Question

Three electromagnetic waves arrive at a point from the same direction. They are polarized parallel to an x axis, and their electric fields at the point are E1 = A sin(ωt) for wave 1, E2 = B sin(ωt + φ2) for wave 2, and E3 = C sin(ωt + φ3) for wave 3. Here t is time. The constants are A = 3.0 µV/m, B = 4.0 µV/m, C = 5.0 µV/m, ω = 7.0 × 106 rad/s, φ2 = π/4, and φ3 = π/2. What is the amplitude (in µV/m) of the electric field that results from the interference of the three waves?

Answer 1

Answer:

E₁ = 9.759${\mu\frac{V}{m}}$sin(7.0 × 10⁶ t)

Explanation:

Given:

E₁ = 3sin(7.0 × 10⁶ t)

E₂ = 4sin(7.0 × 10⁶ t + 45°)

E₃ = 5sin(7.0 × 10⁶ t + 90°)

Now,

the net vertical component of Electric field E is

$E_v$= 3sin(0°) + 4sin(0° + 45°) + 5sin(0° + 90°)

or

$E_v$ = 7.828 ${\mu\frac{V}{m}}$

Now,

the net horizontal component of Electric field E is

$E_h$ = 3cos(0°) + 4cos(0° + 45°) + 5cos(0° + 90°)

or

$E_h$ = 5.828 ${\mu\frac{V}{m}}$

Therefore, the resultant Electric field is

$E_{R} =\sqrt{E_v^2+E_h^2}$

or

$E_{R} =\sqrt{7.828^2+5.828^2}$

or

$E_{R} =9.759\ {\mu\frac{V}{m}}$

Now, the phase angle is given as:

$\phi=\tan^{-1}\frac{0}{9.759} =$

hence,

E₁ = 9.759${\mu\frac{V}{m}}$sin(7.0 × 10⁶ t)