What is the maximum value of the $\vec { B }$ field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency sinusoidal electromagnetic waves uniformly in all directions? (ε₀ = 8.85 × 10^-12 C²/N ∙ m²,μ₀ = 4π × 10^-7 T ∙ m/A,c = 3.0 × 10⁸ m/s) A)0.10 μT B)0.40 μT C)0.50 μT D)0.60 μT E)0.80 μT

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Quizplus · Accepted Answer

The intensity of the electromagnetic wave is given by:$$I = \frac{P}{4\pi r^2}$$where P is the power radiated by the light bulb, and r is the distance from the light bulb. The magnetic field amplitude is given by:$$B = \sqrt{\frac{\mu_0 I}{2}}$$Substituting the given values, we get:$$B = \sqrt{\frac{(4\pi 	imes 10^{-7} 	ext{ T} \cdot 	ext{m/A})(100 	ext{ W})}{4\pi (2.5 	ext{ m})^2(2)}} = 0.10 \mu	ext{T}$$

What Is the Maximum Value of The B⃗\vec { B }B

What Is the Maximum Value of The $\vec { B }$