Question

Fields from a Light Bulb. We can reasonably model a 75 - W incandescent light bulb as a sphere 6.0 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in ) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

Short answer

a. The intensity of visible-light at the surface of the bulb 330$W/m^2$ .

b. The amplitudes of the electric and magnetic fields at the surface, for a sinusoidal wave with intensity $330W/m^2$are$500V/m$ and $1.7\times {10}^{-6}T$respectively.

Step-by-step solution

Define the intensity and define the formulas.

The power transported per unit area is known as the intensity .

The formula used to calculate the intensity ( l ) is:

$l=\frac{P}{A}$

Where, A is area measured in the direction perpendicular to the energy and P is the power in watts.

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

$\begin{array}{r}{E}_{max}=\sqrt{\frac{2l}{{\epsilon }_{0}C}}\\ B_{max}=\frac{E_{max}}{c}\end{array}$

Where,${\epsilon }_0=8.85\times {10}^{-12}C/N.m^2$ and is the speed of light that is equal to $3.0\times {10}^{8}m/s$.

Determine the intensity of visible-light.

Given that,

$$\begin{gathered} P=75W \\ r=3\times {10}^{-2}m \end{gathered}$$

Given that, only 5% of the energy goes to visible-light. So, we use the 5% of power ( P ) .

The formula used to calculate the intensity ( l ) is:

$\begin{array}{r}I=\frac{P}{A}\\ =\frac{5\mathrm{\%}P}{4\pi \left(3\times {10}^{-2}\right)}\\ =\frac{\left(0.05\right)\times 75}{113\times {10}^{-4}}\\ =331\mathrm{W}/{\mathrm{m}}^2\end{array}$

Hence, the intensity of visible-light at the surface of the bulb$330W/m^2$ .

Determine the amplitudes of electric and magnetic fields.

The amplitude of electric field is:

$E_{max}=\sqrt{\frac{2I}{{\epsilon }_{0}c}}$

Substitute the values

$\begin{array}{r}{E}_{max}=\sqrt{\frac{2\times \left(331\right)}{\left(8.85\times {10}^{-12}\right)\left(3\times {10}^8\right)}}\\ =500\mathrm{V}/\mathrm{m}\end{array}$

The amplitude of magnetic field is:

$B_{max}=\frac{E_{max}}{c}$

Substitute the values

$\begin{array}{r}{B}_{max}=\frac{500}{3\times {10}^8}\\ =1.7\times {10}^{-6}\mathrm{T}\end{array}$

Hence, the amplitudes of the electric and magnetic fields at the surface, for a sinusoidal wave with intensity $330W/m^2$are 500 V/m and$1.7\times {10}^{-6}T$ respectively.