Question

A plane monochromatic radio wave (? = 0.3 m) travels in vacuum along the positive x-axis, with a time-averaged intensity I = 45.0 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

Concepts and reason

The concepts required to solve this problem are the intensity, the electric field, and the magnetic field.

Initially, calculate the electric field using the expression of the intensity of the radiation. Then, using the relation of the electric field and the magnetic field, calculate the magnetic field. Later, write the expression of the magnetic field in the positive z-axis. Then, find the wave number using the wavelength. Using the frequency calculate the period and using the period calculate the angular frequency. Finally, substitute the calculated values in the equation of the magnetic field.

Fundamentals

The expression for the intensity of radiation is,

$I = \frac{1}{2}c{\varepsilon _0}{E_0}^2$

Here, I is the intensity, ${\varepsilon _0}$ is the permittivity, ${E_0}$ is the electric field, and c is the speed of light.

The expression of the sinusoidal electromagnetic waves for the magnetic field B moving in the positive x direction is,

$B = {B_0}\cos \left( {kx - \omega t} \right)$

Here, ${B_0}$ is the amplitude of the magnetic field, $k$ is the wave number, $\omega$ is the angular frequency, and t is the time.

The expression of the speed of the wave is,

$v = \frac{\omega }{k}$

Here, v is the speed of the wave, $\omega$ is the angular frequency, and k is the wave number.

The relation between the electric field and the magnetic field is,

$\frac{{{E_0}}}{{{B_0}}} = c$

Here, ${E_0}$ is the electric field, ${B_0}$ is the magnetic field, and c is the speed of light.

The expression of the wave number is,

$k = \frac{{2\pi }}{\lambda }$

Here, $\lambda$ is the wavelength.

The expression of the frequency is,

$f = \frac{c}{\lambda }$

Here, f is the frequency, c is the speed of light, and $\lambda$ is the wavelength.

The expression of the angular frequency is,

$\omega = 2\pi f$

Here, f is the frequency.

The expression for the intensity of radiation is,

$I = \frac{1}{2}c{\varepsilon _0}{E_0}^2$

Rearrange the expression for ${E_0}$ .

${E_0} = \sqrt {\frac{{2I}}{{c{\varepsilon _0}}}}$

Substitute $45.0{\rm{ W/}}{{\rm{m}}^2}$ for I, $3 \times {10^8}{\rm{ m/s}}$ for c, and $8.85 \times {10^{ - 12}}{\rm{ }}{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}$ for ${\varepsilon _0}$ .

$\begin{array}{c}\\{E_0} = \sqrt {\frac{{2\left( {45.0{\rm{ W/}}{{\rm{m}}^2}} \right)}}{{\left( {3 \times {{10}^8}{\rm{ m/s}}} \right)\left( {8.85 \times {{10}^{ - 12}}{\rm{ }}{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}} \right)}}} \\\\ = 184.115{\rm{ V/m}}\\\end{array}$

The relation between the electric field and the magnetic field is,

$\frac{{{E_0}}}{{{B_0}}} = c$

Rearrange the expression for ${B_0}$ .

${B_0} = \frac{{{E_0}}}{c}$

Substitute 184.115 V/m for ${E_0}$ , and $3 \times {10^8}{\rm{ m/s}}$ for c.

$\begin{array}{c}\\{B_0} = \frac{{184.115{\rm{ V/m}}}}{{3 \times {{10}^8}{\rm{ m/s}}}}\\\\{B_0} = 6.14 \times {10^{ - 7}}{\rm{ T}}\\\end{array}$

The expression of the wave number is,

$k = \frac{{2\pi }}{\lambda }$

Substitute 0.3m for $\lambda$ .

$\begin{array}{l}\\k = \frac{{2\pi }}{{0.3{\rm{ m}}}}\\\\ = 20.93{\rm{ rad/m}}\\\end{array}$

The expression of the frequency is,

$f = \frac{c}{\lambda }$

Substitute 0.3m for $\lambda$ and $3 \times {10^8}{\rm{ m/s}}$ for c.

$\begin{array}{c}\\f = \frac{{3 \times {{10}^8}{\rm{ m/s}}}}{{0.3{\rm{ m}}}}\\\\ = 1 \times {10^9}{\rm{ }}{{\rm{s}}^{ - 1}}\\\end{array}$

The expression of the angular frequency is,

$\omega = 2\pi f$

Substitute $1 \times {10^9}{\rm{ }}{{\rm{s}}^{ - 1}}$ for f.

$\begin{array}{c}\\\omega = 2\pi \left( {1 \times {{10}^9}{\rm{ }}{{\rm{s}}^{ - 1}}} \right)\\\\ = 2\pi \times {10^9}{\rm{ rad/s}}\\\end{array}$

The expression of the sinusoidal electromagnetic waves for the magnetic field B moving in the positive z direction is,

$B = {B_0}\cos \left( {kx - \omega t} \right)$

Substitute $6.14 \times {10^{ - 7}}{\rm{ T}}$ for ${B_0}$ , $20.93{\rm{ rad/m}}$ for k, $2\pi \times {10^9}{\rm{ rad/s}}$ for $\omega$ , and 1.5 ns for t.

$\begin{array}{c}\\B = \left( {6.14 \times {{10}^{ - 7}}{\rm{ T}}} \right)\cos \left( {\left( {20.93{\rm{ rad/m}}} \right)\left( 0 \right) - \left( {2\pi \times {{10}^9}{\rm{ rad/s}}} \right)\left( {1.5{\rm{ ns}}\left( {\frac{{{{10}^{ - 9}}{\rm{ s}}}}{{1{\rm{ ns}}}}} \right)} \right)} \right)\\\\ = \left( {6.14 \times {{10}^{ - 7}}{\rm{ T}}} \right)\cos \left( { - \left( {2\pi \times {{10}^9}{\rm{ rad/s}}} \right)\left( {1.5{\rm{ ns}}\left( {\frac{{{{10}^{ - 9}}{\rm{ s}}}}{{1{\rm{ ns}}}}} \right)} \right)} \right)\\\\ = - 6.14 \times {10^{ - 7}}{\rm{ T}}\\\end{array}$

Ans:

The magnetic field at the origin is $- 6.14 \times {10^{ - 7}}{\rm{ T}}$ .