Poynting vector: The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m². What is the maximum value of the magnetic field in the wave? (c = 3.0 × 10⁸ m/s, μ₀ = 4π × 10^-7 T ∙ m/A, ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A) 77.9 nT B) 55.1 nT C) 38.9 nT D) 108 nT E) 156 nT

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Quizplus · Accepted Answer

The average value of the Poynting vector is given by:$$S_{avg} = \frac{1}{2}ε_0 c E_0^2$$where:* S_avg is the average value of the Poynting vector* ε_0 is the permittivity of free space* c is the speed of light* E_0 is the amplitude of the electric fieldThe maximum value of the magnetic field is given by:$$B_0 = \frac{E_0}{c}$$where:* B_0 is the maximum value of the magnetic fieldSubstituting the given values into the above equations, we get:$$S_{avg} = \frac{1}{2}(8.85 × 10^{-12} C^2/N \cdot m^2)(3.0 × 10^8 m/s)(E_0^2)$$$$B_0 = \frac{E_0}{3.0 × 10^8 m/s}$$Solving for E_0, we get:$$E_0 = \sqrt{\frac{2S_{avg}}{ε_0 c}} = \sqrt{\frac{2(0.724 W/m^2)}{8.85 × 10^{-12} C^2/N \cdot m^2(3.0 × 10^8 m/s)}} = 2.45 × 10^3 V/m$$Substituting this value into the equation for B_0, we get:$$B_0 = \frac{2.45 × 10^3 V/m}{3.0 × 10^8 m/s} = 77.9 nT$$Therefore, the correct answer is A.

Poynting Vector: the Magnitude of the Poynting Vector of a Planar