Question

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

Answer 1

Answer

given,

Power of bulb,P = 100 W

distance from bulb, r = 2.5 m

Intensity of bulb,

$I = \dfrac{P}{A}$

P is the power of bulb

A is the area

$I = \dfrac{P}{4\pi r^2}$

$I = \dfrac{100}{4\pi\times 2.5^2}$

I = 1.274 W/m²

The relation between intensity I and E_max

$I = \dfrac{E_{max}^2}{2\mu_0 c}$

where c is the speed of light  

          μ₀ = 4 π x 10⁻⁷

$E_{max}=\sqrt{I(2\mu_0 c)}$

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

E_{max} = 30.99 V/m

now, maximum magnetic field  

 $B_{max}=\dfrac{E_{max}}{c}$

 $B_{max}=\dfrac{30.99}{3\times 10^8}$

 $B_{max}= 1.033 x 10⁻⁷\ T$