Question

Find the rms (a) electric and (b)magnetic fields at a point 2.00 m from a lightbulb that radiates 90.0 W of light uniformly in all directions.

Answer 1

a) rms of electric field =

`E_(rms)`= 25.97 V/m

b) rms of magnetic field

`B_(rms)` = 8.655 × 10⁻⁸

`B_(rms)` = 86.55nT

Step-by-step explanation:

given

power p = 90.0W

distance d = 2.00m

Intensity =
`(power)/(area)`

I =
`(p)/(A)`

A =
`4\pi d^(2)`

I =
`(p)/(4\pi d^(2) )`

I =
`(90)/(4\pi(2^(2)) )`

I = 1.79 W/m²

a)
`I_(ave)` = ε₀ ×
`E^(2) _(rms)` × c

where ε₀ is permittivity of free space = 8.85×10⁻¹²,
`E^(2) _(rms)` is the root mean value and c is speed of light = 3×10⁸m/s

1.79 = 8.85×10⁻¹² ×
`E^(2) _(rms)` × 3×10⁸

`E^(2) _(rms)` =
`(1.79)/(8.85x10^(-12) x 3x10^(8) )`

`E^(2) _(rms)`= 674.1996

`E_(rms)`= 25.97 V/m

b)for rems magnetic field

`E_(rms)`= c
`B_(rms)`

`B_(rms)` =
`(E_(rms) )/(c)`

`B_(rms)` =
`(25.97 V/m)/(3x10^(8) )`

`B_(rms)` = 8.655 × 10⁻⁸

`B_(rms)` = 86.55nT