NEET Electromagnetic Waves Previous Year Questions with Solutions

For a non-magnetic lossless dielectric with relative permittivity epsilon_r = 9 and mu_r = 1:

• • The speed of the wave inside the medium is exactly 10^8 m/s.
• • Statement (1) is factually TRUE (hence not the answer to an "incorrect choice" question).

From the wave equation E_x = E_0 \sin(kz - \omega t), angular frequency is omega = 2*pi*10^6 rad/s, so frequency f is:

• • Wavelength inside the medium: \lambda_{\text{medium}} = \frac{v}{f} = \frac{10^8\text{ m/s}}{10^6\text{ Hz}} = 100\text{ m}.
• • Statement (2) asserts that wavelength inside the medium is 300 m (which is the wavelength in free space c/f = 300 m).
• • Therefore, Statement (2) is definitely INCORRECT.

By Maxwell's equations, the relationship between peak amplitudes of E and B in a medium is:

• • The magnetic field must be B_y = B_0 \sin(kz - \omega t) = \left(\frac{E_0}{v}\right) \sin(kz - \omega t).
• • Statement (3) mistakenly writes: B_y = \left(\frac{B_0}{v}\right) \sin(kz - \omega t), which divides B_0 by v again, producing incorrect units of [Tesla / (m/s)].
• • Therefore, Statement (3) is ALSO factually and dimensionally INCORRECT.
• • During answer key challenges, NTA accepted both challenges and declared BOTH Option (2) and Option (3) as correct answers (Key: B,C).

Microwaves have frequencies tuned to the resonant rotational frequency of water molecules in food:

• • Used universally in domestic and commercial microwave ovens.

UV radiation damages microbial DNA, acting as a powerful germicide:

• • Standard technology in domestic RO/UV water purifiers.

Evaluate the remaining two high and low frequency bands:

• • High-energy Gamma rays destroy malignant tumor cells in radiotherapy.
• • Radio waves propagate over long distances for broadcast communication.

Combine all pairings:

• • Matches Option (3) exactly.

In a vacuum / free space, the ratio of electric field amplitude to magnetic field amplitude equals the speed of light c = ω / k:

• • B₀ = E₀ / c = 60 / (3 × 10⁸) = 2 × 10⁻⁷ T.

The direction of electromagnetic wave propagation is given by the unit vector along the Poynting vector Ŝ = Ê × B̂. Here, the wave travels in the negative x-direction (-î) and E is polarized along the z-axis (k̂):

• • Because k̂ × ĵ = -î, the magnetic field must be along the positive y-axis: B_y.

The magnetic field oscillates in phase with the electric field in a plane EM wave:

• • Corresponds uniquely to Option (1).

For circular parallel plates of radius R, the electric field between the plates is E = σ / ε₀. Since dσ/dt is constant, the time derivative of the electric field is uniform:

• • Displacement current density J_d = ε₀ (dE/dt) is uniform throughout the cylindrical region between the plates.

Construct a circular Ampèrian loop of radius r ≤ R coaxial with the plates. Applying the Ampère-Maxwell law:

• • The magnetic field grows linearly from zero on the central axis (r = 0) up to the edge r = R.

For an Ampèrian loop of radius r ≥ R, the enclosed displacement current is the full total displacement current I_d = ε₀ π R² (dE/dt):

• • Because B increases for r ≤ R and decreases for r ≥ R, it attains its global maximum at r = R (the peripheral boundary).
• • Thus B is non-zero everywhere and reaches maximum at the imaginary cylindrical surface connecting the peripheries, matching Option (3).

Between the plates of a parallel plate capacitor, no actual physical charge carriers cross the vacuum or dielectric gap. However, as the capacitor charges with current I = dQ/dt, the charge Q on the plates increases, causing a time-varying electric field E and electric flux Φ_E between the plates:

Maxwell defined the displacement current I_d as the rate of change of electric flux scaled by permittivity ε₀:

As positive charge accumulates on the positive plate and negative charge on the negative plate, the electric field E points from positive to negative plate. Since E is increasing, dE/dt and therefore the displacement current density J_d = ε₀ dE/dt is directed from positive to negative plate, continuing seamlessly in the same direction as the conduction current I entering the capacitor.

According to Maxwell’s electromagnetic theory: - A stationary charge produces only a static electric field. - A charge moving with uniform velocity (constant speed in a straight line) constitutes a steady current and produces a steady magnetic field, but the energy remains localized with the charge and is not radiated into space. - An accelerated charge (such as an oscillating electron in an LC circuit or antenna) radiates energy continuously into space in the form of electromagnetic waves.

In free space, electromagnetic waves travel at the speed of light c, which is rigorously derived from Maxwell’s equations as 1/√(μ₀ ε₀):

Electromagnetic waves are transverse in nature: the electric field vector E, magnetic field vector B, and propagation vector k are mutually perpendicular to each other:

In free space, the time-averaged energy density of the electric field u_E equals the time-averaged energy density of the magnetic field u_B:

The statement which is NOT a property of electromagnetic waves is (2).

According to Maxwell’s equations for a plane transverse electromagnetic wave propagating in vacuum, the magnitudes of the mutually perpendicular oscillating electric (E⃗) and magnetic (B⃗) fields are in phase and related at every instant by the speed of light c:

Rearranging the equation to solve for the peak magnetic field amplitude B₀:

Substitute the given electric field amplitude E₀ = 48 V m⁻¹ and the speed of light c = 3 × 10⁸ m s⁻¹:

Convert to canonical scientific notation:

In a transverse plane electromagnetic wave propagating in free space: 1. The electric field vector $\vec{E}$ and magnetic field vector $\vec{B}$ are perpendicular to each other: $\vec{E} \cdot \vec{B} = 0$. 2. Both $\vec{E}$ and $\vec{B}$ are perpendicular to the wave propagation direction $\hat{k} = \hat{i}$ (so they lie in the $y$-$z$ plane). 3. The direction of energy flow (Poynting vector) is along $\vec{E} \times \vec{B}$:

For $\vec{E} = -\hat{j} + \hat{k}$ and $\vec{B} = -\hat{j} - \hat{k}$: - Check orthogonality: $\vec{E} \cdot \vec{B} = (-1)(-1) + (1)(-1) = 1 - 1 = 0$ (orthogonal). - Compute cross product using standard cyclic unit vector rules ($\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{j} = -\hat{i}$):

Options (1) and (4) have parallel vectors $\vec{E} \parallel \vec{B}$, yielding $\vec{E} \times \vec{B} = \vec{0}$. Option (3) has antiparallel vectors $\vec{E} = -\vec{B}$, also yielding $\vec{E} \times \vec{B} = \vec{0}$. None of these can propagate an electromagnetic wave.

Maxwell defined displacement current $I_d$ in terms of the rate of change of electric flux $\Phi_E$ between the capacitor plates:

For a parallel plate capacitor of area $A$ and separation $d$, the uniform electric field is $E = V/d$. The electric flux is $\Phi_E = E A = (V/d)A = V (A/d)$:

Substitute $V(t) = V_0 \sin(\omega t)$ into the displacement current expression:

In an electromagnetic wave traveling in free space, energy is stored in both the oscillating electric and magnetic fields. The time-averaged energy densities are given by: $$\langle u_E \rangle = \frac{1}{2} \varepsilon_0 E_{\text{rms}}^2 = \frac{1}{4} \varepsilon_0 E_0^2$$ $$\langle u_B \rangle = \frac{1}{2} \frac{B_{\text{rms}}^2}{\mu_0} = \frac{B_0^2}{4 \mu_0}$$

From Maxwell's equations, the ratio of electric field to magnetic field amplitude equals the speed of light in vacuum: $$E_0 = c B_0 \implies E_{\text{rms}} = c B_{\text{rms}}$$ Also, the speed of light is related to vacuum permittivity and permeability by: $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \implies c^2 = \frac{1}{\mu_0 \varepsilon_0}$$

Substitute $$E_{\text{rms}} = c B_{\text{rms}}$$ into the electric energy density expression: $$\langle u_E \rangle = \frac{1}{2} \varepsilon_0 (c B_{\text{rms}})^2 = \frac{1}{2} \varepsilon_0 c^2 B_{\text{rms}}^2$$ Substituting $$c^2 = \frac{1}{\mu_0 \varepsilon_0}$$: $$\langle u_E \rangle = \frac{1}{2} \varepsilon_0 \left(\frac{1}{\mu_0 \varepsilon_0}\right) B_{\text{rms}}^2 = \frac{B_{\text{rms}}^2}{2 \mu_0} = \langle u_B \rangle$$

The intensity of an electromagnetic wave is the energy transported per unit area per unit time, given by $$I = u \cdot c$$. Since $$\langle u_E \rangle = \langle u_B \rangle$$, the intensity contributions are identical: $$I_E = \langle u_E \rangle c = \langle u_B \rangle c = I_B$$ $$\frac{I_E}{I_B} = \frac{1}{1} = 1 : 1$$

The radiant flux (or intensity $$I$$) of an electromagnetic wave is defined as the radiant energy falling per unit area per unit time perpendicular to the beam: $$I = \frac{U}{A \cdot t} = \frac{P}{A}$$ where $$U$$ is total energy, $$A$$ is surface area, and $$t$$ is time duration.

Given: - Flux $$I = 20\text{ W cm}^{-2}$$ - Area $$A = 20\text{ cm}^2$$ $$P = I \cdot A = (20\text{ W cm}^{-2}) \times (20\text{ cm}^2) = 400\text{ W} = 400\text{ J s}^{-1}$$ (Notice that the units of $$\text{cm}^2$$ cancel immediately, avoiding tedious conversion to $$\text{m}^2$$).

Given time span $$t = 1\text{ minute} = 60\text{ seconds}$$. $$U = P \cdot t = (400\text{ J s}^{-1}) \times (60\text{ s}) = 24000\text{ J} = 24 \times 10^3\text{ J}$$ This confirms Option (3).