Question

A plane monochromatic radio wave (λ = 0.3 m) travels in vacuum along the positive x-axis, with a time-averaged intensity I = 45 W/m2. Suppose at time t = 0, the electric field at the origin is measured to be directed along the positive y-axis with a magnitude equal to its maximum value. What is Bz, the magnetic field at the origin, at time t = 1.5 ns?

Answer 1

Answer:

magnetic field =  - 6.137 × $10^{-7}$ T

Explanation:

given data

radio wave λ = 0.3 m

intensity I = 45 W/m²

time t = 0

time t = 1.5 ns

to find out

What is Bz

solution

we use here equation of intensity of radiation of electric field that is express as

intensity = $\frac{1}{2}c\epsilon_oE_o^2$    .................1

here we know intensity 45 and c is 3 × $10^{8}$ m/s² and ∈o is 8.85  × $10^{-12}$ C²/N.m² and E will be here

E = $\sqrt{\frac{2I}{c\epsilon_o} }$     ...............2

E = $\sqrt{\frac{2*45}{3*10^8 * 8.85*10^{-12}}}$

E = 184.15

and

E = $c*B_o$      .................3

184.15 =  $3.*10^8*B_o$  

$B_o$ = 6.137 × $10^{-7}$ T

and in z axis magnetic field will be here as

B = $B_o* cos(kx - \omega t )$      ..............4

here k = $\frac{2\pi }{\lambda}$    

k =  $\frac{2\pi }{0.3}$ = 20.94

and

f =  $\frac{c}{\lambda}$  

f = $\frac{3*10^8}{0.3}$ = $10^{9}$        

so here

time t = $\frac{1}{f}$      

t = $\frac{1}{f}$   = $10^{-9}$  

so ω = $\frac{2\pi }{t}$

ω = $\frac{2\pi }{10^{-9}}$

ω = 6.28 × $10^{9}$

so now from equation 4

B = $B_o* cos(kx - \omega t )$

B = $6.137*10^{-7} *cos(0 - 6.28*10^9*(1.5*10^{-9})$

B =  - 6.137 × $10^{-7}$

so magnetic field =  - 6.137 × $10^{-7}$ T