Question

What is the maximum value of the electric field at a
distance2.5 m from a 100-W light bulb?

Answer 1

given,

Power of bulb,P = 100 W

distance from bulb, r = 2.5 m

Intensity of bulb,

`I = (P)/(A)`

P is the power of bulb

A is the area

`I = (P)/(4\pi r^2)`

`I = (100)/(4\pi* 2.5^2)`

I = 1.274 W/m²

The relation between intensity I and E_max

`I = (E_(max)^2)/(2\mu_0 c)`

where c is the speed of light

μ₀ = 4 π x 10⁻⁷

`E_(max)=√(I(2\mu_0 c))`

`E_(max)=\sqrt{1.274* (2* 4\pi * 10^(-7)* 3 * 10^8)}`

E_{max} = 30.99 V/m

now, maximum magnetic field

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

`B_(max)=(30.99)/(3* 10^8)`

`B_(max)= 1.033 x 10⁻⁷\ T`