Q.Poynting vector $\mathbf{S}$ is defined as a vector whose magnitude is equal to the wave intensity and whose direction is along the direction of wave propagation. Mathematically, it is given by $\mathbf{S} = \dfrac{1}{\mu_0}\mathbf{E} \times \mathbf{B}$. Show the nature of $S$ vs $t$ graph.

Showing the 12 most recent of 38 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The unification of electromagnetism and optics is based on the discovery that (A) light is an electromagnetic wave (B) light travels with a speed equal to speed of sound (C) light wave consists of electrons (D) light waves are deflected by electric and magnetic fields

   ›Reveal solutionSolution

   The unification of electromagnetism and optics is based on James Clerk Maxwell's discovery that light is an electromagnetic wave, a prediction derived from his fundamental equations. The correct option is (A).

   Concept and Intuition

   Before the mid-19th century, electricity, magnetism, and light were considered distinct phenomena. Electricity dealt with charges and currents, magnetism with magnets and their forces, and optics with the behavior of light. While there were some observed connections, such as Oersted's discovery that electric currents produce magnetic fields, a comprehensive theoretical framework linking all three was missing.

   The breakthrough came with James Clerk Maxwell. He synthesized the existing laws of electricity and magnetism (Ampere's law, Faraday's law, Gauss's laws for electricity and magnetism) and added a crucial term, the "displacement current," to Ampere's law. This modification was necessary for the consistency of the theory and to ensure charge conservation.

   The profound consequence of Maxwell's complete set of equations was the prediction of the existence of electromagnetic waves. These waves consist of oscillating electric and magnetic fields propagating through space, even in a vacuum. When Maxwell calculated the speed at which these theoretical waves should travel, he found it to be approximately $3 \times 10^8 \text{ m/s}$. This value was remarkably close to the experimentally measured speed of light. This striking agreement led to the revolutionary conclusion that light itself is an electromagnetic wave. This discovery unified the previously separate fields of electromagnetism and optics into a single, coherent theory.

   Step-by-step explanation

   1. Maxwell's Equations and Prediction of EM Waves: James Clerk Maxwell formulated a set of four fundamental equations that describe the behavior of electric and magnetic fields and their interactions. From these equations, he mathematically demonstrated that changing electric fields produce changing magnetic fields, and vice versa. This interdependence leads to the propagation of disturbances in the form of electromagnetic waves.

   2. Calculation of Wave Speed: Maxwell derived an expression for the speed of these predicted electromagnetic waves in a vacuum:

      $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$

      where $\mu_0$ is the permeability of free space and $\epsilon_0$ is the permittivity of free space.

      When he substituted the known experimental values for $\mu_0$ and $\epsilon_0$, he obtained a speed of approximately $3 \times 10^8 \text{ m/s}$.

   3. Comparison with Speed of Light: At the time, the speed of light had been measured experimentally by various scientists (e.g., Fizeau, Foucault) to be very close to $3 \times 10^8 \text{ m/s}$. The astonishing agreement between the theoretically predicted speed of electromagnetic waves and the experimentally measured speed of light was not a mere coincidence. …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If a plane electromagnetic wave of intensity $9 \times 10^5 \ \mathrm{Wm}^{-2}$ incidents normally on a perfectly absorbing surface of area $2 \ \mathrm{m}^2$ for a time of $180 \ \mathrm{s}$, then the average force exerted by the electromagnetic wave on the surface during this time is (A) $3 \ \mathrm{mN}$ (B) $12 \ \mathrm{mN}$ (C) $6 \ \mathrm{mN}$ (D) $9 \ \mathrm{mN}$

   ›Reveal solutionSolution

   Electromagnetic waves carry momentum, and when they are absorbed by a surface, they exert a force due to the transfer of this momentum. For a perfectly absorbing surface, the radiation pressure is $I/c$, leading to an average force of $\boxed{6 \ \mathrm{mN}}$.

   Concept and Intuition

   Electromagnetic (EM) waves, such as light, not only carry energy but also momentum. When an EM wave interacts with a surface, it transfers some or all of its momentum to that surface. According to Newton's second law, a change in momentum over time results in a force. This force exerted by an EM wave is known as radiation pressure.

   Here's the core idea:

   1. Momentum of EM Waves: For an EM wave, energy $E$ and momentum $p$ are directly related by the speed of light $c$: $E = pc$. This means $p = E/c$.

   2. Intensity and Energy Flow: The intensity $I$ of an EM wave is defined as the average power per unit area. It represents the rate at which energy flows through a unit area perpendicular to the direction of wave propagation.

   3. Momentum Transfer: When an EM wave is incident on a surface, it transfers momentum.

      • If the surface is perfectly absorbing, all the incident momentum is transferred to the surface.
      • If the surface is perfectly reflecting, the momentum is not only absorbed but also reversed in direction, leading to twice the momentum transfer compared to a perfectly absorbing surface.

   4. Radiation Pressure: The force per unit area exerted by the EM wave is called radiation pressure ($P_{rad}$). For a perfectly absorbing surface, the radiation pressure is given by:

      $P_{rad} = \frac{I}{c}$

      This formula arises because the rate of energy flow per unit area is $I$, and since $p=E/c$, the rate of momentum flow per unit area (which is the pressure) must be $I/c$.

      ›Proof

      Let's quickly derive $P_{rad} = I/c$.

      Consider an electromagnetic wave incident normally on a perfectly absorbing surface of area $A$.

      In a time interval $\Delta t$, the energy incident on the surface is $E = I A \Delta t$.

      Since the energy and momentum of an EM wave are related by $E = pc$, the momentum incident on the surface in time $\Delta t$ is $p = E/c = (I A \Delta t)/c$.

      For a perfectly absorbing surface, this entire momentum is transferred to the surface.

      The force $F$ exerted on the surface is the rate of change of momentum:

      $F = \frac{\Delta p}{\Delta t} = \frac{(I A \Delta t)/c}{\Delta t} = \frac{I A}{c}$.

      The radiation pressure $P_{rad}$ is the force per unit area:

      $P_{rad} = \frac{F}{A} = \frac{I A / c}{A} = \frac{I}{c}$.

   The problem asks for the average force. Since the intensity is constant, the radiation pressure and thus the force exerted will also be constant. Therefore, the average force is simply this constant force. The time duration given ($180 \ \mathrm{s}$) confirms that the wave is incident for a period, but it does not affect the magnitude of the constant force.

   Step-by-step Solution

   1. Identify the given values:
      • Intensity of the electromagnetic wave, $I = 9 \times 10^5 \ \mathrm{Wm}^{-2}$.
      • Area of the surface, $A = 2 \ \mathrm{m}^2$. …

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The magnetic energy stored per unit volume in a solenoid with 1000 turns per metre carrying a current of 0.7 A is (A) $0.154~\mathrm{J~m^{-3}}$ (B) $0.616~\mathrm{J~m^{-3}}$ (C) $0.308~\mathrm{J~m^{-3}}$ (D) $0.924~\mathrm{J~m^{-3}}$

   ›Reveal solutionSolution

   The magnetic energy density in a solenoid is given by $u = \frac{B^2}{2\mu_0}$, where $B = \mu_0 n I$. For $n = 1000$ turns/m and $I = 0.7$ A, the result is $0.308~\mathrm{J/m^3}$, which corresponds to option (C).

   The key idea here is that the energy stored in a magnetic field is distributed throughout the volume where the field exists. For a long solenoid, the field is nearly uniform inside and zero outside, so the total energy is simply the energy density times the volume. The problem asks for the energy per unit volume, so we only need to compute the energy density $u$.

   The relevant formula comes from the energy stored in an inductor: $U = \frac{1}{2} L I^2$. For a solenoid of length $\ell$, cross-sectional area $A$, and $n$ turns per meter, the inductance is $L = \mu_0 n^2 A \ell$. Then the total energy is $U = \frac{1}{2} \mu_0 n^2 A \ell I^2$. Dividing by the volume $V = A \ell$ gives the energy density:

   $u = \frac{U}{V} = \frac{1}{2} \mu_0 n^2 I^2.$

   But we can also express this in terms of the magnetic field $B = \mu_0 n I$, giving the more general result:

   $u = \frac{B^2}{2\mu_0}.$

   This is the magnetic analog of the electric field energy density $\frac{1}{2} \epsilon_0 E^2$. It tells us that wherever there is a magnetic field, energy is stored in the field itself.

   Now let’s compute step by step.

   1. Find the magnetic field inside the solenoid. For an ideal solenoid, $B = \mu_0 n I$. Here $n = 1000$ turns/m and $I = 0.7$ A.

   $B = (4\pi \times 10^{-7}) \times 1000 \times 0.7 = 4\pi \times 10^{-7} \times 700 = 2.8\pi \times 10^{-4}~\mathrm{T}.$

   2. Write the energy density formula.

   $u = \frac{B^2}{2\mu_0}.$

   3. Substitute $B$ and simplify.

   $u = \frac{(2.8\pi \times 10^{-4})^2}{2 \times 4\pi \times 10^{-7}}.$

   First square the numerator:

   $(2.8\pi \times 10^{-4})^2 = (2.8^2)(\pi^2)(10^{-8}) = 7.84 \pi^2 \times 10^{-8}.$

   So

   $u = \frac{7.84 \pi^2 \times 10^{-8}}{8\pi \times 10^{-7}} = \frac{7.84 \pi}{8} \times 10^{-1}.$ …

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4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The wavelength range of the electromagnetic waves suitable for radar systems used in aircraft navigation is (A) $0.1\, \text{m to}\ 1\, \text{mm}$ (B) $700\, \text{nm to}\ 400\, \text{nm}$ (C) $300\, \text{nm to}\ 10\, \text{nm}$ (D) $1\, \text{nm to}\ 10^{-3}\, \text{nm}$

   ›Reveal solutionSolution

   Radar systems for aircraft navigation use radio waves, specifically microwaves, with wavelengths typically in the range of about 0.1 m to 1 mm. The correct option is (A).

   The key idea is that radar relies on electromagnetic waves that can travel long distances, penetrate weather, and reflect off large objects like aircraft. These requirements point to the microwave region of the spectrum, not visible light, ultraviolet, or X-rays.

   Why this approach works:

   Radar (Radio Detection And Ranging) was originally named for its use of radio waves. For aircraft navigation, the waves must be long enough to avoid heavy absorption by rain or fog, yet short enough to give reasonable resolution. This balance is struck in the microwave band, which spans roughly from 1 mm to 1 m. The given range 0.1 m to 1 mm fits perfectly.

   1. Eliminate visible light options

      Option (B) is 700 nm to 400 nm — that’s visible light (red to violet). Visible light is blocked by clouds and has very short range in the atmosphere for radar purposes. Not suitable.

   2. Eliminate ultraviolet options

      Option (C) is 300 nm to 10 nm — that’s ultraviolet. UV is strongly absorbed by the atmosphere and cannot penetrate weather. Also not suitable.

   3. Eliminate X-ray / gamma-ray options

      Option (D) is 1 nm to $10^{-3}$ nm — that’s X-rays and gamma rays. These are ionizing, dangerous, and cannot be used for long-range navigation radar. They also don’t reflect well from aircraft in a practical way.

   4. Confirm the correct range …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the peak voltage of message signal is 60% less than the peak voltage of the carrier wave, then the modulation index is (A) 0.3 (B) 0.4 (C) 0.6 (D) 0.2

   ›Reveal solutionSolution

   The modulation index is the ratio of the peak message voltage to the peak carrier voltage. If the message peak is 60% less than the carrier peak, it is 40% of the carrier peak, so the modulation index is 0.4. The correct option is (B).

   Concept & Intuition

   In amplitude modulation (AM), the modulation index $m$ tells us how strongly the message signal varies the carrier amplitude. It is defined as

   $m = \frac{V_m}{V_c}$

   where $V_m$ is the peak voltage of the message (modulating) signal and $V_c$ is the peak voltage of the carrier wave.

   The phrase “60% less than” is the key trap: many students mistakenly subtract 0.6 from 1 and get 0.4, but they must be careful — “less than” means the message peak is smaller by that percentage of the carrier peak. So if the carrier is 100%, the message is only 40% of it.

   Step-by-step reasoning

   1. Interpret the given condition “Peak voltage of message signal is 60% less than the peak voltage of the carrier wave.” This means:

   $V_m = V_c - 0.6\,V_c = 0.4\,V_c$

   The message peak is only 40% of the carrier peak.

   2. Apply the definition of modulation index $m = \frac{V_m}{V_c} = \frac{0.4\,V_c}{V_c} = 0.4$ …

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6. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.A resistor of resistance $20\sqrt{10}\ \Omega$ is connected to an alternating source of voltage $V = (A \sin \omega t + 40 \cos \omega t)\ \text{V}$. If the rms value of the current through the resistor is $500\ \text{mA}$, then the value of $A$ is (A) $40$ (B) $20$ (C) $40\sqrt{2}$ (D) $20\sqrt{2}$

   ›Reveal solutionSolution

   To find the unknown amplitude $A$, we first determine the RMS voltage of the given composite AC source. Then, we use Ohm's law for AC circuits, $V_{rms} = I_{rms} R$, to relate the RMS voltage to the given RMS current and resistance. Solving the resulting equation yields $A = \boxed{20}$.

   The problem asks us to find the value of $A$, which is an amplitude in a given alternating voltage expression. We are provided with the resistance of the circuit and the RMS value of the current flowing through it. The core idea here is to correctly determine the RMS voltage of the source when it's expressed as a sum of sine and cosine functions, and then apply Ohm's law for AC circuits.

   For a purely resistive circuit, Ohm's law holds true for RMS values, meaning $V_{rms} = I_{rms} R$. The challenge lies in correctly calculating $V_{rms}$ from the given voltage expression $V = (A \sin \omega t + 40 \cos \omega t)\ \text{V}$.

   Any sinusoidal voltage can be represented in the form $V = V_0 \sin(\omega t + \phi)$, where $V_0$ is the peak voltage. When a voltage is given as a sum of two sinusoidal components with the same frequency but different phases, like $V = V_1 \sin \omega t + V_2 \cos \omega t$, its peak voltage $V_0$ is found by treating $V_1$ and $V_2$ as orthogonal components (similar to vector addition or phasor addition).

   For an alternating voltage $V = V_1 \sin \omega t + V_2 \cos \omega t$, the peak voltage $V_0$ is given by:

   $V_0 = \sqrt{V_1^2 + V_2^2}$

   The RMS voltage $V_{rms}$ is then related to the peak voltage by:

   $V_{rms} = \frac{V_0}{\sqrt{2}}$

   Let's apply these concepts to solve the problem.

   1. Identify the given quantities and convert units:

      • Resistance, $R = 20\sqrt{10}\ \Omega$.
      • Voltage source, $V = (A \sin \omega t + 40 \cos \omega t)\ \text{V}$.
      • RMS current, $I_{rms} = 500\ \text{mA}$. We need to convert this to Amperes: $I_{rms} = 500 \times 10^{-3}\ \text{A} = 0.5\ \text{A}$.

   2. Determine the peak voltage of the source:

      The given voltage is in the form $V = V_1 \sin \omega t + V_2 \cos \omega t$, where $V_1 = A$ and $V_2 = 40$.

      Using the formula for peak voltage:

      $V_0 = \sqrt{A^2 + 40^2}$

      $V_0 = \sqrt{A^2 + 1600}$

   3. Calculate the RMS voltage of the source:

      The RMS voltage is related to the peak voltage by $V_{rms} = \frac{V_0}{\sqrt{2}}$.

      Substituting the expression for $V_0$:

      $V_{rms} = \frac{\sqrt{A^2 + 1600}}{\sqrt{2}}$ …

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7. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.If $\mu$ and $\varepsilon$ represent the permeability and permittivity of a medium respectively, then the physical quantity having the dimensions of $\sqrt{\frac{\mu}{\varepsilon}}$ is (A) Inductance (B) Impedance (C) Speed (D) Capacitance

   ›Reveal solutionSolution

   The quantity $\sqrt{\frac{\mu}{\varepsilon}}$ has the dimensions of impedance (resistance), because it arises from the ratio of electric to magnetic field amplitudes in an electromagnetic wave, and its SI unit is the ohm.

   The key here is to connect $\mu$ and $\varepsilon$ to something familiar. You know that in an electromagnetic wave, the speed $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ in vacuum. But here we have $\sqrt{\frac{\mu}{\varepsilon}}$, which is the reciprocal of that — so it’s not speed. Instead, think about the ratio of the electric field $E$ to the magnetic field $H$ in a wave: that ratio has the unit of impedance (ohms). Indeed, for a plane wave in a medium, $E/H = \sqrt{\mu/\varepsilon}$, which is called the intrinsic impedance of the medium. So the dimensions must match those of resistance or impedance.

   Let’s verify dimensionally.

   1. Recall the dimensions of $\mu$ and $\varepsilon$.

      From Faraday’s law and Ampere’s law, we know:

      • $\mu$ (permeability) appears in $B = \mu H$, and from $F = qvB$, we get $[B] = [M T^{-2} A^{-1}]$. Since $H$ has dimensions $[L^{-1} A]$, we find $[\mu] = [M L T^{-2} A^{-2}]$.
      • $\varepsilon$ (permittivity) appears in $D = \varepsilon E$, and from Coulomb’s law $F = \frac{1}{4\pi\varepsilon} \frac{q_1 q_2}{r^2}$, we get $[\varepsilon] = [M^{-1} L^{-3} T^4 A^2]$.

   2. Form the ratio $\mu / \varepsilon$.

   $\left[\frac{\mu}{\varepsilon}\right] = \frac{[M L T^{-2} A^{-2}]}{[M^{-1} L^{-3} T^4 A^2]} = [M^2 L^4 T^{-6} A^{-4}]$

   3. Take the square root. …

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8. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.If E and B are the magnitudes of the electric and magnetic fields respectively of a plane electromagnetic wave and $\omega$ is its angular frequency, then the wavelength of the wave is (A) $\dfrac{\pi E}{\omega B}$ (B) $\dfrac{2\pi E}{\omega B}$ (C) $\dfrac{E}{2\pi \omega B}$ (D) $\dfrac{E}{\pi \omega B}$

   ›Reveal solutionSolution

   The magnitudes of the electric and magnetic fields in a plane electromagnetic wave are related by the wave's speed, which is also related to its angular frequency and wavelength. Combining these relationships, the wavelength is found to be $\boxed{\dfrac{2\pi E}{\omega B}}$.

   A plane electromagnetic wave consists of oscillating electric ($\vec{E}$) and magnetic ($\vec{B}$) fields that are perpendicular to each other and to the direction of wave propagation. These fields oscillate in phase, meaning they reach their maximum and minimum values at the same time and location. The wave travels at the speed of light, $c$, in a vacuum.

   The core idea here is to connect the given quantities ($E$, $B$, $\omega$) to the wavelength ($\lambda$) using the fundamental properties of electromagnetic waves:

   1. The ratio of the magnitudes of the electric and magnetic fields gives the speed of the wave.
   2. The speed of any wave is also related to its frequency and wavelength.
   3. Angular frequency is directly related to linear frequency.

   By establishing these relationships, we can derive an expression for the wavelength.

   1. Relate E and B to the speed of light:

      For a plane electromagnetic wave propagating in a vacuum, the ratio of the peak (or RMS) magnitudes of the electric field ($E$) and magnetic field ($B$) is equal to the speed of light ($c$).

      $c = \frac{E}{B}$

   2. Relate speed, frequency, and wavelength:

      The speed of any wave ($c$) is the product of its frequency ($f$) and its wavelength ($\lambda$).

      $c = f\lambda$

   3. Relate angular frequency to linear frequency: …

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9. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The dielectric constant of a medium is 8 and its relative permeability is 200. If an electromagnetic wave of frequency 100 MHz travels in this medium, then its wavelength is (A) 15 m (B) 15 cm (C) 7.5 m (D) 7.5 cm

   ›Reveal solutionSolution

   The wavelength in a medium is given by $\lambda = \frac{c}{f \sqrt{\varepsilon_r \mu_r}}$. With $\varepsilon_r = 8$, $\mu_r = 200$, and $f = 100$ MHz, the result is $\lambda \approx 0.075$ m = 7.5 cm, so the correct option is (D).

   The key concept here is that electromagnetic waves travel slower in a material medium than in vacuum, and the reduction factor depends on both the electric and magnetic properties of the medium. The wavelength shrinks proportionally to the speed reduction.

   Why this approach works:

   In vacuum, an EM wave’s speed is $c = 3 \times 10^8$ m/s, and wavelength is $\lambda_0 = c/f$. In a medium with dielectric constant $\varepsilon_r$ and relative permeability $\mu_r$, the wave speed becomes $v = c / \sqrt{\varepsilon_r \mu_r}$. Since frequency remains unchanged (it’s set by the source), the wavelength in the medium is $\lambda = v/f = \lambda_0 / \sqrt{\varepsilon_r \mu_r}$. So we just compute the factor $\sqrt{\varepsilon_r \mu_r}$ and divide the vacuum wavelength by it.

   1. Find the vacuum wavelength Frequency $f = 100$ MHz = $100 \times 10^6$ Hz = $10^8$ Hz. Vacuum wavelength:

   $\lambda_0 = \frac{c}{f} = \frac{3 \times 10^8}{10^8} = 3 \text{ m}.$

   2. Compute the refractive index (or speed-reduction factor) The medium’s refractive index is $n = \sqrt{\varepsilon_r \mu_r}$. Here $\varepsilon_r = 8$ and $\mu_r = 200$, so

   $n = \sqrt{8 \times 200} = \sqrt{1600} = 40.$

   This means the wave travels 40 times slower than in vacuum. …

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10. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.For an aperture of $5 \times 10^{-3}$ m and a monochromatic light of wavelength $\lambda$, the distance for which ray optics becomes a good approximation is 50 m, then $\lambda =$ (A) 5000 Å (B) 6000 Å (C) 5400 Å (D) 6500 Å

    ›Reveal solutionSolution

    The key idea is that ray optics is a good approximation when the Fresnel distance $Z_F \approx a^2/\lambda$ is less than or equal to the given distance. Solving $50 = (5\times10^{-3})^2 / \lambda$ gives $\lambda = 5\times10^{-7}\,\text{m} = 5000\,\text{Å}$, so option (A) is correct.

    Concept and intuition

    When light passes through an aperture, diffraction effects become significant if the propagation distance is small compared to the Fresnel distance $Z_F = a^2/\lambda$. For distances much larger than $Z_F$, the wave nature of light is less noticeable and ray optics (straight-line propagation) becomes a good approximation. The problem gives the distance at which ray optics is “good” — that distance is exactly the Fresnel distance. So we set $Z_F = 50$ m and solve for $\lambda$.

    Step-by-step reasoning

    1. Recall the Fresnel distance formula For an aperture of size $a$ (here $a = 5 \times 10^{-3}$ m) and wavelength $\lambda$, the Fresnel distance is

    $Z_F = \frac{a^2}{\lambda}.$

    This is the distance beyond which diffraction effects are small and ray optics is valid.

    2. Set the given distance equal to $Z_F$ The problem states that ray optics becomes a good approximation at 50 m. That means

    $50 = \frac{(5 \times 10^{-3})^2}{\lambda}.$

    3. Solve for $\lambda$ Compute $a^2$:

    $(5 \times 10^{-3})^2 = 25 \times 10^{-6} = 2.5 \times 10^{-5} \,\text{m}^2.$

    Then …

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11. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.If electromagnetic waves of power 600 W incident on a non-reflecting surface, then the total force acting on the surface is (A) $12 \times 10^{-6}$ N (B) $9 \times 10^{-9}$ N (C) $6 \times 10^{-6}$ N (D) $2 \times 10^{-6}$ N

    ›Reveal solutionSolution

    The force from electromagnetic radiation on a perfectly absorbing surface is given by $F = P/c$, where $P$ is the power and $c$ is the speed of light. For $P = 600\ \text{W}$, the force is $2 \times 10^{-6}\ \text{N}$, so the correct option is (D).

    ---

    Concept and Intuition

    When electromagnetic waves hit a surface, they carry momentum. If the surface absorbs the radiation completely (a "non-reflecting" or perfectly absorbing surface), all that momentum is transferred to the surface. The rate of change of momentum equals the force. For light, the momentum per unit time (i.e., force) is simply the power divided by the speed of light:

    $F = \frac{P}{c}$

    This is a direct consequence of the relation $E = pc$ for photons (energy = momentum × speed of light). So, the larger the power, the larger the force — but because $c$ is huge ($3 \times 10^8\ \text{m/s}$), the force is tiny for everyday powers.

    ---

    Step-by-step solution

    1. Identify the type of surface

       The problem says "non-reflecting surface." This means the surface absorbs all incident radiation — no reflection. For a perfectly absorbing surface, the radiation pressure is $p = \frac{I}{c}$, where $I$ is intensity (power per area). The total force is then pressure times area, which simplifies to $F = \frac{P}{c}$.

    2. Write down the given data

       • Power, $P = 600\ \text{W}$
       • Speed of light, $c = 3 \times 10^8\ \text{m/s}$

    3. Apply the formula

       $F = \frac{P}{c} = \frac{600}{3 \times 10^8}$ …

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12. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.The amplitude of the electric field associated with a light beam of intensity $\dfrac{15}{\pi} \ \mathrm{Wm}^{-2}$ is (A) $120 \ \mathrm{NC}^{-1}$ (B) $15 \ \mathrm{NC}^{-1}$ (C) $60 \ \mathrm{NC}^{-1}$ (D) $30 \ \mathrm{NC}^{-1}$

    ›Reveal solutionSolution

    The key idea is that the intensity of an electromagnetic wave is related to the square of the electric field amplitude by $I = \frac{1}{2} \varepsilon_0 c E_0^2$. Solving for $E_0$ with the given intensity yields $E_0 = 60\ \mathrm{NC^{-1}}$, so the correct option is (C).

    Concept and Intuition

    Light is an electromagnetic wave. Its intensity (power per unit area) is the time-averaged energy flux. For a sinusoidal plane wave, the electric field oscillates as $E = E_0 \sin(\omega t - kx)$, and the magnetic field is related by $B = E/c$. The Poynting vector $\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}$ gives the instantaneous power flow. Averaging over a cycle, the intensity becomes $I = \frac{1}{2} \varepsilon_0 c E_0^2$. This formula is the direct bridge between the measurable intensity and the field amplitude we want.

    Step-by-step solution

    1. Recall the intensity–amplitude relation For an electromagnetic wave in vacuum, the time-averaged intensity is

    $I = \frac{1}{2} \varepsilon_0 c E_0^2,$

    where $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{F\,m^{-1}}$ (permittivity of free space), $c = 3 \times 10^8\ \mathrm{m\,s^{-1}}$ (speed of light), and $E_0$ is the amplitude of the electric field.

    2. Insert the given intensity The problem states $I = \frac{15}{\pi}\ \mathrm{W\,m^{-2}}$. So

    $\frac{15}{\pi} = \frac{1}{2} \varepsilon_0 c E_0^2.$

    3. Solve for $E_0^2$ Multiply both sides by 2:

    $\frac{30}{\pi} = \varepsilon_0 c E_0^2.$

    Then

    $E_0^2 = \frac{30}{\pi \varepsilon_0 c}.$

    4. Substitute the constants $\varepsilon_0 c = (8.85 \times 10^{-12}) \times (3 \times 10^8) = 2.655 \times 10^{-3}$. So

    $E_0^2 = \frac{30}{\pi \times 2.655 \times 10^{-3}}.$

    Compute the denominator: $\pi \times 2.655 \times 10^{-3} \approx 8.34 \times 10^{-3}$.

    Then

    $E_0^2 \approx \frac{30}{8.34 \times 10^{-3}} \approx 3597.$

    5. Take the square root …

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