Question

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

Answer 1

Answer:

$-5.78414\times 10^{-6}\ T$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

$\lambda$ = Wavelength = 0.3 m

T = Time period

f = Frequency

$E_0$ = Electric field

Intensity of electric field is given by

$I=\dfrac{1}{2}c\epsilon_0E_0^2\\\Rightarrow E_0=\sqrt{\dfrac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\dfrac{2\times 45}{3\times 10^{8}\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=184.11492\ N/C$

Magnetic field is given by

$B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{184.11492}{3\times 10^8}\\\Rightarrow B_0=6.13716\times 10^{-7}\ T$

$k=\dfrac{2\pi}{\lambda}\\\Rightarrow k=\dfrac{2\pi}{0.3}\\\Rightarrow k=20.94\ /m$

$f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^{8}}{0.3}\\\Rightarrow f=1\times 10^{9}\ Hz$

$T=\dfrac{1}{f}\\\Rightarrow T=10^{-9}$

$\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{10^{-9}}\\\Rightarrow \omega=6283185307.17958\ rad/s$

Magnetic field in the z direction is given by (x=0)

$B_z=B_0(kx-\omega t)\\\Rightarrow B_z=6.13716\times 10^{-7}\times (0-6283185307.17958\times 1.5\times 10^{-9})\\\Rightarrow B_z=-5.78414\times 10^{-6}\ T$

The magnetic field at the origin is $-5.78414\times 10^{-6}\ T$