JEE Physics Electromagnetic Waves MCQs Set 07

Answer: (a) continuous oscillation (acceleration) of electric charge

Explanation:
1. Steady charges give only a static electric field and steady currents give a constant magnetic field.
2. Only accelerated or oscillating charges produce time-varying fields that regenerate each other and radiate.
3. So the source is the oscillation of electric charge.

Teacher's Note:
1. Remember: no acceleration, no radiation.
2. The wave frequency equals the oscillation frequency of the source charge.

Answer: (b) \( \frac{1}{\sqrt{\mu_0\varepsilon_0}} \)

Explanation:
1. Maxwell's equations give the wave equation with speed \( c=\frac{1}{\sqrt{\mu_0\varepsilon_0}} \).
2. Numerically \( c\approx 3\times10^{8}\ \text{m/s} \).

Teacher's Note:
1. Check the units: the result must be a speed, so the square root of a product in the denominator is needed.
2. This value of c matched the measured speed of light.

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

Explanation:
1. The energy flow (Poynting vector) is \( \vec S=\frac{1}{\mu_0}\vec E\times\vec B \).
2. The wave travels in the direction of the energy flow.

Teacher's Note:
1. Use the right-hand rule on E then B.
2. E and B are both perpendicular to the direction of travel.

Answer: (b) microwaves

Explanation:
1. Microwaves and radio waves penetrate the atmosphere and can be directed in narrow beams.
2. They are used for radar and communication, so microwaves is the choice among the options.

Teacher's Note:
1. Short wavelength radio waves are called microwaves.
2. They also heat water molecules in a microwave oven.

Answer: (d) beta rays

Explanation:
1. Heat rays (infrared), X-rays and gamma rays are electromagnetic.
2. Beta rays are streams of fast electrons, which are particles.

Teacher's Note:
1. Alpha and beta rays are particles.
2. Gamma rays are the most energetic electromagnetic waves.

Answer: (b) in phase and perpendicular to each other

Explanation:
1. E and B reach their maximum and zero values together, so they are in phase.
2. They are perpendicular to each other and to the propagation direction.

Teacher's Note:
1. Phase and direction are different ideas.
2. B0 = E0/c at every instant.

Answer: (d) (i) and (iii)

Explanation:
1. Wave number \( k=\omega/c=10^7/(3\times10^8)=0.033 \) rad/m, so (ii) is false.
2. Wavelength \( \lambda=2\pi/k=188.4 \) m, so (i) is true; the amplitude 10 V/m makes (iii) true.
3. The phase \( \omega t+kx \) has a plus sign, so the wave moves along -x and (iv) is false.

Teacher's Note:
1. Compute k = omega / c first.
2. Sign of the kx term decides the direction.

Answer: (b) \( k/\omega \)

Explanation:
1. Since \( \omega=ck \), \( k/\omega=1/c \) is a constant.
2. k and omega each change with wavelength.

Teacher's Note:
1. Any ratio that equals 1/c does not depend on wavelength.
2. In vacuum all wavelengths travel at the same speed.

Answer: (d) \( i/2 \)

Explanation:
1. The electric field between the plates is uniform and exists over the whole area A.
2. The displacement current through the whole area A equals i.
3. Through half the area it is \( i/2 \).

Teacher's Note:
1. Displacement current is proportional to the flux of the changing field.
2. Equal areas of the same uniform field carry equal share.

Answer: (b) \( 8\times10^{-12} \) J

Explanation:
1. The beam moves at c, so a 60 cm length passes in \( t=0.6/(3\times10^{8})=2\times10^{-9} \) s.
2. Energy \( =Pt=4\times10^{-3}\times2\times10^{-9}=8\times10^{-12} \) J.

Teacher's Note:
1. Energy in the length equals power times the time light takes to cross it.
2. Convert mW and cm to SI first.

Answer: (d) current density

Explanation:
1. The displacement current density is \( \varepsilon_0\,dE/dt \).
2. Multiplying by area would give current, so the quantity itself has the dimension of current per unit area.

Teacher's Note:
1. Displacement current i_d = eps0 d(flux)/dt.
2. Divide by area to get current density.

Answer: (b) \( \sqrt{\frac{\varepsilon\mu}{\varepsilon_0\mu_0}} \)

Explanation:
1. Speed in vacuum \( c=1/\sqrt{\mu_0\varepsilon_0} \) and in the medium \( v=1/\sqrt{\mu\varepsilon} \).
2. Refractive index \( n=c/v=\sqrt{\frac{\mu\varepsilon}{\mu_0\varepsilon_0}} \).

Teacher's Note:
1. n is the ratio of speeds, c over v.
2. A medium slows the wave because mu and eps are larger.

Answer: (d) 2.68 V/m

Explanation:
1. Radiated power \( =3 \) W; intensity at 5 m \( I=\frac{3}{4\pi(5)^2}=9.55\times10^{-3} \) W/m².
2. From \( I=\tfrac12\varepsilon_0cE_0^2 \): \( E_0=\sqrt{\frac{2I}{\varepsilon_0c}}\approx2.68 \) V/m.

Teacher's Note:
1. Spread the power over a sphere of area 4 pi r squared.
2. Then use the intensity relation.

Answer: (b) \( 10^{-8} \) J/m³

Explanation:
1. Average energy density \( u=\tfrac12\varepsilon_0E_{rms}^2=\tfrac14\varepsilon_0E_0^2 \).
2. Substituting \( E_0=50 \): \( u=\tfrac14(8.85\times10^{-12})(2500)\approx5.5\times10^{-9} \) J/m³, which is nearly \( 10^{-8} \)

Teacher's Note:
1. Electric and magnetic parts contribute equally.
2. Average of E squared is half its peak square.

Answer: (d) \( 0.25\times10^{4} \)

Explanation:
1. \( N=\frac{1\ \text{mm}}{4000\ \text{A}}=\frac{10^{-3}}{4\times10^{-7}}=2500=0.25\times10^{4} \).

Teacher's Note:
1. Convert both lengths to metres.
2. Number of waves = length / wavelength.

Answer: (b) none of these

Explanation:
1. Electromagnetic waves carry no charge.
2. Hence neither an electric nor a magnetic field can deflect them.

Teacher's Note:
1. Deflection by fields needs a charged particle.
2. Light bends only through media or gravity, not by such fields.

Answer: (d) \( 1.5\times10^{-2} \) m

Explanation:
1. \( \lambda=\frac{c}{f}=\frac{3\times10^{8}}{2\times10^{10}}=1.5\times10^{-2} \) m.

Teacher's Note:
1. Wavelength = speed / frequency.
2. Amplitude does not affect the wavelength.

Answer: (b) a pyrometer

Explanation:
1. A pyrometer measures radiated heat (infrared) from hot bodies.
2. A spectrometer analyses spectra, a photometer measures light intensity.

Teacher's Note:
1. Infrared is also called heat radiation.
2. Match the instrument to the quantity measured.

Answer: (d) gamma rays

Explanation:
1. Beta and alpha rays are particles, with de Broglie wavelengths far larger than a gamma photon of comparable energy.
2. Among electromagnetic waves, gamma rays have shorter wavelengths than X-rays.

Teacher's Note:
1. Gamma rays sit at the top of the electromagnetic spectrum.
2. Higher frequency means shorter wavelength.

Answer: (b) \( \hat X\parallel\vec E,\ \hat k\parallel\vec E\times\vec B \)

Explanation:
1. Polarization of a wave is the direction of its electric field, so \( \hat X\parallel\vec E \).
2. Propagation is along the Poynting vector direction \( \vec E\times\vec B \).

Teacher's Note:
1. Polarization is defined by E.
2. Order matters in the cross product.

Answer: (d) Statement-1 is true, Statement-2 is false

Explanation:
1. Radio waves have longer wavelength, so they diffract more: Statement-1 is true.
2. Longer wavelength means lower frequency, so radio waves have lower frequency than microwaves: Statement-2 is false.

Teacher's Note:
1. Diffraction increases with wavelength.
2. Frequency and wavelength vary inversely.

Answer: (b) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

Explanation:
1. Infrared (heat) therapy treats muscular strain: (a)-(i).
2. Radio waves are used for broadcasting: (b)-(ii).
3. X-rays image bones: (c)-(iii).
4. The ozone layer absorbs ultraviolet: (d)-(iv).

Teacher's Note:
1. Link each wave to its best-known use.
2. Eliminate options one pairing at a time.

Answer: (c) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

Explanation:
1. 700 nm to 1 mm is infrared, produced by vibrating atoms and molecules.
2. 1 nm to 400 nm is ultraviolet, from transitions of inner shell electrons.
3. Below \( 10^{-3} \) nm is gamma radiation, from radioactive decay.
4. 1 mm to 0.1 m is microwave, produced by a magnetron.

Teacher's Note:
1. Identify each band by its wavelength first.
2. Then recall its production method.

Answer: (d) \( 35.2\times10^{-12} \) J/m³

Explanation:
1. \( u_E=\tfrac14\varepsilon_0E_0^2=\tfrac14(8.8\times10^{-12})(16)=3.52\times10^{-11}=35.2\times10^{-12} \) J/m³.

Teacher's Note:
1. The time average of E squared is E0 squared over 2.
2. Then u = (1/2) eps0 times that.

Answer: (a) G, X, U, V, I, M, R

Explanation:
1. Wavelength from smallest to largest: gamma, X-rays, ultraviolet, visible, infrared, microwave, radio.
2. That sequence is G, X, U, V, I, M, R.

Teacher's Note:
1. Reverse the order of frequency.
2. Remember the spectrum from gamma to radio.

Answer: (b) \( (-2\hat i-3\hat j) \) and \( (3\hat i-2\hat j) \)

Explanation:
1. E and B must be perpendicular, and \( \vec E\times\vec B \) must point along +z.
2. Option 1: dot product 8, not perpendicular.
3. Options 2, 3, 4 are perpendicular; the z component of \( \vec E\times\vec B \) is \( E_xB_y-E_yB_x \) = +13 (option 2), -25 (option 3), -5 (option 4).
4. Only option 2 gives +z.

Teacher's Note:
1. Two conditions: perpendicular and cross product along z.
2. Compute the z component of the cross product.

Answer: (c) D, B, A, C

Explanation:
1. Photon energy \( E=hf \) increases with frequency.
2. Frequency order: radio, yellow, blue, X-ray, so D, B, A, C.

Teacher's Note:
1. Higher frequency means higher energy per photon.
2. Radio waves have the lowest.

Answer: (d) \( \frac{\varepsilon_{r_1}}{\varepsilon_{r_2}}=\frac14 \)

Explanation:
1. In air \( k=2\pi\nu/c \). In the medium the phase is \( 2kz-kct \), so wave number \( 2k \) and the same angular frequency \( kc \).
2. Speed in the medium \( v=\frac{kc}{2k}=\frac{c}{2} \), so \( n=2 \).
3. For a non-magnetic medium \( n^2=\varepsilon_{r_2}/\varepsilon_{r_1}=4 \), so \( \varepsilon_{r_1}/\varepsilon_{r_2}=1/4 \).

Teacher's Note:
1. Find the speed from omega over wave number.
2. Use n squared equals the permittivity ratio.