A plane electromagnetic wave is moving in free space with velocity c = 3 × 10⁸ m/s and its electric field is given as
$\vec{E}$ = 54 sin(kz - ωt) ĵ V/m,
where ĵ is the unit vector along y-axis. The magnetic field vector $\vec{B}$ of the wave is :
Options

Question

A plane electromagnetic wave is moving in free space with velocity c = 3 × 10⁸ m/s and its electric field is given as

$\vec{E}$ = 54 sin(kz - ωt) ĵ V/m,

where ĵ is the unit vector along y-axis. The magnetic field vector $\vec{B}$ of the wave is :

Options

G1 Physics · Accepted Answer

The phase kz-ωt shows that the wave propagates in the +z direction. The electric field is along +y: E⃗ ∥ ĵ. For an electromagnetic wave, E⃗ × B⃗ must point in the direction of propagation, i.e. +k̂. Since ĵ × (-î) = +k̂, the magnetic field must be along -î. Its magnitude is B₀ = E₀/c = 54/(3×10⁸) = 1.8×10⁻⁷ T. Therefore, B⃗ = -1.8×10⁻⁷ sin(kz-ωt) î T. Correct option: A.

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