JEE Main · Physics → Steady

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

Year 2026 2025 2024 Total
Questions 10 12 9 31
Practice all Electromagnetic Waves questions → View full JEE Main weightage & trends →

A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time) E = 9.6 j V/m. The value of magnetic field at this point is:

Solution & Explanation

Related Formula

The relation between the amplitudes of electric field E and magnetic field B in an EM wave is:

c = (E)/(B)

where:

  • c = 3 × 10⁸ m/s is the speed of light.

    • The directions are related by the vector cross product:

E × B = v

where v is the direction of propagation of the EM wave.

Core Logic

Given:

  • Wave propagation direction, v = i (along X-direction)
  • Electric field vector direction, E = j
  • Electric field magnitude, E = 9.6 V/m
Step 1: Calculate Magnetic Field Magnitude

Using the magnitude relation:

B = (E)/(c) = (9.6)/(3 × 10⁸) = 3.2 × 10⁻⁸ T
Step 2: Determine Magnetic Field Direction

Using the direction cross-product rule:

E × B = v j × B = i

Since we know that j × k = i, the direction of the magnetic field must be k:

B = k

Combining the magnitude and direction:

B = 3.2 × 10⁻⁸ k T
Pattern Recognition

A propagation along +x with electric field along +y strictly mandates the magnetic field must point along +z (+y × +z = +x). This instantly eliminates Option 2 and Option 3, leaving only magnitude verification.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Reference Study Guides

Electromagnetic Waves

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_2024_29_january_evening

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%)