JEE Physics Electromagnetic Waves MCQs Set 01

Answer: (c) \( \vec E \times \vec B \)

Explanation:
1. The wave travels along the Poynting vector \( \vec S \propto \vec E \times \vec B \).
2. It is perpendicular to both \( \vec E \) and \( \vec B \), so options 1 and 2 are wrong.

Teacher's Note:
1. Use the right-hand rule: curl from \( \vec E \) to \( \vec B \) and the thumb gives the direction of travel.
2. The wave is transverse.

Answer: (b) Charge

Explanation:
1. EM waves carry energy and momentum and are used to send information.
2. The wave is made of fields only and has no net charge, so it cannot transport charge.

Teacher's Note:
1. Radiation pressure shows that EM waves carry momentum.
2. Fields are neutral, so no charge flows with the wave.

Answer: (b) \( 3\times10^{8} \) m/s

Explanation:
1. Compare with \( \sin(kx-\omega t) \): \( k=3\pi\times10^{6} \), \( \omega=9\pi\times10^{14} \).
2. \( v=\omega/k=9\times10^{14}/(3\times10^{6})=3\times10^{8} \) m/s.

Teacher's Note:
1. In the form sin(kx - wt) the speed is simply w/k.
2. The result equals the speed of light, as expected.

Answer: (a) 666 nm

Explanation:
1. \( k=3\pi\times10^{6} \) m\(^{-1}\).
2. \( \lambda=2\pi/k=2/(3\times10^{6})=6.67\times10^{-7} \) m = 666 nm.

Teacher's Note:
1. Do not forget the factor of pi inside the bracket when reading k.
2. 666 nm lies in the visible region.

Answer: (c) Equally the electric and magnetic fields

Explanation:
1. The electric and magnetic energy densities are \( \tfrac12\varepsilon_0E^2 \) and \( B^2/2\mu_0 \).
2. Since \( E=cB \), the two are equal, so each carries half of the total energy.

Teacher's Note:
1. Energy is shared equally between the two fields at every instant.
2. Total energy density is epsilon0 E^2.

Answer: (b) \( \dfrac{B^2}{2\mu_0} \)

Explanation:
1. Energy stored per unit volume in a magnetic field is \( u_B=B^2/2\mu_0 \).
2. Options 1 is an energy, not a density, and the others have wrong dependence on B.

Teacher's Note:
1. Magnetic energy density has B squared in the numerator and mu0 in the denominator.
2. Compare with electric: epsilon0 E^2 / 2.

Answer: (d) \( \tfrac12\varepsilon_0E^2 \)

Explanation:
1. Electric energy per unit volume is \( u_E=\tfrac12\varepsilon_0E^2 \).
2. Options 1 and 2 are total energies of a capacitor, not densities.

Teacher's Note:
1. Energy density has units J/m^3.
2. Capacitor energy formulas give total energy, not density.

Answer: (a) Lower

Explanation:
1. Greenhouse gases trap outgoing infrared radiation and keep the surface warm.
2. Without them the heat would escape and the surface would be much colder.

Teacher's Note:
1. Greenhouse effect warms the Earth, about 33 K on average.
2. Without the atmosphere the Earth would be too cold for life.

Answer: (b) Stratosphere

Explanation:
1. Most ozone lies in the stratosphere, roughly 15 to 35 km above the ground.
2. This is where it absorbs harmful ultraviolet radiation.

Teacher's Note:
1. Troposphere is the weather layer, stratosphere has the ozone layer.
2. Remember S for stratosphere and S for shield.

Answer: (c) Absorbs ultraviolet rays

Explanation:
1. Ozone absorbs most of the harmful ultraviolet radiation coming from the Sun.
2. Radio signals are reflected by the ionosphere, not by ozone.

Teacher's Note:
1. Ultraviolet causes skin cancer and cataract.

Answer: (b) Super high frequency band

Explanation:
1. The standard radio bands are HF (3 to 30 MHz), VHF (30 to 300 MHz), UHF (300 to 3000 MHz) and SHF (3 to 30 GHz).
2. The range 3 GHz to 30 GHz is SHF.

Teacher's Note:
1. Each band is a factor of ten wide.
2. SHF is used in radar and satellite links.

Answer: (d) High frequency band

Explanation:
1. The HF band extends from 3 MHz to 30 MHz.
2. MF is 0.3 to 3 MHz and VHF is 30 to 300 MHz.

Teacher's Note:
1. Learn the bands in sequence: VLF, LF, MF, HF, VHF, UHF, SHF.
2. Short-wave radio lies in HF.

Answer: (b) They travel with the same speed in all media

Explanation:
1. The speed in a medium is \( v=1/\sqrt{\mu\varepsilon} \) and differs from medium to medium.
2. Only in vacuum is it c.

Teacher's Note:
1. Speed in vacuum is the same for all EM waves but not in a medium.
2. Frequency stays fixed while the wavelength changes.

Answer: (c) Transverse

Explanation:
1. \( \vec E \) and \( \vec B \) oscillate perpendicular to the direction of travel.
2. The wave progresses, so it is not stationary.

Teacher's Note:
1. Transverse waves can be polarised, which proves the nature of light.
2. Sound in air is longitudinal.

Answer: (c) Warm at night

Explanation:
1. Greenhouse gases absorb and re-radiate infrared radiation from the ground.
2. This prevents rapid cooling and keeps the surface warm at night.

Teacher's Note:
1. Think of the atmosphere as a blanket.
2. Without it nights would be extremely cold.

Answer: (d) \( \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} \)

Explanation:
1. Maxwell's equations give \( c=1/\sqrt{\mu_0\varepsilon_0} \).
2. Putting the values gives about \( 3\times10^{8} \) m/s.

Teacher's Note:
1. This relation linked light with electromagnetism.
2. Check units: 1/sqrt(mu0 eps0) gives m/s.

Answer: (a) Infrared rays

Explanation:
1. The Earth radiates heat as infrared, which greenhouse gases such as CO2 and water vapour absorb.
2. The trapped radiation warms the lower atmosphere.

Teacher's Note:
1. Infrared is heat radiation.
2. UV is absorbed higher up, by ozone.

Answer: (a) Microwave, infrared, ultraviolet, gamma rays

Explanation:
1. Wavelength decreases as frequency increases along the spectrum.
2. The order of increasing frequency is microwave, infrared, ultraviolet, gamma.

Teacher's Note:
1. Radio, micro, infrared, visible, UV, X-ray, gamma is increasing frequency.
2. Wavelength is the reverse.

Answer: (c) 2 m\(^{-1}\)

Explanation:
1. \( \omega=6\times10^{8} \) rad/s.
2. \( k=\omega/c=6\times10^{8}/(3\times10^{8})=2 \) m\(^{-1}\).

Teacher's Note:
1. In a vacuum k = omega / c.
2. Always read omega from the coefficient of t.

Answer: (b) The frequency of the microwaves matches the natural frequency of the water molecules

Explanation:
1. Water molecules are polar and have a natural rotational frequency in the microwave range.
2. Efficient energy absorption occurs when the two frequencies match.

Teacher's Note:
1. Matching frequencies give maximum absorption.
2. Microwaves are not heat waves in themselves, the energy must be absorbed.

Answer: (b) X-rays

Explanation:
1. \( \lambda=hc/E=12400\ \text{eV}\,\AA/15000\ \text{eV}\approx0.83\ \AA \).
2. This wavelength belongs to X-rays.

Teacher's Note:
1. Photon energies of keV order are X-rays.
2. Gamma rays are usually above 100 keV.

Answer: (a) \( \dfrac{2E}{c} \)

Explanation:
1. Incident momentum is \( E/c \) and the reflected momentum is \( -E/c \).
2. The change is \( 2E/c \).

Teacher's Note:
1. For total absorption the transfer is E/c.
2. For total reflection it is doubled.

Answer: (d) 4.9 \( \mu \)T

Explanation:
1. \( I=P/A=9\times10^{-3}/(\pi\times10^{-6})=2.86\times10^{3} \) W/m\(^2\).
2. \( E_0=\sqrt{2I/(c\varepsilon_0)}\approx1.47\times10^{3} \) V/m.
3. \( B_0=E_0/c=4.9\times10^{-6} \) T.

Teacher's Note:
1. The area uses the radius 1 mm, not 2 mm.
2. Use I = (1/2) epsilon0 c E0^2.

Answer: (c) \( 4.66\times10^{-6} \) N/m\(^2\)

Explanation:
1. A black body absorbs completely, so \( p=I/c \).
2. \( p=1400/(3\times10^{8})=4.66\times10^{-6} \) N/m\(^2\).

Teacher's Note:
1. Absorbing surface: I/c. Reflecting surface: 2I/c.
2. The pressure of sunlight is very small.

Answer: (b) \( 3/2 \)

Explanation:
1. The wave speed is \( v=2\times10^{8} \) m/s.
2. \( n=c/v=3\times10^{8}/(2\times10^{8})=3/2 \).

Teacher's Note:
1. Read the speed from z - vt.
2. Refractive index is c divided by v.

Answer: (d) \( H_0=E_0\sqrt{\varepsilon_0/\mu_0} \)

Explanation:
1. \( B_0=E_0/c \) and \( H_0=B_0/\mu_0 \).
2. So \( H_0=E_0/(c\mu_0)=E_0\sqrt{\varepsilon_0/\mu_0} \).

Teacher's Note:
1. Impedance of free space is E0/H0 = sqrt(mu0/eps0), about 377 ohm.
2. Do not confuse B and H.