The filament of a light bulb is cylindrical with length l = 20 mm and radius r = 0.05 mm. The filament is maintained at a temperature T = 5000 K by an electric current. The filament behaves \approx imately as a black body, emitting radiation isotropically.

(A) The total power radiated by the filament can be found using the Stefan-Boltzmann law, which states that the power radiated per unit surface area by a black body is proportional to the fourth power of its temperature. Mathematically, it can be expressed as: {eq}P = sigma A T^4 {/eq} where {eq}sigma = 5.67 times 10^{-8} \text{ W/m}^2\text{ K}^4 {/eq} is the Stefan-Boltzmann constant and A is the surface area of the filament. Substituting the given values, we have: {eq}begin{aligned} P &= sigma A T^4 \ &= (5.67 times 10^{-8} \text{ W/m}^2\text{ K}^4) (2pi r l) (5000 \text{ K})^4 \ &approx 1.11 \text{ W} end{aligned} {/eq} Therefore, the total power radiated by the filament is \approx imately {eq}1.11 \text{ W} {/eq}. (B) The wavelength at which the filament radiates the most power can be found using Wien's displacement law, which gives the wavelength corresponding to the peak of the black body radiation curve. Mathematically, it can be expressed as: {eq}lambda_{\text{ max}} = \frac{b}{T} {/eq} where {eq}b = 2.898 times 10^{-3} \text{ m}cdot\text{ K} {/eq} is the Wien's displacement constant. Substituting the given values, we have: {eq}begin{aligned} lambda_{\text{ max}} &= \frac{b}{T} \ &= \frac{2.898 times 10^{-3} \text{ m}cdot\text{ K}}{5000 \text{ K}} \ &approx 5.80 times 10^{-7} \text{ m} end{aligned} {/eq} Therefore, the maximum wavelength the filament radiates the most power is \approx imately {eq}5.80 times 10^{-7} \text{ m} {/eq}, which corresponds to the visible range of the electromagnetic spectrum, specifically the yellow-green region.