Question

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 foes largely to nonvisible infrared radiation.(a) What is the visible light intensity (in W/m^2) 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?

Answer 1

Answer:

a

$I  =  6637 \  W/m^2$

b

$E_{max} =  500 \ N/m$

And

$B_{max} =  1.67*10^{-6} \  T$

Explanation:

From the question we are told that  

   The  power is  $P =  75 \  W$

   The  diameter is  $d =  6.0 \ cm =  0.06 \  m$

Generally the radius is mathematically represented as

      $r =  \frac{d}{2}$

=>   $r =  \frac{ 0.06}{2}$

=>   $r =  0.03 \  m$

Generally the area of the sphere is mathematically evaluated as

          $A = 4 \pi r^2$

=>      $A = 4 * 3.142 *  (0.03)^2$

=>       $A =  0.0113 \ m^2$

Generally the total  Intensity of the  incandescent light bulb is mathematically represented as

     $I= \frac{P}{A}$

=>    $I  =  \frac{75}{ 0.0113}$

=>   $I  =  6637 \  W/m^2$

Given that 5%  of the energy goes to visible  light

Then the intensity that goes visible light is  

    $I_v  =  0.05 * 6637$

    $I_v  =  332 \ W/m^2$

The amplitude of the electric field at the surface is mathematically represented as

      $E_{max} =  \sqrt{\frac{2 * I_v}{\epsilon_o *  c } }$

=>   $E_{max} =  \sqrt{\frac{2 * 332}{ 8.85*10^{-12} *  3.0*10^8} }$

=>     $E_{max} =  500 \ N/m$

The amplitude of the magnetic  field at the surface is mathematically represented as

     $B_{max} =  \frac{E_{max}}{c}$

=>     $B_{max} =  \frac{ 500}{3.0*10^8}$

=>   $B_{max} =  1.67*10^{-6} \  T$