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

a

`I  =  6637 \  W/m^2`

b

`E_(max) =  500 \ N/m`

And

`B_(max) =  1.67*10^(-6) \  T`

Step-by-step 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 =  (d)/(2)`

=>
`r =  ( 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= (P)/(A)`

=>
`I  =  (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{(2 * I_v)/(\epsilon_o *  c ) }`

=>
`E_(max) =  \sqrt{(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) =  (E_(max))/(c)`

=>
`B_(max) =  ( 500)/(3.0*10^8)`

=>
`B_(max) =  1.67*10^(-6) \  T`