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 15 on this concept.

1. COMEDK 2025Set 2025-E1 markMCQ
   Q.A plane electromagnetic wave with frequency 40 MHz travels in free space. At a particular point in space and time, the magnetic field is $2 \times 10^{-8} T$. What will be the electric field at this point? (A) $16 \mathrm{Vm}^{-1}$ (B) $6 \mathrm{Vm}^{-1}$ (C) $8 \mathrm{Vm}^{-1}$ (D) $18 \mathrm{Vm}^{-1}$

   ›Reveal solutionSolution

   In free space, the electric and magnetic fields of an electromagnetic wave are related by $E = cB$. Given $B = 2 \times 10^{-8}\,\mathrm{T}$ and $c = 3 \times 10^8\,\mathrm{m/s}$, the electric field is $6\,\mathrm{V/m}$, so the correct option is (B).

   The key concept here is the intrinsic relationship between the electric and magnetic fields in an electromagnetic wave in free space. Unlike in circuits or static fields, where $E$ and $B$ are independent, in a traveling EM wave they are locked together: their magnitudes are proportional, and the constant of proportionality is the speed of light $c$. This comes directly from Maxwell’s equations — specifically, Faraday’s law and Ampère’s law — which show that a changing magnetic field creates an electric field, and vice versa, with the ratio fixed by $c = 1/\sqrt{\mu_0 \varepsilon_0}$.

   The frequency given (40 MHz) is a red herring here: it tells us the wave is in the radio band, but the instantaneous relation $E = cB$ holds at every point and time for a plane wave in vacuum, regardless of frequency. So we don’t need the frequency at all.

   Let’s work it through:

   1. Recall the fundamental relation for a plane electromagnetic wave in free space:

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

   where $c = 3 \times 10^8\,\mathrm{m/s}$ is the speed of light. This holds because the wave’s energy is equally shared between the fields, and the wave equation forces the ratio.

   2. Plug in the given magnetic field:

   $B = 2 \times 10^{-8}\,\mathrm{T}$

   So

   $E = c \cdot B = (3 \times 10^8) \times (2 \times 10^{-8})$

   3. Calculate:

   $E = 6 \,\mathrm{V/m}$

   4. Check the options: (A) 16, (B) 6, (C) 8, (D) 18 — so (B) matches.

   Watch out

   A common mistake is to try using the frequency with formulas like $E = hf$ (photon energy) or to think the fields are independent. But here it’s a classical wave, and the frequency only affects wavelength, not the instantaneous $E/B$ ratio.

   Tip

   If you ever forget the relation, remember that the units work out: $E$ in V/m, $B$ in T, and $c$ in m/s gives V/m = (m/s)·T, which is exactly right because 1 T = 1 V·s/m².

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. COMEDK 2025Set 2025-M1 markMCQ
   Q.The electric and magnetic fields associated with an electromagnetic wave propagating along +z axis, can be represented by (A) $E=E_0 i, \quad B=B_0 j$ (B) $E=E_0 i, \quad B=B_0 k$ (C) $E=E_0 k, \quad B=B_0 j$ (D) $E=E_0 j, \quad B=B_0 k$

   ›Reveal solutionSolution

   For an electromagnetic wave propagating along the +z axis, the electric and magnetic fields must be perpendicular to each other and to the direction of propagation. The only option satisfying this is (A): $E = E_0 \hat{i}$, $B = B_0 \hat{j}$.

   The key concept here is the mutual perpendicularity of $\vec{E}$, $\vec{B}$, and the direction of propagation in an electromagnetic wave. In free space, electromagnetic waves are transverse: the electric field, magnetic field, and wave vector $\vec{k}$ (pointing in the propagation direction) form a right-handed orthogonal set. This means:

   • $\vec{E} \perp \vec{k}$
   • $\vec{B} \perp \vec{k}$
   • $\vec{E} \perp \vec{B}$
   • And the direction of $\vec{E} \times \vec{B}$ gives the direction of propagation.

   Let’s check each option step by step.

   1. Identify the propagation direction. The wave propagates along the +z axis, so the wave vector $\vec{k}$ is along $\hat{k}$ (the unit vector in the z-direction).

   2. Check option (A): $E = E_0 \hat{i}$, $B = B_0 \hat{j}$.

      • $\vec{E}$ is along x-axis ($\hat{i}$), $\vec{B}$ is along y-axis ($\hat{j}$).
      • Both are perpendicular to $\hat{k}$ (z-axis) because $\hat{i} \cdot \hat{k} = 0$ and $\hat{j} \cdot \hat{k} = 0$.
      • $\vec{E} \times \vec{B} = E_0 B_0 (\hat{i} \times \hat{j}) = E_0 B_0 \hat{k}$, which points along +z. This matches the propagation direction.
      • So (A) is valid.

   3. Check option (B): $E = E_0 \hat{i}$, $B = B_0 \hat{k}$.

      • $\vec{E}$ is along x-axis, $\vec{B}$ is along z-axis.
      • $\vec{B}$ is parallel to the propagation direction, not perpendicular. This violates the transverse nature of EM waves in free space.
      • Invalid.

   4. Check option (C): $E = E_0 \hat{k}$, $B = B_0 \hat{j}$.

      • $\vec{E}$ is along z-axis, parallel to propagation. Again, not transverse.
      • Invalid.

   5. Check option (D): $E = E_0 \hat{j}$, $B = B_0 \hat{k}$.

      • $\vec{E}$ is along y-axis, $\vec{B}$ is along z-axis.
      • $\vec{B}$ is parallel to propagation — invalid for the same reason.

   Watch out

   A common mistake is to forget that both fields must be perpendicular to the propagation direction. Even if one field is perpendicular, if the other is parallel, the wave cannot be a transverse electromagnetic wave in free space.

   Tip

   A quick mental check: For propagation along +z, the electric field must lie in the xy-plane, and the magnetic field must also lie in the xy-plane, but rotated 90° from the electric field. The cross product $\vec{E} \times \vec{B}$ should then point along +z.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

3. COMEDK 2024Set 2024-E1 markMCQ
   Q.Column - I lists the waves of the electromagnetic spectrum. Column - II gives approximate frequency range of these waves. Match Column - I and Column - II and choose the correct match from the given choices. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0lax{text-align:left;vertical-align:top} Column I Column II (A) Radiowaves (P) $10^{18} \text { to } 10^{20} \mathrm{~Hz}$ (B) Microwaves (P) $10^{11} \text { to } 5 \times 10^{14} \mathrm{~Hz}$ (C) Infrared (R) $10^4 \text { to } 10^8 \mathrm{~Hz}$ (D) X-rays (S) $10^9 \text { to } 10^{12} \mathrm{~Hz}$ (A) (A)-(R) (B)-(P) (C)-(S) (D)-(Q) (B) (A)-(R) (B)-(S) (C)-(Q) (D)-(P) (C) (A)-(R) (B)-(S) (C)-(P) (D)-(Q) (D) (A)-(R) (B)-(Q) (C)-(S) (D)-(P)

   ›Reveal solutionSolution

   A-R, B-S, C-Q, D-P by increasing frequency.

   Match each band by frequency:

   • Radiowaves (A): lowest frequency $10^4$–$10^8$ Hz = R.
   • Microwaves (B): $10^9$–$10^{12}$ Hz = S.
   • Infrared (C): $10^{11}$–$5\times10^{14}$ Hz = Q (the entry misprinted as a second P).
   • X-rays (D): highest, $10^{18}$–$10^{20}$ Hz = P.

   So A-R, B-S, C-Q, D-P.

   ✓Final answer

   The correct option is (B) — (A)-(R) (B)-(S) (C)-(Q) (D)-(P)

4. COMEDK 2023Set 2023-E1 markMCQ
   Q.What feature of the infrared waves make it useful for the haze photography? (A) Since it is invisible (B) Since it has large wave length (C) Since it is absorbed by the medium (D) Since it has high frequency

   ›Reveal solutionSolution

   Infrared's long wavelength means little scattering by haze particles, so it passes through and produces clearer distant images.

   Scattering of light by small particles (Rayleigh scattering) falls off strongly with increasing wavelength ($\propto 1/\lambda^4$). Infrared radiation has a much larger wavelength than visible light, so it is scattered far less by the fine particles of haze, mist and smoke. It therefore penetrates the haze and reaches the camera, allowing clear photographs of distant objects. Its invisibility, absorption or frequency are not the operative reason — the long wavelength (low scattering) is.

   ✓Final answer

   The correct option is (B) — Since it has large wave length

5. COMEDK 2023Set 2023-M1 markMCQ
   Q.The ratio of amplitude of magnetic field to the amplitude of electric field of an electromagnetic wave propagating in vacuum is (A) reciprocal of speed of light in vacuum (B) the speed of light in vacuum (C) proportional to frequency of the electromagnetic wave (D) inversely proportional to the frequency of the electromagnetic wave

   ›Reveal solutionSolution

   Maxwell's relation $E_0=cB_0$ makes $B_0/E_0=1/c$, i.e. the reciprocal of the speed of light, independent of frequency.

   For a plane electromagnetic wave in vacuum, the field amplitudes satisfy:

   $\frac{E_0}{B_0}=c\quad\Rightarrow\quad \frac{B_0}{E_0}=\frac{1}{c}.$

   This ratio is a constant equal to $1/c$; it does not depend on the frequency.

   ✓Final answer

   The correct option is (A) — reciprocal of speed of light in vacuum

6. COMEDK 2023Set 2023-M1 markMCQ
   Q.A plane electromagnetic wave of frequency $20 \mathrm{~MHz}$ travels through a space along $x$-direction. If the electric field vector at a certain point in space is $6 \mathrm{~Vm}^{-1}$, then what is the magnetic field vector at that point? (A) $2 \times 10^{-8} \mathrm{~T}$ (B) $\frac{1}{2} \times 10^{-8} \mathrm{~T}$ (C) $2 \mathrm{~T}$ (D) $\frac{1}{2} \mathrm{~T}$

   ›Reveal solutionSolution

   In an EM wave $E$ and $B$ are related by $B=E/c$, giving $B=6/(3\times10^8)=2\times10^{-8}\,$T.

   For a plane electromagnetic wave, the field magnitudes satisfy

   $B=\frac{E}{c}=\frac{6\ \text{V m}^{-1}}{3\times10^{8}\ \text{m s}^{-1}}=2\times10^{-8}\ \text{T}.$

   (The frequency is not needed — it does not affect the $E/B$ ratio.)

   ✓Final answer

   The correct option is (A) — $2 \times 10^{-8} \mathrm{~T}$

7. KCET 2022Set B-31 markMCQ
   Q.A fully charged capacitor ‘C’ with initial charge ‘q_0’ is connected to a coil of self inductance ‘L’ at t=0. The time at which the energy is stored equally between the electric and the magnetic field is (A) $\pi \sqrt{LC}$ (B) $\dfrac{\pi}{4} \sqrt{LC}$ (C) $2\pi \sqrt{LC}$ (D) $\sqrt{LC}$

   ›Reveal solutionSolution

   In an LC oscillation, energy sloshes between capacitor and inductor. Equal sharing happens when the charge on the capacitor is $q_0/\sqrt{2}$, which occurs at $t = \frac{\pi}{4}\sqrt{LC}$.

   The problem is about an ideal LC circuit — a capacitor charged to $q_0$ connected to an inductor at $t=0$. No resistance, so total energy is conserved. At any instant, the capacitor stores electric energy $U_E = \frac{q^2}{2C}$ and the inductor stores magnetic energy $U_B = \frac{1}{2} L i^2$. Their sum is constant: $\frac{q_0^2}{2C}$.

   The question asks: when are these two equal? That means $U_E = U_B$, so each is half the total energy. That gives $\frac{q^2}{2C} = \frac{1}{2} \cdot \frac{q_0^2}{2C}$, i.e. $q^2 = \frac{q_0^2}{2}$, so $q = \pm \frac{q_0}{\sqrt{2}}$.

   Now we need the time when the charge first reaches this value. The charge on the capacitor in an LC circuit oscillates sinusoidally. Since the capacitor is fully charged at $t=0$ and then begins to discharge, the charge follows a cosine function:

   $q(t) = q_0 \cos(\omega t)$

   where $\omega = \frac{1}{\sqrt{LC}}$ is the angular frequency of the LC oscillation.

   1. Set $q(t) = \frac{q_0}{\sqrt{2}}$.

   $q_0 \cos(\omega t) = \frac{q_0}{\sqrt{2}} \quad\Rightarrow\quad \cos(\omega t) = \frac{1}{\sqrt{2}}$

   2. The smallest positive angle whose cosine is $1/\sqrt{2}$ is $\pi/4$ radians. So:

   $\omega t = \frac{\pi}{4}$

   3. Substitute $\omega = 1/\sqrt{LC}$:

   $\frac{t}{\sqrt{LC}} = \frac{\pi}{4} \quad\Rightarrow\quad t = \frac{\pi}{4} \sqrt{LC}$

   Watch out

   A common mistake is to think equal energy means $q = q_0/2$ (half the charge). But energy depends on $q^2$, so half energy means $q = q_0/\sqrt{2}$, not $q_0/2$.

   Tip

   The cosine starts at 1 and decreases. The first time it hits $1/\sqrt{2}$ is at $\pi/4$ rad. If the question asked for the second time (when charge is rising again), it would be $3\pi/4$ rad, but the first occurrence is the intended answer.

   ✓Final answer

   The correct option is (B) $\dfrac{\pi}{4} \sqrt{LC}$.

8. COMEDK 2022Set 20221 markMCQ
   Q.Speed of electromagnetic wave in a medium having relative permittivity $\varepsilon_r$ and relative permeability $\mu_r$ is (speed of light in air, $c=3\times10^8$ m/s) (A) ${1 \over {\sqrt {{\mu _r}{\varepsilon _r}} }}$ (B) ${c \over {\sqrt {{\mu _r}{\varepsilon _r}} }}$ (C) $c\sqrt {{{{\mu _r}} \over {{\varepsilon _r}}}}$ (D) ${c \over {{\mu _r}{\varepsilon _r}}}$

   ›Reveal solutionSolution

   (Option A is dimensionally wrong — it is a pure number, not a speed; the refractive index is n = √(μ_r ε_r), and v = c/n.)

   Concept: Speed of an EM wave in a medium, v = 1/√(με), where μ = μ₀μ_r and ε = ε₀ε_r.

   v = 1/√(μ₀μ_r ε₀ε_r) = [1/√(μ₀ε₀)] × 1/√(μ_r ε_r) = c/√(μ_r ε_r).

   (Option A is dimensionally wrong — it is a pure number, not a speed; the refractive index is n = √(μ_r ε_r), and v = c/n.)

   ✓Final answer

   The correct option is (B) — ${c \over {\sqrt {{\mu _r}{\varepsilon _r}} }}$

   ANSWER: B

9. COMEDK 2022Set 20221 markMCQ
   Q.A 30 mW laser beam has a cross-sectional area of 15 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by [Permuittivity of space, $\varepsilon_0=9\times10^{-12}$ Speed of light, $c=3\times10^8$ m/s] (A) 1.22 kV/m (B) 12 kV/m (C) 10 kV/m (D) 201 kV/m

   ›Reveal solutionSolution

   Solve for E0: E0 = sqrt( 2I / (epsilon0 c) ) = sqrt( 2 x 2000 / (9 x 10^-12 x 3 x 10^8) ) = sqrt( 4000 / (2.7 x 10^-3) ) = sqrt( 1.481 x 10^6 ) = 1.22 x 10^3 V/m = 1.22 kV/m

   Concept: intensity of an EM wave in terms of the peak electric field:

   I = (1/2) epsilon0 c E0^2

   Intensity of the beam:

   I = P/A = 30 x 10^-3 W / (15 x 10^-6 m^2) = 2000 W/m^2

   Solve for E0:

   E0 = sqrt( 2I / (epsilon0 c) )

   = sqrt( 2 x 2000 / (9 x 10^-12 x 3 x 10^8) )

   = sqrt( 4000 / (2.7 x 10^-3) )

   = sqrt( 1.481 x 10^6 )

   = 1.22 x 10^3 V/m = 1.22 kV/m

   ✓Final answer

   The correct option is (A) — 1.22 kV/m

   ANSWER: A

10. COMEDK 2021Set 20211 markMCQ
    Q.The correct arrangement in increasing order of wavelength of X-rays, UV rays, microwave is (A) microwave, X-rays, UV rays (B) UV rays, X-rays, microwave (C) X-rays, UV rays, microwave (D) microwave, UV rays, X-rays

    ›Reveal solutionSolution

    Increasing wavelength: X-rays < UV rays < microwave.

    Concept: electromagnetic spectrum ordering.

    Typical wavelengths:

    • X-rays: ~1e-10 m (0.01-10 nm)
    • Ultraviolet: ~1e-8 m (10-400 nm)
    • Microwave: ~1e-2 m (1 mm - 1 m)

    Increasing wavelength: X-rays < UV rays < microwave.

    ✓Final answer

    The correct option is (C) — X-rays, UV rays, microwave

    ANSWER: C

11. COMEDK 2021Set 20211 markMCQ
    Q.Which of the following waves are used to treatment of muscles ache? (A) Ultraviolet (B) Infrared (C) Microwave (D) X-rays

    ›Reveal solutionSolution

    Infrared radiation has a heating effect (it excites molecular vibrations) and is used in heat lamps / IR therapy to relieve muscular aches and sprains. UV is used for sterilisation, microwaves for cooking and radar, X-rays for imaging.

    Concept: uses of electromagnetic waves.

    Infrared radiation has a heating effect (it excites molecular vibrations) and is used in heat lamps / IR therapy to relieve muscular aches and sprains. UV is used for sterilisation, microwaves for cooking and radar, X-rays for imaging.

    ✓Final answer

    The correct option is (B) — Infrared

    ANSWER: B

12. COMEDK 2021Set 2021-B1 markMCQ
    Q.The radiations used in treatment of muscles ache are (A) Microwave (B) Ultraviolet (C) Infrared (D) X-rays

    ›Reveal solutionSolution

    Infrared radiation is the heat radiation used to treat muscle aches.

    Infrared waves are absorbed by tissue and produce a warming effect, which relaxes muscles, dilates blood vessels and eases pain. This is the basis of infrared physiotherapy lamps used for muscular aches. Microwaves and ultraviolet are not used this way for muscle therapy, and X-rays are ionizing and used for imaging, not for heating muscles.

    ✓Final answer

    The correct option is (C) — Infrared