hat{j} = -\hat{i}$ confirms that the electric field must be oriented along the $\hat{k}$ axis, leading to the expression $\vec{E} = B_0 c \sin(kx + \omega t) \hat{k}$.

For an electron to pass through perpendicular electric and magnetic fields without deflection, the net Lorentz force must be zero. This requires the electrostatic force $qE$ to be exactly balanced by the magnetic force $qvB$, which establishes the velocity selector condition $E = vB$.

The speed of the electron is determined from its kinetic energy, which is 728 eV. Converting this energy into Joules by multiplying by $1.6 \times 10^{-19}$ and equating it to the kinetic energy formula $\frac{1}{2}mv^2$, we use the mass of the electron $m = 9.1 \times 10^{-31} \text{ kg}$. Solving for the velocity yields $v = 16 \times 10^6 \text{ m/s}$.

Substituting the calculated speed $v = 16 \times 10^6 \text{ m/s}$ and the given magnetic field strength $B = 12 \text{ mT}$, or $12 \times 10^{-3} \text{ T}$, into the expression $E = vB$ gives an electric field magnitude of $192 \times 10^3 \text{ V/m}$. Expressed in terms of kilovolts per meter, the required field strength is 192 kVm$^{-1}$.

The propagation direction of an electromagnetic wave is determined by the cross product of the electric and magnetic field vectors, following the relation $\hat{E} \times \hat{B} = \hat{c}$. Since the phase term indicates the wave travels along the positive $x$-axis, the propagation unit vector is $\hat{c} = \hat{i}$, while the magnetic field oscillates along the $y$-axis represented by $\hat{j}$. Applying the vector cross product $\hat{E} \times \hat{j} = \hat{i}$ confirms that the electric field must oscillate along the $-\hat{k}$ direction. Finally, the relationship between field amplitudes is governed by $E_0 = c B_0$, where the wave speed $c = v \lambda$, yielding an electric field amplitude of $v \lambda B_0$. Integrating these components results in the expression $\vec{E} = -v \lambda B_0 \sin \left(2 \pi v t - \frac{2 \pi x}{\lambda}\right) \hat{k}$.

The magnetic field $\vec{B}$ of an electromagnetic wave is related to the electric field $\vec{E}$ and the direction of propagation $\hat{n}$ by the relation $\vec{B} = \frac{1}{c}(\hat{n} \times \vec{E})$, where $c$ is the speed of light. Given the frequency $\nu = \frac{3}{2\pi} \times 10^{12} \text{ Hz}$, the angular frequency is $\omega = 2\pi\nu = 3 \times 10^{12} \text{ rad/s}$, and the magnitude of the propagation vector is $k = \frac{\omega}{c} = \frac{3 \times 10^{12}}{3 \times 10^8} = 10^4 \text{ rad/m}$. With the propagation unit vector $\hat{n} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$ and the electric field $\vec{E} = E_0 \hat{k} \cos(\vec{k} \cdot \vec{r} - \omega t)$, the propagation vector is $\vec{k} = 10^4 \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$. The direction of the magnetic field is determined by the cross product $\hat{n} \times \hat{k} = \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \times \hat{k} = \frac{1}{\sqrt{2}}(\hat{i} \times \hat{k} + \hat{j} \times \hat{k}) = \frac{1}{\sqrt{2}}(-\hat{j} + \hat{i}) = \frac{\hat{i}-\hat{j}}{\sqrt{2}}$. Substituting these into the expression for $\vec{B}$ results in $\vec{B} = \frac{E_0}{c} \frac{\hat{i}-\hat{j}}{\sqrt{2}} \cos\left[ 10^4 \frac{\hat{i}+\hat{j}}{\sqrt{2}} \cdot \vec{r} - (3 \times 10^{12}) t \right]$.

In an electromagnetic wave propagating through free space, the magnitude of the magnetic field relates to the electric field through the speed of light using the relationship $|\vec{B}| = \frac{|\vec{E}|}{c}$. Taking the speed of light as $3 \times 10^8 \text{ m/s}$, we calculate the magnitude by dividing the electric field strength of $6.6 \text{ V/m}$ by this constant, which yields $2.2 \times 10^{-8} \text{ T}$.

To determine the orientation of the magnetic field, recall that the direction of electromagnetic wave propagation is given by the cross product of the electric and magnetic fields, denoted as $\vec{E} \times \vec{B}$. Given the wave travels in the $x$-direction (unit vector $\hat{i}$) and the electric field oscillates along the $y$-direction (unit vector $\hat{j}$), the relationship becomes $\hat{j} \times \vec{B} = \hat{i}$. Based on the properties of unit vector cross products, $\hat{j} \times \hat{k} = \hat{i}$, which confirms that the magnetic field must oscillate along the $z$-axis. Combining this direction with our previously calculated magnitude results in a magnetic field vector of $2.2 \times 10^{-8} \hat{k} \text{ T}$.

Electromagnetic waves emerge from the coupling of changing electric and magnetic fields, where a time-varying electric field induces a magnetic field and a changing magnetic field generates an electric field, forming a self-propagating disturbance. This fundamental principle correctly identifies the source of electromagnetic radiation. However, the speed of these waves within a material medium is defined by the medium's permittivity and permeability, represented by the equation $v = \frac{1}{\sqrt{\varepsilon_0 \varepsilon_r \mu_0 \mu_r}}$, which renders the expression provided in the second statement incorrect.

In an electromagnetic wave propagating through free space, the magnitudes of the electric field $\vec{E}$ and magnetic field $\vec{B}$ are related to the speed of light $c$ by the equation $E/B = c$. Given the electric field magnitude is 9.6 V/m and the speed of light is $3 \times 10^8$ m/s, the magnitude of the magnetic field is found by $B = E/c$, resulting in $3.2 \times 10^{-8}$ T. The direction of the magnetic field is determined by the cross product relation $\hat{B} = \hat{v} \times \hat{E}$, where $\hat{v}$ is the unit vector of propagation. Since the wave travels along the x-direction ($\hat{i}$) and the electric field is in the y-direction ($\hat{j}$), the magnetic field must lie along the z-direction ($\hat{k}$), calculated as $\hat{i} \times \hat{j} = \hat{k}$, yielding the final vector $\vec{B} = 3.2 \times 10^{-8}\hat{k}$ T.

Gauss's Law of electrostatics establishes that the total electric flux through a closed surface is determined by the enclosed charge, represented by the relationship $\oint \vec{E} \cdot d\vec{s} = \frac{q}{\epsilon_0}$. Conversely, Gauss's Law of magnetism asserts the non-existence of magnetic monopoles by stating that the net magnetic flux over any closed surface is always zero, expressed as $\oint \vec{B} \cdot d\vec{A} = 0$. Faraday's Law of induction describes how a time-varying magnetic field induces an electric field, which is mathematically formulated as $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$. Finally, the Ampere-Maxwell Law links the magnetic field around a closed loop to both conduction current and the rate of change of electric flux, shown as $\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$.

The energy contained within an electromagnetic wave is determined by the energy density associated with its oscillating electric field. For a wave defined by the equation $\vec{E} = 20 \sin \omega (t - \frac{x}{c}) \hat{j} \, \text{NC}^{-1}$, the amplitude of the electric field is $E_0 = 20 \, \text{NC}^{-1}$. The volumetric energy density for such an electric field is expressed by the formula $u = \frac{1}{2} \varepsilon_0 E_0^2$. Substituting the value of the permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2$ and the square of the amplitude gives $u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (20)^2$, which simplifies to $0.5 \times 8.85 \times 10^{-12} \times 400 = 200 \times 8.85 \times 10^{-12} \, \text{J m}^{-3}$. To find the total energy contained in the specified volume $V = 5 \times 10^{-4} \, \text{m}^3$, we multiply this energy density by the volume, resulting in $200 \times 8.85 \times 10^{-12} \times 5 \times 10^{-4}$. Calculating this product yields $1000 \times 8.85 \times 10^{-16}$, which equates to $8.85 \times 10^{-13} \, \text{J}$.