Q.A plane electromagnetic wave of frequency $25\ \text{MHz}$ travels in free space along the $x$-direction. At a particular point in space and time, $\mathbf{E} = 6.3\ \hat{j}\ \text{V/m}$. What is $\mathbf{B}$ at this point?

Showing the 12 most recent of 76 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Solar energy flux of $2000\ Wm^{-2}$ incident normally on a solar plate of $1\ m^2$ surface area in 1 hour. Then the momentum received by solar plate is (A) $24\times10^3\ kgms^{-1}$ (B) $24\times10^{-3}\ kgms^{-1}$ (C) $48\times10^{-3}\ kgms^{-1}$ (D) $72\times10^3\ kgms^{-1}$

   ›Reveal solutionSolution

   A solar plate absorbs the incident radiant energy; using $p=E/c$ with the total energy received in 1 hour gives $24\times10^{-3}$ kg·m/s.

   Concept and Intuition

   Electromagnetic radiation carries momentum along with energy, related by $p=E/c$ for a fully absorbing surface (and $p=2E/c$ for a fully reflecting one). A 'solar plate' (like a solar panel) is designed to absorb sunlight to convert it to usable energy, so we use the absorption relation.

   Step-by-Step Solution

   1. Total energy incident in 1 hour: $E = \text{flux}\times\text{area}\times\text{time} = 2000\times1\times3600 = 7.2\times10^6$ J.
   2. For a fully absorbing surface, momentum delivered equals $p=\dfrac{E}{c}$.
   3. $p = \dfrac{7.2\times10^6}{3\times10^8} = 2.4\times10^{-2} = 24\times10^{-3}$ kg·m/s.

   Common Mistakes

   • Using $p=2E/c$ (the reflecting-surface formula) instead of $p=E/c$ for an absorbing plate.
   • Forgetting to convert the time from hours to seconds before computing total energy.

   ✓Final answer

   The correct option is (B) — $24\times10^{-3}\ kgms^{-1}$.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Electromagnetic waves do not transport (A) Charge (B) Energy (C) Momentum (D) Information

   ›Reveal solutionSolution

   EM waves transport energy, momentum, and information, but never charge — charge is a property of the source particles, not of the propagating field.

   Concept and Intuition

   An electromagnetic wave is a self-sustaining oscillation of electric and magnetic fields that propagates through space (or vacuum) without needing a material medium. It is generated by accelerating charges but the wave itself is a field pattern, not a stream of charged particles. Because it carries energy density $u = \varepsilon_0 E^2$ (time-averaged, plus the magnetic contribution) and momentum density $p = u/c$, it exerts radiation pressure and can do work on absorbing surfaces. It can also encode information (as amplitude, frequency, or phase modulation, e.g. radio and light signals). But charge, being a conserved property intrinsic to particles, is never "carried" by the field wave itself — the field doesn't transport net charge from one place to another.

   Step-by-Step Solution

   1. Energy: EM waves transport energy — this is why sunlight warms your skin and radio waves can power a receiver (Poynting vector $S = \dfrac{1}{\mu_0}E\times B$ describes this energy flux).
   2. Momentum: EM waves carry momentum, causing measurable radiation pressure (e.g., comet tails, solar sails).
   3. Information: modulated EM waves (radio, microwave, optical) are the basis of all wireless communication — they clearly transport information.
   4. Charge: the wave itself is just oscillating $E$ and $B$ fields; there is no charge associated with or transported by the wave.

   Common Mistakes

   • Confusing the source of EM waves (accelerating charges) with the wave itself carrying charge — the wave is charge-neutral by nature.
   • Assuming that because EM waves interact with charges, they must carry charge — interaction is not the same as transport.

   ✓Final answer

   The correct option is (A) — Charge.

   ANSWER: A

3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Which of the following statements are correct about electromagnetic waves?

   i) Electromagnetic waves are produced by accelerating charges

   ii) Electromagnetic waves do not transport charge

   iii) Energy of electromagnetic waves is shared equally between electric and magnetic fields

   iv) Electromagnetic waves travel with same speed in all media (A)

   (i) and

   (iv) only (B)

   (ii) and

   (iii) only (C) (i),

   (ii) and

   (iv) only (D) (i),

   (ii) and

   (iii) only

   ›Reveal solutionSolution

   Tests basic properties of EM waves — production, charge/energy transport, and propagation speed. Statements (i),

   (ii),

   (iii) are correct;

   (iv) is false because EM wave speed depends on the medium.

   Concept and Intuition

   Electromagnetic waves are a self-sustaining oscillation of mutually perpendicular electric and magnetic fields, generated whenever a charge accelerates (this is the classical mechanism behind all radio, light, X-ray emission). Because the wave is a field disturbance, not a stream of matter, it transports energy and momentum but never net electric charge. In the wave, $E=cB$ at every instant, and since energy density $u_E=\tfrac12\epsilon_0E^2$ and $u_B=\tfrac{B^2}{2\mu_0}$, using $c=1/\sqrt{\mu_0\epsilon_0}$ shows $u_E=u_B$ always — the energy is shared equally between the two fields. The speed, however, is a property of the medium: $v=1/\sqrt{\mu\epsilon}=c/n$, so it changes from medium to medium (that's exactly why refraction happens).

   Step-by-Step Solution

   1. (i) Accelerating charge → radiates EM wave: correct (fundamental production mechanism).
   2. (ii) EM wave transports energy/momentum, not charge: correct.
   3. (iii) $u_E = u_B$ on time-average in an EM wave: correct.
   4. (iv) Speed $=c/n$, varies with medium: incorrect.
   5. So the correct set is (i), (ii), (iii) only.

   Common Mistakes

   • Assuming EM waves always travel at $c$ regardless of medium (only true in vacuum).
   • Thinking the magnetic field's energy contribution is negligible compared to the electric field's — it is exactly equal.

   ✓Final answer

   The correct option is (D) — (i), (ii) and (iii) only.

   ANSWER: D

4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Electromagnetic radiations are emitted from a 15W point source. The peak value of the magnetic field at a distance of 5m from the source is (A) $4\times10^{-8}$ T (B) $2\times10^{-8}$ T (C) $2\times10^{-7}$ T (D) $4\times10^{-7}$ T

   ›Reveal solutionSolution

   This tests the relation between the intensity of EM radiation from a point source and the peak magnetic field of the wave.

   Concept and Intuition

   A point source radiates power $P$ isotropically, so at distance $r$ the intensity (power per unit area) is $I=\dfrac{P}{4\pi r^2}$. This intensity is the time-averaged Poynting flux of the electromagnetic wave, which in terms of the peak electric/magnetic fields is $I = \dfrac{E_0 B_0}{2\mu_0} = \dfrac{cB_0^2}{2\mu_0}$ (using $E_0=cB_0$). Solving this for $B_0$ connects the source power directly to the wave's peak magnetic field at that distance.

   Step-by-Step Solution

   1. Intensity at $r=5$ m: $I = \dfrac{P}{4\pi r^2} = \dfrac{15}{4\pi(5)^2} = \dfrac{15}{100\pi} \approx 0.04775\,\text{W/m}^2$.
   2. From $I = \dfrac{cB_0^2}{2\mu_0}$, solve for $B_0$: $B_0 = \sqrt{\dfrac{2\mu_0 I}{c}}$.
   3. Substitute: $2\mu_0 I = 2(4\pi\times10^{-7})(0.04775) \approx 1.2\times10^{-7}$.
   4. Divide by $c$: $\dfrac{1.2\times10^{-7}}{3\times10^8} = 4\times10^{-16}$.
   5. $B_0 = \sqrt{4\times10^{-16}} = 2\times10^{-8}\,\text{T}$.

   Common Mistakes

   • Using $4\pi r$ instead of $4\pi r^2$ for the area of the sphere over which power spreads.
   • Forgetting the factor of 2 in the average-intensity formula (confusing peak intensity with average/rms intensity).

   ✓Final answer

   The correct option is (B) — $2\times10^{-8}$ T.

   ANSWER: B

5. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If the rms value of the electric field of an electromagnetic wave is $360\sqrt\pi\ NC^{-1}$, the average energy density of the electric field of the wave is (A) $3.6\times10^{-6}\ Jm^{-3}$ (B) $3.6\times10^{-9}\ Jm^{-3}$ (C) $1.8\times10^{-6}\ Jm^{-3}$ (D) $1.8\times10^{-9}\ Jm^{-3}$

   ›Reveal solutionSolution

   Direct substitution into $u_E=\tfrac12\epsilon_0E_{rms}^2$ gives $1.8\times10^{-6}\ J\,m^{-3}$.

   Concept and Intuition

   In an electromagnetic wave, the instantaneous energy density stored in the electric field is $u_E=\tfrac12\epsilon_0E^2$. Averaged over a cycle, since $E_{rms}^2\equiv\langle E^2\rangle$ by definition, the average energy density is simply $\langle u_E\rangle=\tfrac12\epsilon_0E_{rms}^2$ — no additional averaging factor is needed because the rms value already carries the time-average.

   Step-by-Step Solution

   1. $E_{rms}=360\sqrt\pi$ N/C $\Rightarrow E_{rms}^2=360^2\times\pi=129600\pi\approx4.0715\times10^{5}\ (\text{N/C})^2$.
   2. $\langle u_E\rangle=\tfrac12\epsilon_0E_{rms}^2=0.5\times8.85\times10^{-12}\times4.0715\times10^{5}$.
   3. $=0.5\times8.85\times4.0715\times10^{-7}\approx0.5\times36.03\times10^{-7}$.
   4. $\approx1.8\times10^{-6}\ \mathrm{J\,m^{-3}}$.

   Common Mistakes

   • Applying an extra factor of $\tfrac12$ for "time-averaging" on top of already using $E_{rms}$ (double counting).
   • Confusing the peak value $E_0$ with $E_{rms}$ (they're related by $E_0=\sqrt2\,E_{rms}$).

   ✓Final answer

   The correct option is (C) — $1.8\times10^{-6}\ Jm^{-3}$.

   ANSWER: C

6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The electric field intensity produced by the radiations coming from 100W bulbs at 3m distance is E. The electric field intensity produced by the radiations coming from 50W bulbs at the same distance is (A) $\dfrac{E}{2}$ (B) 2E (C) $\dfrac{E}{\sqrt{2}}$ (D) $\sqrt{2}E$

   ›Reveal solutionSolution

   Since intensity $\propto E^2$ and intensity $\propto$ power at fixed distance, the field scales as $\sqrt{\text{power}}$; halving the bulb's power scales $E$ by $1/\sqrt2$. Answer: (C).

   Concept and Intuition

   A bulb radiates (roughly) isotropically, so the average intensity (power per unit area) at distance $r$ is $I=\dfrac{P}{4\pi r^2}$. For an electromagnetic wave, the (time-averaged) intensity is proportional to the square of the electric-field amplitude, $I\propto E^2$. Combining these, at a fixed distance, $E\propto\sqrt{P}$.

   Step-by-Step Solution

   1. At distance $3\,$m, for the 100 W bulb: $I_1=\dfrac{100}{4\pi(3)^2}\propto E^2$, given $E$.
   2. For the 50 W bulb at the same distance: $I_2=\dfrac{50}{4\pi(3)^2}$.
   3. $\dfrac{I_2}{I_1}=\dfrac{50}{100}=\dfrac12$, and since $I\propto E^2$: $\dfrac{E_2^2}{E_1^2}=\dfrac12\Rightarrow E_2=\dfrac{E_1}{\sqrt2}=\dfrac{E}{\sqrt2}$.

   Common Mistakes

   • Assuming $E\propto P$ directly (linear), forgetting that intensity — not field — is what's directly proportional to power; the field only goes as the square root.
   • Bringing the distance into the ratio unnecessarily; since the distance is unchanged, it cancels out and never matters here.

   ✓Final answer

   The correct option is (C) — $\dfrac{E}{\sqrt2}$.

   ANSWER: C

7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A light of energy flux $9\ W\ cm^{-2}$ is incident for 20 minutes on a black surface of area $100\ cm^2$ then the maximum average force exerted on the surface is (velocity of light, $C = 3 \times 10^8\ ms^{-1}$) (A) $3 \times 10^{-3}\ N$ (B) $3 \times 10^{-6}\ N$ (C) $3 \times 10^{-8}\ N$ (D) 10 N

   ›Reveal solutionSolution

   Radiation pressure force on a fully absorbing surface is $F=P/c$, independent of exposure duration. Answer: $3\times10^{-6}\ N$.

   Concept and Intuition

   Light carries momentum, and when it strikes a surface it exerts a force. For a black (fully absorbing) surface, all the incident momentum is transferred, giving $F = P/c$ (half of what a perfectly reflecting surface would experience, since reflection reverses momentum and transfers twice as much). This force is a steady-state quantity set by the power (energy per second) hitting the surface — the fact that the light shines for 20 minutes doesn't change the instantaneous/average force, it only affects total energy delivered.

   Step-by-Step Solution

   1. Incident power: $P = \text{(energy flux)} \times \text{(area)} = 9\ W\,cm^{-2} \times 100\ cm^2 = 900\ W$.
   2. For a black (absorbing) surface, force $F = \dfrac{P}{c}$.
   3. $F = \dfrac{900}{3\times10^8} = 300\times10^{-8} = 3\times10^{-6}\ N$.
   4. The 20-minute duration is a distractor — force is a rate quantity (power/c), not affected by how long the light shines.

   Common Mistakes

   • Using the reflecting-surface formula $F=2P/c$ for a black (absorbing) surface.
   • Trying to bring in the time duration to compute total impulse and mistakenly calling it 'force'.

   ✓Final answer

   The correct option is (B) — $3\times10^{-6}\ N$.

   ANSWER: B

8. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.The magnetic field in a plane electromagnetic wave is $B_Y = 2 \times 10^{-7} \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)$ (all quantities are in SI units). The correct expression for electric field of the wave is (A) $E_Z = 60\sin(0.5 \times 10^3 x - 1.5 \times 10^{11} t)$ (B) $E_Y = 2 \times 10^{-7}\sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)$ (C) $E_Y = 2 \times 10^{-7}\cos(1.5 \times 10^3 x + 0.5 \times 10^{11} t)$ (D) $E_Z = 60\sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)$

   ›Reveal solutionSolution

   Tests reconstructing the electric field of a plane EM wave from its given magnetic field, using $E_0=cB_0$, the same-phase relationship, and the right-hand-rule direction consistent with the wave's propagation direction.

   Concept and Intuition

   In a plane EM wave, $\vec E$, $\vec B$, and the direction of propagation are mutually perpendicular and always in a fixed right-handed relationship — knowing any two of "direction of $B$," "direction of propagation," and "direction of $E$" fixes the third. The magnitudes are locked together by $E_0=cB_0$ at every instant, and both fields oscillate exactly in phase (same argument of sine/cosine) for a wave with no phase lag between them.

   Step-by-Step Solution

   1. Given: $B_Y = 2\times10^{-7}\sin(0.5\times10^3 x + 1.5\times10^{11}t)$. The phase is $(kx+\omega t)$ with $k=0.5\times10^3$ and $\omega=1.5\times10^{11}$ both positive — a phase of the form $(kx+\omega t)$ describes a wave moving in the $-\hat x$ direction (constant phase requires $x$ to decrease as $t$ increases).
   2. Amplitude relation: $E_0 = cB_0 = (3\times10^8)(2\times10^{-7}) = 60\,\text{(SI units, V/m)}$.
   3. Since $B$ is along $\hat y$ and the wave travels along $-\hat x$, the mutual perpendicularity/right-hand-rule (propagation direction $\parallel \hat E\times\hat B$) requires $\hat E$ to be along $\hat z$ (checking: $\hat z\times\hat y = -\hat x$, matching the $-\hat x$ propagation direction with both amplitudes positive and in phase).
   4. $E$ and $B$ oscillate in phase (same sinusoidal argument, not phase-shifted and not a cosine), so $E_Z$ carries the same phase $(0.5\times10^3x+1.5\times10^{11}t)$ as $B_Y$.
   5. Putting it together: $E_Z = 60\sin(0.5\times10^3x+1.5\times10^{11}t)$, which is option (D).

   Common Mistakes

   • Flipping the propagation direction (reading $kx+\omega t$ as $+\hat x$ instead of $-\hat x$), which would flip the required sign/axis of $E$ and select the wrong option.
   • Forgetting $E_0=cB_0$ and instead keeping the same numeric coefficient as $B_0$ (as in the distractor option that repeats $2\times10^{-7}$) — the amplitudes of $E$ and $B$ in an EM wave are never numerically equal in SI units.

   ✓Final answer

   The correct option is (D) — $E_Z = 60\sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)$.

   ANSWER: D

9. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.In an electromagnetic wave, the angle and phase difference between electric and magnetic fields are respectively (A) $0°, 90°$ (B) $0°, 0°$ (C) $90°, 0°$ (D) $90°, 90°$

   ›Reveal solutionSolution

   $E$ and $B$ in an EM wave are perpendicular to each other (and to the direction of propagation) and oscillate perfectly in phase.

   Concept and Intuition

   Maxwell's equations show that a changing $E$ field creates a $B$ field and vice versa, with both fields perpendicular to each other and to the propagation direction (forming a right-handed triad $\vec E,\vec B,\vec c$). Crucially, both fields reach their maxima and zeros at the same instants and same locations — they are in phase, unlike, say, current and voltage across a capacitor.

   Step-by-Step Solution

   1. Spatial relationship: $\vec E \perp \vec B$, both perpendicular to the direction of wave propagation $\Rightarrow$ angle $=90°$.
   2. Temporal relationship: both fields vary as $\sin(kx-\omega t)$ with the same phase, so phase difference $=0°$.

   Common Mistakes

   • Confusing this with an LC oscillator, where energy sloshes between E-type and B-type storage out of phase; in a propagating EM wave, $E$ and $B$ are in phase.

   ✓Final answer

   The correct option is (C) — $90°, 0°$.

   ANSWER: C

10. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
    Q.If the average energy density of the electric field of an electromagnetic wave is $u_E$ and the average energy density of the magnetic field of the wave is $u_B$, then (c – speed of light in vacuum) (A) $u_B = c^2 u_E$ (B) $u_B = c\,u_E$ (C) $u_E = u_B$ (D) $u_E = c\,u_B$

    ›Reveal solutionSolution

    A core fact about electromagnetic waves: the energy carried by the electric and magnetic fields is always equally split.

    Concept and Intuition

    Electromagnetic waves have $E=cB$ at every instant. Plugging this into the standard energy density expressions shows the two contributions are always numerically equal — a beautiful symmetry in how EM waves carry energy, unlike static fields where the two need not match.

    Step-by-Step Solution

    1. Energy density in the electric field: $u_E=\dfrac12\epsilon_0 E^2$.
    2. Energy density in the magnetic field: $u_B=\dfrac{B^2}{2\mu_0}$.
    3. Using $E=cB$ and $c^2=\dfrac{1}{\mu_0\epsilon_0}$: $u_E=\dfrac12\epsilon_0(cB)^2=\dfrac12\epsilon_0 c^2 B^2=\dfrac12\epsilon_0\cdot\dfrac{1}{\mu_0\epsilon_0}B^2=\dfrac{B^2}{2\mu_0}=u_B$.
    4. Hence $u_E=u_B$ always, for the average (or instantaneous, in vacuum) energy densities of an EM wave.

    Common Mistakes

    • Assuming the electric field "dominates" simply because $E=cB$ makes $E$ numerically much larger than $B$ — this ignores the very different constants ($\epsilon_0$ vs $1/\mu_0$) multiplying each square.
    • Confusing this equal-energy-density result with the (different) statement about the Poynting vector's magnitude.

    ✓Final answer

    The correct option is (C) — $u_E = u_B$.

    ANSWER: C

11. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.In a plane electromagnetic wave, the magnetic field is given by $\vec{B} = 3\times10^{-7}\sin(100\pi x + 10^{12} t)\ \mathrm{T}$, then the wavelength of the wave is (In the equation x is in metre and t is in second) (A) 0.02 m (B) 0.2 m (C) 0.4 m (D) 0.04 m

    ›Reveal solutionSolution

    This tests reading the wave number directly off an electromagnetic wave's field equation to get the wavelength. Answer: 0.02 m.

    Concept and Intuition

    Any travelling wave written as $\sin(kx\pm\omega t)$ has its spatial periodicity encoded in the wave number $k$, related to wavelength by $k=\dfrac{2\pi}{\lambda}$. For an electromagnetic wave, this applies directly to the argument of the sine function in the magnetic (or electric) field expression, regardless of the field's amplitude or the frequency term.

    Step-by-Step Solution

    1. Given: $\vec{B}=3\times10^{-7}\sin(100\pi x+10^{12}t)$ T.
    2. Compare with the standard form $B_0\sin(kx+\omega t)$: here $k=100\pi\ \text{rad/m}$.
    3. Wavelength: $\lambda=\dfrac{2\pi}{k}=\dfrac{2\pi}{100\pi}=\dfrac{2}{100}=0.02$ m.

    Common Mistakes

    • Using the coefficient of $t$ (angular frequency $10^{12}$) instead of the coefficient of $x$ to find wavelength — that coefficient gives the period/frequency, not wavelength.
    • Forgetting to cancel the $\pi$ when dividing $2\pi$ by $100\pi$.

    ✓Final answer

    The correct option is (A) — 0.02 m.

    ANSWER: A

12. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.A plane electromagnetic wave of frequency 25 MHz propagates in vacuum along positive x-direction. At a particular point in space and time, if the electric field is $6.3\hat{j}\ \mathrm{V\,m^{-1}}$, then the magnitude of the magnetic field of the wave at this point at the same time is (A) $2.1\times10^{-8}\ \mathrm{T}$ (B) $4.2\times10^{-8}\ \mathrm{T}$ (C) $6.3\times10^{-8}\ \mathrm{T}$ (D) $8.4\times10^{-8}\ \mathrm{T}$

    ›Reveal solutionSolution

    The magnitudes of E and B fields in an electromagnetic wave in vacuum are always related by $E_0 = cB_0$; solving gives $B_0 = 2.1\times10^{-8}$ T.

    Concept and Intuition

    In a plane electromagnetic wave travelling through vacuum, the electric and magnetic field magnitudes at every point and instant satisfy $E = cB$, where $c$ is the speed of light. This comes directly from Maxwell's equations and holds regardless of frequency, since it is a statement about the wave's intrinsic impedance-like relationship in free space.

    Step-by-Step Solution

    1. Given: $E = 6.3\ \text{V/m}$ (the value of the $\hat{j}$ component, i.e. its magnitude), $c = 3\times10^8\ \text{m/s}$.
    2. Apply $B = E/c$.
    3. $B = \dfrac{6.3}{3\times10^8} = 2.1\times10^{-8}\ \text{T}$.

    Common Mistakes

    • Trying to use the given frequency (25 MHz) in the calculation — it is a distractor; $E=cB$ doesn't need frequency.
    • Multiplying by $c$ instead of dividing.

    ✓Final answer

    The correct option is (A) — $2.1\times10^{-8}$ T.

    ANSWER: A