Q.If the amplitude of the magnetic field is $3\times10^{-6}$ T, then amplitude of the electric field for an electromagnetic wave is

Showing the 12 most recent of 32 on this concept.

1. CBSE 2026Set 55/3/11 markMCQ
   Q.A plane electromagnetic wave travels through a medium and the magnetic field associated with it is given by $B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)\,\text{T}$, where $x$ is in metres and $t$ is in seconds. The velocity of the wave is : (A) $2.0\times10^{8}\ \text{ms}^{-1}$ (B) $4.5\times10^{7}\ \text{ms}^{-1}$ (C) $3.5\times10^{7}\ \text{ms}^{-1}$ (D) $2.5\times10^{8}\ \text{ms}^{-1}$

   ›Reveal solutionSolution

   The wave velocity is found from the ratio $\omega/k$ in the given sinusoidal form. Here $\omega = 3\times10^{10}$ rad/s and $k = 150$ rad/m, giving $v = \omega/k = 2.0\times10^{8}$ m/s. The correct option is (A).

   The magnetic field is given as $B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)$ T. This is a standard travelling wave expression of the form $B = B_0 \sin(\omega t - kx)$, where $\omega$ is the angular frequency and $k$ is the wave number. For any wave, the phase velocity is $v = \omega/k$. That is the direct route — no need to involve permittivity, permeability, or refractive index unless the medium is specified differently. Here the medium is simply given by the wave parameters themselves.

   1. Identify $\omega$ and $k$ from the equation.

      The term multiplying $t$ is $\omega = 3\times10^{10}$ rad/s.

      The term multiplying $x$ is $k = 150$ rad/m.

   2. Apply the wave velocity formula:

   $v = \frac{\omega}{k} = \frac{3\times10^{10}}{150}$

   3. Simplify:

   $v = \frac{3\times10^{10}}{1.5\times10^{2}} = 2\times10^{8}\ \text{m/s}$

   Watch out

   A common mistake is to confuse $k$ with wavelength $\lambda$ or to use $v = f\lambda$ without first finding $f$ and $\lambda$ correctly. Here $f = \omega/(2\pi)$ and $\lambda = 2\pi/k$, so $v = f\lambda = (\omega/2\pi)(2\pi/k) = \omega/k$ — same result. But the direct ratio is faster and less error-prone.

   Tip

   In an exam, whenever you see a wave written as $\sin(\omega t - kx)$ or $\cos(\omega t - kx)$, immediately read off $\omega$ and $k$ and compute $v = \omega/k$. This works for any sinusoidal wave — mechanical or electromagnetic.

   ✓Final answer

   The velocity of the wave is $2.0\times10^{8}$ m/s, which corresponds to option (A).

2. CBSE 2026Set A1 markMCQ
   Q.The dimensions of B0^2/μ0 will be the same as that of (A) energy density (B) work (C) momentum (D) electric flux

   ›Reveal solutionSolution

   Magnetic energy density = B²/2μ₀, so B²/μ₀ carries the dimensions of energy density.

   The energy stored per unit volume in a magnetic field is $u_B = \dfrac{B^2}{2\mu_0}$.

   Apart from the numerical factor ½, the quantity $\dfrac{B^2}{\mu_0}$ therefore has the dimensions of energy density (J/m³).

   Checking: $[B] =$ T $=$ kg·s⁻²·A⁻¹, $[\mu_0] =$ kg·m·s⁻²·A⁻². Then $\dfrac{B^2}{\mu_0}$ = kg·m⁻¹·s⁻² = J/m³ = energy density. ✓

   ✓Final answer

   (A) energy density.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The velocity of electromagnetic waves in vacuum is:

   (a) c = 1/√(μ₀ε₀)

   (b) c = √(μ₀ε₀)

   (c) c = √(μ₀/ε₀)

   (d) c = √(ε₀/μ₀)

   ›Reveal solutionSolution

   Maxwell's equations predict electromagnetic waves travelling in vacuum at speed $c=1/\sqrt{\mu_0\varepsilon_0}$, which numerically matches the measured speed of light.

   Starting from Maxwell's equations in free space (no charges or currents), the wave equations for $\vec E$ and $\vec B$ both take the standard wave-equation form with wave speed $v=1/\sqrt{\mu_0\varepsilon_0}$. Substituting $\mu_0=4\pi\times10^{-7}\ \text{T m/A}$ and $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ gives $v\approx3\times10^{8}\ \text{m/s}$, exactly the speed of light — this agreement is what led Maxwell to conclude that light itself is an electromagnetic wave. The other options mix up the reciprocal/square-root arrangement and do not have the right dimensions of speed.

   ✓Final answer

   (a) $c=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.In a plane electromagnetic wave the electric field oscillates sinusoidally with a frequency $2.5\times10^{10}Hz$ and amplitude $480V/m$. The amplitude of the oscillating magnetic field will be –

   (a) $1.52\times10^{-8}\ weber/m^2$

   (b) $1.52\times10^{-7}\ weber/m^2$

   (c) $1.6\times10^{-6}\ weber/m^2$

   (d) $1.6\times10^{-7}\ weber/m^2$

   ›Reveal solutionSolution

   $B_0 = E_0/c$.

   In a plane electromagnetic wave, the amplitudes of the electric and magnetic fields are related by $B_0 = \dfrac{E_0}{c}$:

   $B_0 = \frac{480}{3\times10^8} = 1.6\times10^{-6}\ \text{T} = 1.6\times10^{-6}\ \text{weber/m}^2$

   ✓Final answer

   (c) $1.6\times10^{-6}$ weber/m$^2$.

5. CBSE 2025Set 55/4/11 markMCQ
   Q.The amplitude of the electric field in an electromagnetic wave in free space is $1000$ Vm$^{-1}$. The amplitude of the magnetic field in this electromagnetic wave is: (A) $3.0 \times 10^{-3}$ T (B) $3.33 \times 10^{-8}$ T (C) $3.0 \times 10^{11}$ T (D) $3.33 \times 10^{-6}$ T

   ›Reveal solutionSolution

   In an electromagnetic wave, the electric and magnetic field amplitudes are related by the speed of light: $E_0 = c B_0$. With $E_0 = 1000$ V/m, we find $B_0 = 3.33 \times 10^{-6}$ T.

   Why the fields are linked by $c$

   An electromagnetic wave is a self-sustaining oscillation of electric and magnetic fields that propagate together through space. Maxwell's equations demand a precise relationship between these two fields: at every instant and every point in the wave, the ratio of the electric field amplitude to the magnetic field amplitude equals the speed of light in that medium.

   In free space, this relationship is beautifully simple:

   $E_0 = c B_0$

   where $E_0$ is the amplitude of the electric field, $B_0$ is the amplitude of the magnetic field, and $c = 3 \times 10^8$ m/s is the speed of light in vacuum. This isn't arbitrary—it emerges directly from the wave equations derived from Maxwell's laws. The electric and magnetic fields are perpendicular to each other and to the direction of propagation, oscillating in phase, with their amplitudes locked in this ratio.

   Finding the magnetic field amplitude

   We're given the electric field amplitude and need to find the magnetic field amplitude.

   1. Write down the fundamental relationship:

   $E_0 = c B_0$

   2. Rearrange to solve for $B_0$:

   $B_0 = \frac{E_0}{c}$

   3. Substitute the given values:
      • $E_0 = 1000$ V/m
      • $c = 3 \times 10^8$ m/s

   $B_0 = \frac{1000}{3 \times 10^8}$

   4. Simplify the fraction:

   $B_0 = \frac{1000}{3 \times 10^8} = \frac{10^3}{3 \times 10^8} = \frac{1}{3} \times 10^{-5}$

   $B_0 = 0.333... \times 10^{-5} = 3.33 \times 10^{-6} \text{ T}$

   Tip

   A quick sanity check: magnetic field amplitudes in EM waves are always much smaller than electric field amplitudes (numerically) because $c$ is such a large number. If you ever get $B_0 > E_0$ in magnitude, something has gone wrong.

   ✓Final answer

   The correct option is (D) $3.33 \times 10^{-6}$ T.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.A plane electromagnetic wave travels in free space along x-direction. At a point in space and time, $\vec{E} = 9.0\hat{j}$ V/m. Magnitude of $\vec{B}$ at this point is-

   (a) $3\times10^{-8}$ T

   (b) $9\times10^{-8}$ T

   (c) $27\times10^{9}$ T

   (d) $2.1\times10^{-8}$ T

   ›Reveal solutionSolution

   In a plane EM wave in free space, $B = E/c$.

   For a plane electromagnetic wave travelling in free space, the magnitudes of the electric and magnetic fields are related by

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

   Here $E = 9.0$ V/m and $c = 3\times10^{8}$ m/s, so

   $B = \frac{9.0}{3\times10^{8}} = 3\times10^{-8}\ \text{T}$

   ✓Final answer

   (a) $3\times10^{-8}$ T

7. CBSE 2025Set D1 markMCQ
   Q.Unit of √(μ0/ε0) is (A) newton/coulomb (B) ohm (C) henry (D) farad

   ›Reveal solutionSolution

   √(μ₀/ε₀) has the dimensions of resistance; it is the intrinsic impedance of free space, about 377 ohm.

   The ratio of the electric to magnetic field amplitude of an electromagnetic wave in vacuum is c = E/B, and the quantity

   $Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}$

   is the impedance of free space. Numerically

   $Z_0 = \sqrt{\frac{4\pi\times10^{-7}}{8.85\times10^{-12}}} \approx 377\,\Omega$

   Since impedance is measured in ohms, the unit of √(μ₀/ε₀) is the ohm.

   ✓Final answer

   (B) ohm.

8. CBSE 2025Set ANNUAL1 mark
   Q.The magnetic field in a plane electromagnetic wave is given by $B_y = 2 \times 10^{-7} \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)\ T$. Find the wavelength of the given electromagnetic wave.

   ›Reveal solutionSolution

   $\lambda = 2\pi/k$, with $k=0.5\times10^3$ rad/m read off the wave equation.

   Comparing $B_y = 2\times10^{-7}\sin(0.5\times10^3 x + 1.5\times10^{11}t)$ with the standard form $B_y = B_0\sin(kx+\omega t)$, the wave number is $k = 0.5\times10^3\ \text{rad/m}$. The wavelength is

   $\lambda = \frac{2\pi}{k} = \frac{2\pi}{500} \approx 1.257\times10^{-2}\ \text{m} \approx 1.26\ \text{cm}$

   ✓Final answer

   $\lambda \approx 1.26\times10^{-2}$ m $\approx 1.26$ cm.

9. CBSE 2024Set A1 markMCQ
   Q.The value of (μ₀ε₀)^-1/2 is (A) 3 × 10^8 cm/second (B) 3 × 10^10 cm/second (C) 3 × 10^9 cm/second (D) 3 × 10^8 km/second

   ›Reveal solutionSolution

   1/√(μ₀ε₀) is the speed of light c = 3×10⁸ m/s = 3×10¹⁰ cm/s.

   Maxwell showed the speed of electromagnetic waves in vacuum is

   $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3\times10^8\ \text{m/s}$

   Converting to CGS: 3×10⁸ m/s × 100 cm/m = 3×10¹⁰ cm/s.

   ✓Final answer

   (B) 3 × 10¹⁰ cm/second.

10. CBSE 2024Set ANNUAL1 markMCQ
    Q.If the amplitude of the magnetic field is $3 \times 10^{-6}$ T, then the amplitude of the electric field for a electromagnetic wave is :

    (a) 600 Vm$^{-1}$

    (b) 100 Vm$^{-1}$

    (c) 900 Vm$^{-1}$

    (d) 300 Vm$^{-1}$

    ›Reveal solutionSolution

    Using $E_0 = cB_0$ with $c = 3\times10^8\ \text{ms}^{-1}$ and $B_0 = 3\times10^{-6}$ T gives $E_0 = 900\ \text{Vm}^{-1}$.

    Working

    For an electromagnetic wave travelling in free space, the peak electric and magnetic field amplitudes are related by

    $E_0 = c B_0$

    Given $B_0 = 3\times10^{-6}$ T:

    $E_0 = (3\times10^{8})\times(3\times10^{-6}) = 9\times10^{2} = 900\ \text{Vm}^{-1}$

    ✓Final answer

    The correct option is (c): $E_0 = 900\ \text{Vm}^{-1}$

11. CBSE 2023Set TERM21 markMCQ
    Q.Amplitude of the magnetic field part of a harmonic Electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave ?

    (a) 150 NC-1

    (b) 160 NC-1

    (c) 153 NC-1

    (d) 163 NC-1

    ›Reveal solutionSolution

    In vacuum the ratio of the electric-field amplitude to the magnetic-field amplitude of an EM wave equals the speed of light: $E_0 = cB_0$.

    For a plane electromagnetic wave travelling in vacuum, Maxwell's equations require the electric and magnetic field amplitudes to be related by

    $E_0 = c\,B_0$

    where $c = 3\times10^{8}\ \text{m/s}$ is the speed of light in vacuum.

    Given $B_0 = 510\ \text{nT} = 510\times10^{-9}\ \text{T}$:

    $E_0 = (3\times10^{8})(510\times10^{-9}) = 153\ \text{NC}^{-1}$

    ✓Final answer

    (c) $153\ \text{NC}^{-1}$.

12. CBSE 2023Set ANNUAL1 markMCQ
    Q.If the magnitude of the magnetic field is $3 \times 10^{-6}$ T, then the magnitude of the electric field for a electromagnetic wave is :

    (a) 600 Vm$^{-1}$

    (b) 100 Vm$^{-1}$

    (c) 900 Vm$^{-1}$

    (d) 300 Vm$^{-1}$

    ›Reveal solutionSolution

    Using $E=cB$ for an electromagnetic wave with $c=3\times10^8\ \text{m/s}$ and $B=3\times10^{-6}$ T gives $E=900\ \text{Vm}^{-1}$.

    Working

    For an electromagnetic wave travelling in free space (or air), the magnitudes of the electric and magnetic field amplitudes are related by

    $E = cB$

    where $c = 3\times10^8\ \text{ms}^{-1}$ is the speed of light.

    Given $B = 3\times10^{-6}$ T:

    $E = (3\times10^{8})\times(3\times10^{-6}) = 9\times10^{2} = 900\ \text{Vm}^{-1}$

    ✓Final answer

    The correct option is (c): $E = 900\ \text{Vm}^{-1}$