Question

Assume that a $7.0-\mathrm{cm}$-diameter, $100\mathrm{W}$light bulb radiates all its energy as a single wavelength of visible light. Estimate the electric and magnetic field strengths at the surface of the bulb.

Short answer

At surface of the bulb,

Electric field strength is $2.21\times {10}^3\frac{\mathrm{v}}{\mathrm{m}}$.

Magnetic field strength is$7.3\times {10}^{-6}\mathrm{T}$.

Step-by-step solution

Calculation for intensity

Given info:

Diameter is $7.0\mathrm{cm}$.

Power is$100\mathrm{W}$.

Intensity is,

${\mathrm{I}}_{\mathrm{avg}}=\frac{\mathrm{P}}{\mathrm{A}}$

Where,

Power$=$$100\mathrm{W}$

localid="1649957590468" ${\mathrm{I}}_{\mathrm{avg}}=\frac{100\mathrm{W}}{4\mathrm{\pi }{\left(3.5\times {10}^{-2}\right)}^2{\mathrm{m}}^2}$

${\mathrm{I}}_{\mathrm{avg}}=65\times {10}^2\mathrm{W}/{\mathrm{m}}^2$

Calculation of Electric field strength

Electric field is,

${\mathrm{E}}_{\mathrm{o}}={\left(\frac{2\times {\mathrm{I}}_{\mathrm{avg}}}{\mathrm{c}\times {\mathrm{\epsilon }}_{\mathrm{O}}}\right)}^{1/2}$

localid="1649960680664" ${\mathrm{E}}_{\mathrm{o}}={\left(\frac{2\times 65\times {10}^2{\displaystyle \frac{\mathrm{W}}{{\mathrm{m}}^2}}}{3\times {10}^8{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\times 8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{c}}^2}{{\mathrm{Nm}}^2}}}\right)}^{1/2}$

${\mathrm{E}}_{\mathrm{o}}=2.21\times {10}^3\frac{\mathrm{v}}{\mathrm{m}}$

Calculation for magnetic field strength

Magnetic field is,

${\mathrm{B}}_0=\frac{{\mathrm{E}}_0}{\mathrm{c}}$

${\mathrm{B}}_0=\frac{2.21\times {10}^3{\displaystyle \frac{\mathrm{v}}{\mathrm{m}}}}{3\times {10}^8{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}}$

${\mathrm{B}}_0=7.3\times {10}^{-6}\mathrm{T}$