JEE Physics Electromagnetic Waves MCQs Set 03

Answer: (a) \( 35.2\times10^{-12} \) J/m\( ^3 \)

Explanation:
1. Average electric energy density \( u_E = \frac{1}{2}\varepsilon_0 E_{rms}^2 = \frac{1}{4}\varepsilon_0 E_0^2 \).
2. \( u_E = \frac{1}{4}\times8.8\times10^{-12}\times16 = 35.2\times10^{-12} \) J/m\( ^3 \).
3. The frequency is not needed.

Teacher's Note:
1. The average of \( \cos^2 \) over a cycle is one half, which gives the factor of one quarter.
2. The magnetic field carries the same average energy density.

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

Explanation:
1. Electromagnetic waves are transverse, with \( \vec{E} \) and \( \vec{B} \) perpendicular to each other.
2. The direction of propagation is that of the Poynting vector, \( \vec{E}\times\vec{B} \).

Teacher's Note:
1. The order of the cross product matters.
2. Reversing it would give the opposite direction.

Answer: (a) \( \vec{E} = E_0\hat{i},\ \vec{B} = B_0\hat{j} \)

Explanation:
1. The wave direction is along \( \vec{E}\times\vec{B} = +\hat{k} \), and both fields must be perpendicular to z.
2. Options (b) and (d) have a field along z, so they are not transverse.
3. In option (c), \( \hat{j}\times\hat{i} = -\hat{k} \), which is the wrong direction. In option (a), \( \hat{i}\times\hat{j} = +\hat{k} \), which is correct.

Teacher's Note:
1. Use the right-hand rule for \( \vec{E}\times\vec{B} \).
2. Both fields lie in the plane perpendicular to the wave.

Answer: (c) \( 1.6\times10^{-6} \) Wb m\( ^{-2} \)

Explanation:
1. In vacuum \( \frac{E_0}{B_0} = c \), so \( B_0 = \frac{E_0}{c} \).
2. \( B_0 = \frac{480}{3\times10^{8}} = 1.6\times10^{-6} \) T (Wb m\( ^{-2} \)).

Teacher's Note:
1. The frequency does not enter the ratio of the field amplitudes.
2. The magnetic field of a wave is tiny compared with the electric field in SI units.

Answer: (a) \( \gamma \)-rays

Explanation:
1. Of the electromagnetic radiations listed, gamma rays have the highest frequency and the shortest wavelength, shorter than X rays.
2. Alpha and beta rays are streams of particles, not electromagnetic waves, so they are not ranked by wavelength.

Teacher's Note:
1. Order of EM spectrum by increasing wavelength: gamma, X rays, ultraviolet, visible, infrared, microwaves, radio.
2. Remember that alpha and beta are particles.

Answer: (d) \( \frac{IA}{c}(2 - \eta) \)

Explanation:
1. Incident momentum per second is \( \frac{IA}{c} \).
2. The absorbed part \( \eta \) transfers \( \frac{\eta IA}{c} \), and the reflected part \( (1 - \eta) \) transfers twice its momentum, \( \frac{2(1 - \eta)IA}{c} \).
3. Total force \( F = \frac{IA}{c}[\eta + 2(1 - \eta)] = \frac{IA}{c}(2 - \eta) \).

Teacher's Note:
1. Perfect reflection gives twice the radiation force of perfect absorption.
2. This is radiation pressure at work.

Answer: (c) wavelength is halved and frequency remains unchanged

Explanation:
1. Frequency is fixed by the source and does not change on entering a medium.
2. Speed in the medium is \( v = \frac{c}{\sqrt{\varepsilon_r\mu_r}} = \frac{c}{2} \) for \( \mu_r = 1 \).
3. Since \( \lambda = \frac{v}{\nu} \), the wavelength is halved.

Teacher's Note:
1. Wave speed and wavelength change on entering a medium, but frequency does not.
2. The refractive index here is 2.

Answer: (c) 2.2 \( \mu \)A

Explanation:
1. Displacement current equals the conduction current charging the capacitor, \( I_d = C\frac{dV}{dt} \).
2. \( C = \frac{K\varepsilon_0 A}{d} = \frac{10\times8.85\times10^{-12}\times1}{10^{-3}} = 8.85\times10^{-8} \) F.
3. \( I_d = 8.85\times10^{-8}\times25 \approx 2.2\times10^{-6} \) A = 2.2 \( \mu \)A.

Teacher's Note:
1. Displacement current is what makes Ampere's law consistent between the plates.
2. It has the same value as the charging current in the wires.