A plane electromagnetic wave is moving in free space with velocity c and its electric field is given as

Q. A plane electromagnetic wave is moving in free space with velocity c = 3 × 10⁸ m/s and its electric field is given as E⃗ = 54 sin(kz − ωt) ĵ V/m, where ĵ is the unit vector along y-axis. The magnetic field vector B⃗ of the wave is :

A. −1.8 × 10⁻⁷ sin(kz − ωt) î T

B. +1.8 × 10⁻⁷ sin(kz − ωt) î T

C. 1.4 × 10⁻⁷ sin(kz − ωt) k̂ T

D. 1.4 × 10⁻⁷ sin(kz − ωt) î T

Correct Answer: −1.8 × 10⁻⁷ sin(kz − ωt) î T

Solution

For an electromagnetic wave in free space, the magnitudes of electric and magnetic fields are related by:

E = cB

Given electric field amplitude:

E = 54 V/m

So magnetic field amplitude is:

B = E / c = 54 / (3 × 10⁸) = 1.8 × 10⁻⁷ T

The wave is propagating along +z direction since the phase is (kz − ωt). Electric field is along +y direction.

Using right-hand rule, the magnetic field must be along −x direction so that:

E⃗ × B⃗ → direction of propagation (+z)

Hence,

B⃗ = −1.8 × 10⁻⁷ sin(kz − ωt) î

Therefore, the correct answer is option A.