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The electric field of a uniform plane electromagnetic wave is E = -301.6 sin(kz - omega*t) ax + 452.4 sin(kz - omega*t) ay (V/m). Find the magnetic intensity H of the wave in A/m. (Take c = 3 x 10⁸ m/s, mu0 = 4*pi x 10⁻⁷ N/A².)

1. -0.8 sin(kz - omega*t) ay - 1.2 sin(kz - omega*t) ax
2. +0.8 sin(kz - omega*t) ay + 0.8 sin(kz - omega*t) ax
3. +1.0 x 10⁻⁶ sin(kz - omega*t) ay + 1.5 x 10⁻⁶ sin(kz - omega*t) ax
4. -1.0 x 10⁻⁶ sin(kz - omega*t) ay - 1.5 x 10⁻⁶ sin(kz - omega*t) ax

Correct answer: -0.8 sin(kz - omega*t) ay - 1.2 sin(kz - omega*t) ax

Solution

For a plane wave traveling along +z, H = (1/(mu0*c)) (az x E). The impedance mu0*c = 377 ohm, so 1/(mu0*c) approx 2.654e-3. Compute az x E: az x (-301.6 ax) = -301.6 ay; az x (452.4 ay) = -452.4 ax. Multiply each amplitude by 2.654e-3: -301.6*2.654e-3 approx -0.8 (ay component) and -452.4*2.654e-3 approx -1.2 (ax component). So H = [-0.8 ay - 1.2 ax] sin(kz - omega*t) A/m.

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