Chapter Exam

Unit 15: Electromagnetic Waves — Online MCQ Test

Q: A laser beam has intensity of 4.0 x 10^14 W / m². The amplitude of magnetic field associated with beam is ____ T. (Take ε0 = 8.85 x 10^-12 C²/Nm² and c = 3 x 10^8 m/s)

Answer: D. 1.83 — Using I = (1/2)ε0 E0² c and E0 = B0c, we derive B0 = sqrt(2I / ε0c³). Plugging in values, B0 = 1.83 T.

Q: A plane electromagnetic wave is moving in free space with velocity c = 3 * 10^8 m/s and its electric field is given as E = 54*sin(kz - omega*t) j_hat V/m, where j_hat is the unit vector along the y-axis. The magnetic field vector B of the wave is:

Answer: A. (1.8 * 10^-7) * sin(kz - omega*t) * (-i_hat) T — Using B = (1/c) * (k_hat cross E). Since E is along j_hat and wave propagates along z (k_hat), B = (1/c) * E * (k_hat cross j_hat) = -(E/c) * i_hat. E_0 = 54, so B_0 = 54 / (3 * 10^8) = 1.8 * 10^-7.

Q: A plane electromagnetic wave is moving in free space with velocity c = 3 × 10^8 m/s and its electric field is given as E = 54 sin(kz - ωt) ĵ V/m, where ĵ is the unit vector along y-axis. The magnetic field vector B of the wave is:

Answer: A. 1.8 × 10^-7 sin(kz - ωt) î T — Using the relation B = (1/c)(k̂ × E), where direction of propagation k̂ = k̂, and E is along ĵ. (k̂ × ĵ) = -î. Magnitude B₀ = E₀/c = 54 / (3 × 10⁸) = 1.8 × 10⁻⁷ T. Thus, B = -1.8 × 10⁻⁷ sin(kz - ωt) î T.

Q: The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by, E_y = 20 sin(3*10^6 x - 4.5*10^14 t) V/m. The dielectric constant of the medium is ____ . (speed of light in free space is 3*10^8 m/s)

Answer: C. 4 — Given E_y = 20sin(kx - ωt). The velocity of the wave in the medium is v = ω/k = (4.5*10^14)/(3*10^6) = 1.5*10^8 m/s. The refractive index n = c/v = (3*10^8)/(1.5*10^8) = 2. Since the medium is non-magnetic, relative permeability μ_r = 1. Thus, n^2 = ε_r μ_r = ε_r. The dielectric constant ε_r = n^2 = 2^2 = 4.

Q: A message signal of frequency $3\,\text{kHz}$ is used to modulate a carrier signal of frequency $1.5\,\text{MHz}$. The bandwidth of the amplitude modulated wave is

Answer: D. 6 kHz — For conventional amplitude modulation, bandwidth is twice the highest modulating frequency: $$B=2f_m.$$ Here $f_m=3\,\text{kHz}$, so $$B=2\times 3\,\text{kHz}=6\,\text{kHz}.$$ The carrier frequency does not affect the bandwidth.

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Unit 15: Electromagnetic Waves — Important Questions & Answers (FAQ)

Frequently asked questions from JEE JEE MAINS PHYSICS — Unit 15: Electromagnetic Waves, with answers and explanations. These are sample questions; the exam has its own separate question set.

A laser beam has intensity of 4.0 x 10^14 W / m². The amplitude of magnetic field associated with beam is ____ T. (Take ε0 = 8.85 x 10^-12 C²/Nm² and c = 3 x 10^8 m/s)

A. 2.0
B. 18.3
C. 5.5
D. 1.83 ✓

Answer: D. 1.83
Using I = (1/2)ε0 E0² c and E0 = B0c, we derive B0 = sqrt(2I / ε0c³). Plugging in values, B0 = 1.83 T.

A plane electromagnetic wave is moving in free space with velocity c = 3 * 10^8 m/s and its electric field is given as E = 54*sin(kz - omega*t) j_hat V/m, where j_hat is the unit vector along the y-axis. The magnetic field vector B of the wave is:

A. (1.8 * 10^-7) * sin(kz - omega*t) * (-i_hat) T ✓
B. (1.4 * 10^-7) * sin(kz - omega*t) * k_hat T
C. (1.4 * 10^-7) * sin(kz - omega*t) * i_hat T
D. (1.8 * 10^-7) * sin(kz - omega*t) * i_hat T

Answer: A. (1.8 * 10^-7) * sin(kz - omega*t) * (-i_hat) T
Using B = (1/c) * (k_hat cross E). Since E is along j_hat and wave propagates along z (k_hat), B = (1/c) * E * (k_hat cross j_hat) = -(E/c) * i_hat. E_0 = 54, so B_0 = 54 / (3 * 10^8) = 1.8 * 10^-7.

A plane electromagnetic wave is moving in free space with velocity c = 3 × 10^8 m/s and its electric field is given as E = 54 sin(kz - ωt) ĵ V/m, where ĵ is the unit vector along y-axis. The magnetic field vector B of the wave is:

A. 1.8 × 10^-7 sin(kz - ωt) î T ✓
B. 1.4 × 10^-7 sin(kz - ωt) k̂ T
C. 1.4 × 10^-7 sin(kz - ωt) î T
D. 1.8 × 10^-7 sin(kz - ωt) î T

Answer: A. 1.8 × 10^-7 sin(kz - ωt) î T
Using the relation B = (1/c)(k̂ × E), where direction of propagation k̂ = k̂, and E is along ĵ. (k̂ × ĵ) = -î. Magnitude B₀ = E₀/c = 54 / (3 × 10⁸) = 1.8 × 10⁻⁷ T. Thus, B = -1.8 × 10⁻⁷ sin(kz - ωt) î T.

The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by, E_y = 20 sin(3*10^6 x - 4.5*10^14 t) V/m. The dielectric constant of the medium is ____ . (speed of light in free space is 3*10^8 m/s)

A. 2
B. 3
C. 4 ✓
D. 6

Answer: C. 4
Given E_y = 20sin(kx - ωt). The velocity of the wave in the medium is v = ω/k = (4.5*10^14)/(3*10^6) = 1.5*10^8 m/s. The refractive index n = c/v = (3*10^8)/(1.5*10^8) = 2. Since the medium is non-magnetic, relative permeability μ_r = 1. Thus, n^2 = ε_r μ_r = ε_r. The dielectric constant ε_r = n^2 = 2^2 = 4.

A message signal of frequency $3\,\text{kHz}$ is used to modulate a carrier signal of frequency $1.5\,\text{MHz}$. The bandwidth of the amplitude modulated wave is

A. 6 MHz
B. 3 MHz
C. 3 kHz
D. 6 kHz ✓

Answer: D. 6 kHz
For conventional amplitude modulation, bandwidth is twice the highest modulating frequency: $$B=2f_m.$$ Here $f_m=3\,\text{kHz}$, so $$B=2\times 3\,\text{kHz}=6\,\text{kHz}.$$ The carrier frequency does not affect the bandwidth.

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