JEE Physics Electromagnetic Waves MCQs Set 08

Answer: (b) \( E_0k = B_0\omega \)

Explanation:
1. In vacuum \( \frac{E_0}{B_0} = c = \frac\omega k \).
2. Cross-multiplying, \( E_0k = B_0\omega \).

Teacher's Note:
1. Speed of the wave is ω/k, which equals E₀/B₀ in vacuum.
2. Check units on each relation.

Answer: (d) along the electric field

Explanation:
1. The electric force \( qE \) acts at once along E, whereas the magnetic force \( qv\times B \) needs a non-zero velocity.
2. So the proton begins to move along the electric field.

Teacher's Note:
1. A charge at rest feels no magnetic force.
2. Later motion has a small drift along the propagation direction.

Answer: (a) changes with time (increases or decreases)

Explanation:
1. \( I_d = \epsilon_0\frac{d\phi_E}{dt} \), and the electric flux between the plates is proportional to the charge.
2. So displacement current exists only when the charge is changing.

Teacher's Note:
1. Displacement current needs a changing electric field.
2. In steady state it is zero.

Answer: (c) the electric field and the magnetic field, with energy equally shared between them

Explanation:
1. Energy densities: \( u_E = \tfrac12\epsilon_0E^2 \) and \( u_B = \frac{B^2}{2\mu_0} \). With \( E = cB \) and \( c^2 = \frac1{\mu_0\epsilon_0} \) they are equal.

Teacher's Note:
1. Each field carries half of the total energy.
2. This holds for plane waves in vacuum.

Answer: (c) 377 m

Explanation:
1. \( f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10^{-4}\times4\times10^{-10}}} = 7.96\times10^5 \) Hz.
2. \( \lambda = \frac cf = \frac{3\times10^8}{7.96\times10^5} \approx 377 \) m.

Teacher's Note:
1. Find the oscillation frequency first.
2. λ = c/f.

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

Explanation:
1. The speed of an electromagnetic wave in a medium is c/n, which is less than c.
2. Only in vacuum is the speed c.

Teacher's Note:
1. Speed depends on the medium's permittivity and permeability.
2. Frequency is unchanged in a medium; wavelength changes.

Answer: (d) zero

Explanation:
1. E and B reach their maxima and zero values together; they are in phase.
2. They are perpendicular in space but not in phase.

Teacher's Note:
1. Space perpendicularity should not be confused with phase difference.
2. Both vary as sin(ωt − kx).

Answer: (b) accelerated or decelerated charges

Explanation:
1. Radiation needs a time-varying electric and magnetic field, which requires acceleration of charge.
2. Static charges and charges in uniform motion produce steady fields only.

Teacher's Note:
1. An oscillating charge is the standard source.
2. Uniform motion is a steady current.

Answer: (b) \( \tfrac12\epsilon_0E^2 + \frac{B^2}{2\mu_0} \)

Explanation:
1. Electric energy density \( \tfrac12\epsilon_0E^2 \); magnetic energy density \( \frac{B^2}{2\mu_0} \). The total is their sum.

Teacher's Note:
1. Check dimensions of each term.
2. μ₀ appears in the denominator of the magnetic term.

Answer: (c) \( 3.33\times10^{-9} \) T in the −x direction

Explanation:
1. \( B = \frac Ec = \frac{1}{3\times10^8} = 3.33\times10^{-9} \) T.
2. Direction: \( \hat E\times\hat B = \hat z \). \( \hat j\times(-\hat i) = +\hat k \), so B points along −x.

Teacher's Note:
1. The wave moves along E × B.
2. B is smaller than E by the factor c.

Answer: (a) \( \frac5\pi\sin(10^8t+2z)\,\hat i \)

Explanation:
1. Wave speed \( v = \frac\omega k = \frac{10^8}{2} = 5\times10^7 \) m/s, travelling along −z.
2. \( H = \frac{B}{\mu_0} = \frac{E}{v\mu_0} = \frac{100}{5\times10^7\times4\pi\times10^{-7}} = \frac5\pi \). Direction: \( \hat j\times\hat i = -\hat k \), so H is along x.

Teacher's Note:
1. The (ωt + kz) form means propagation along −z.
2. H is perpendicular to both E and the direction of travel.

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

Explanation:
1. The wave is transverse and the energy flows along the Poynting vector \( \vec E\times\vec B/\mu_0 \).
2. So the direction of propagation is E × B.

Teacher's Note:
1. Right-hand rule: fingers from E to B, thumb gives the direction of propagation.
2. E and B are both perpendicular to the velocity.

Answer: (b) \( 7\times10^{-6} \) N

Explanation:
1. Radiation pressure for complete absorption \( = \frac Ic \). \( F = \frac{IA}{c} = \frac{1400\times1.5}{3\times10^8} = 7\times10^{-6} \) N.

Teacher's Note:
1. Complete reflection would give twice as much force.
2. Pressure = intensity/c.