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Electromagnetic Waves appeared 31 times across 3 years — 3.6% of Physics. This question is from Electric and Magnetic Fields.

Year 2026 2025 2024 Total
Questions 10 12 9 31

The electric field a plane electromagnetic wave is given by : Ey = 69 [ 0.6 × 10³ x - 1.8 × 10¹¹ t ] V/m. The expression for magnetic field associated with this electromagnetic wave is ____ T.

Solution & Explanation

Related Formula
B₀ = (E₀)/(c) c = E × B
Core Logic

The phase of the wave is (0.6 × 10³ x - 1.8 × 10¹¹ t). This indicates the wave propagates in the +x direction, so c = i. The electric field oscillates along the y-axis, so E = j. From B = c × E, we have B = i × j = k. So, the magnetic field is along the z-axis (Bz).

Step 1: Calculate Amplitude of B

Wave speed v = c = (ω)/(k) = 1.8 × 10¹¹0.6 × 10³ = 3 × 10⁸ m/s. The amplitude of the magnetic field is:

B₀ = (E₀)/(c) = (69)/(3 × 10⁸) = 23 × 10⁻⁸ = 2.3 × 10⁻⁷ T

The phase remains exactly the same as the electric field:

Bz = 2.3 × 10⁻⁷ (0.6 × 10³ x - 1.8 × 10¹¹ t)
Pattern Recognition

B₀ = E₀/c gives the magnitude. The vector identity B = v × E gives the direction. Phase part never changes sign or terms between E and B equations.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 7

Q40 jee_main_2024_31_jan_morning Energy Density Of EM Waves
In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 10¹⁰ ~Hz and an amplitude of 50 ~Vm⁻¹. The total average energy density of the electromagnetic field of the wave is : [Use ε₀ = 8.85 × 10⁻¹² C² / Nm² ]
  • A. 1.106 × 10⁻⁸ Jm⁻³
  • B. 4.425 × 10⁻⁸ ~Jm⁻³
  • C. 2.212 × 10⁻⁸ ~Jm⁻³
  • D. 2.212 × 10⁻¹⁰ ~Jm⁻³

Solution

Related Formula
Utotal average = (1)/(2)ε₀ E₀²
Core Logic

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so:

Uavg = UE + UB = 2UE = 2 ( (1)/(4)ε₀ E₀² ) = (1)/(2)ε₀ E₀²

Where E₀ is the amplitude of the electric field.

Step 2: Substitution

Given: E₀ = 50 V/m ε₀ = 8.85 × 10⁻¹² C²/(N· m²)

Uavg = (1)/(2) × (8.85 × 10⁻¹²) × (50)² Uavg = (1)/(2) × 8.85 × 10⁻¹² × 2500 Uavg = 1.10625 × 10⁻⁸ J/m³
Chapter Mix

Class 12 Physics: Electromagnetic Waves

More Electromagnetic Waves Questions — jee_main_2026_21_jan_morning

Practice all Electromagnetic Waves previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)