Electromagnetic waves practice · Medium

Option B (Cellular phone communication) is correct because Ultra High Frequency (UHF) radio waves are commonly used for cellular phone communication.

Option C is correct because the energy (and therefore power) of an X-ray photon is directly proportional to its frequency, as described by the equation E=hv.

Option C (Is same for all of them) is correct because the speed of electromagnetic waves in a vacuum is a constant, independent of the source or frequency.

Option D ($\mathrm{v}_{\mathrm{s}}=\mathrm{v}_{\mathrm{x}}=\mathrm{v}_{\mathrm{m}}$) is correct because all electromagnetic waves travel at the same speed (speed of light) in a vacuum.

Option A ($1.8 \times {10^{ - 8}}\mathrm{\;C}{\mathrm{s}^{ - 1}}$) is correct because the displacement current is equal to the rate of change of charge on the capacitor plates.

Option A ($\frac{{{\mathrm{\mu }_0}{i_D}r}}{{2\mathrm{\pi }{R^2}}}$) is correct because it represents the magnetic field inside the capacitor plates, derived using Ampere-Maxwell's law.

Option D ($3776 \Omega$) is incorrect. The wave impedance of free space is approximately 377 ohms. The provided solution incorrectly calculates the impedance.

Option B is correct because the Doppler effect causes a redshift (increase in wavelength) for objects moving away, shifting the spectrum towards the infrared region.

Option A is correct. Energy is inversely proportional to wavelength, and the order of increasing energy is Radio wave < Yellow light < Blue light < X-ray.

Option B (1.732) is correct because the refractive index is the square root of the product of relative permeability and dielectric constant.

Option D ($-24\,\,\widehat x\,\,\mathrm{V/m}$) is correct because the electric field is calculated using the relationship E = Bc and the direction is determined by the direction of wave propagation.

Option B ($\frac{{{B^2}}}{{2{\mu _0}}}$) is correct because the average magnetic energy density of an electromagnetic wave in free space is given by this formula.

Option B ($\frac{1}{{\sqrt {\mathrm{\mu \varepsilon }} }}$) is correct because Maxwell's equations relate the speed of light to the permeability and permittivity of the medium.

Option D (None of these) is correct because a wave with a frequency of $6 \times {10^{15}}\;Hz$ falls within the ultraviolet (UV) range of the electromagnetic spectrum.

Option B (Statement 1 is true & Statement 2 is true but is not correct explanation of Statement 1) is correct because both statements are individually true, but the pressure exerted by electromagnetic waves isn't a direct explanation for the equality of electric and magnetic energy densities.

Option D (Gamma - rays) is correct because gamma rays have the shortest wavelength in the electromagnetic spectrum among the given options.

Option A is correct because the energy density of an electromagnetic wave is proportional to the square of the electric field, leading to a frequency doubling.

Option D ($\frac{{B}^2 {AL}}{2 \mu_0}$) is correct because it represents the energy density multiplied by the volume of the region.

Option B is correct because the ratio of the energy density of an electromagnetic wave to the energy density due to the magnetic field is 2:1.

Option D is correct because the average value of both the electric and magnetic fields over a complete cycle in a plane electromagnetic wave is zero.

Option C is correct because the total energy density of electromagnetic waves in vacuum is the sum of electric and magnetic energy densities, which equals $\epsilon_0 E^2$.

Option C ($\frac{1}{C}$) is correct because the ratio of the magnetic field to the electric field in an EM wave is the inverse of the speed of light.

Option A ($8 \mu \mathrm{A}$; Yes) is correct because the conduction current is equal to the displacement current in a capacitor, and its value is calculated using the given parameters.

Option C (Statement I is false, statement II is true) is correct because all electromagnetic waves have the same speed in vacuum, but different frequencies.

Option D is correct because the average electric energy density is half the total average energy density in an electromagnetic wave.

Option B ($\frac{I A}{c}$) is correct because the momentum of photons per unit time is equal to the energy per unit time divided by the speed of light.

Option A ($40 \mathrm{~m}$ to $25 \mathrm{~m}$) is correct because it represents the calculated wavelength range corresponding to the given frequency band.

Option B (X-rays) is correct because electromagnetic waves with energy around 15 keV fall within the X-ray portion of the electromagnetic spectrum.

Option C (A charge moving in a circular orbit) is correct because accelerating charges, such as those moving in a circular orbit, produce electromagnetic waves.

Option B ($\sqrt {{\mu _0}{\varepsilon _0}} {E_0} = {B_0}$) is correct because the speed of light, $c$, is related to the electric field amplitude, $E_0$, and magnetic field amplitude, $B_0$, by $c = E_0/B_0$ and $c = 1/\sqrt{{\mu_0}{\varepsilon_0}}$.