Chapter 22, Problem 36QAP

To determine

(a)

The amplitude of the electric field.

Answer to Problem 36QAP

The amplitude of the magnetic field is given by,

$B_o=0.7\text{μT}$

Explanation of Solution

Given:

$B(x,t)=(0.7\mu \text{T})\sin [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t]$

Formula used:

The magnetic field is given,

$B(x,t)=B_o\cos [kx-\omega t+\varphi ]$

Where,

$B(x,t)$ =Magnetic field $B_o$ = Amplitude of magnetic field $k$ = Wave number

$\omega$ =Angular frequency

$\varphi$ =Phase difference

Calculation:

The magnetic field is given,

$B(x,t)=B_o\cos [kx-\omega t+\varphi ]\mathrm{...}(1)$

$\begin{array}{l}B(x,t)=(0.7\mu \text{T})\sin [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t]\\ B(x,t)=(0.7\mu \text{T})\cos [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t+\varphi ]\mathrm{...}(2)\end{array}$

Comparing equation (1) and (2) we get,

$B_o=0.7\text{μT}$

Thus, the amplitude of the magnetic field is given by,

$B_o=0.7\text{μT}$

To determine

(b)

The speed of the electromagnetic wave.

Answer to Problem 36QAP

The speed of the electromagnetic wave is, $c=3\times {10}^8\text{m/s}$

Explanation of Solution

Given:

$B(x,t)=(0.7\mu \text{T})\sin [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t]$

Formula used:

The speed of electromagnetic wave is is,

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

Where,

$c$ =Speed of electromagnetic wave$k$ = Wave number

$\omega$ =Angular frequency

Calculation:

The magnetic field is given,

$B(x,t)=B_o\cos [kx-\omega t+\varphi ]\mathrm{...}(1)$

$\begin{array}{l}B(x,t)=(0.7\mu \text{T})\sin [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t]\\ B(x,t)=(0.7\mu \text{T})\cos [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t+\varphi ]\mathrm{...}(2)\end{array}$

Comparing equation (1) and (2) we get,

$\begin{array}{l}{B}_o=0.7\text{μT}\\ \omega =\text{2}\text{.40}\pi \times {\text{10}}^{15}{\text{s}}^{-1}\\ k\text{=8}\pi \times {\text{10}}^6{\text{m}}^{-1}\end{array}$

Thus, the speed of the magnetic field is given by,

$\begin{array}{l}c=\frac{\omega }{k}\\ c=\frac{\text{2}\text{.40}\pi \times {\text{10}}^{15}}{\text{8}\pi \times {\text{10}}^6}\\ c=3\times {10}^8\text{m/s}\end{array}$

To determine

(c)

The frequency of the electromagnetic wave.

Answer to Problem 36QAP

The frequency of the electromagnetic wave is, $f=1.200\times {10}^{14}\text{Hz}$

Explanation of Solution

Given:

$B(x,t)=(0.7\mu \text{T})\sin [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t]$

Formula used:

The angular frequency of electromagnetic wave is is,

$\omega =2\pi f$

Where,

$\omega$ =Angular frequency

$f$ =Frequency

Calculation:

The magnetic field is given,

$B(x,t)=B_o\cos [kx-\omega t+\varphi ]\mathrm{...}(1)$

$\begin{array}{l}B(x,t)=(0.7\mu \text{T})\sin [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t]\\ B(x,t)=(0.7\mu \text{T})\cos [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t+\varphi ]\mathrm{...}(2)\end{array}$

Comparing equation (1) and (2) we get,

$\begin{array}{l}{B}_o=0.7\text{μT}\\ \omega =\text{2}\text{.40}\pi \times {\text{10}}^{15}{\text{s}}^{-1}\\ k\text{=8}\pi \times {\text{10}}^6{\text{m}}^{-1}\end{array}$

The angular frequency of electromagnetic wave is is,

$\begin{array}{l}\omega =2\pi f\\ \text{2}\text{.40}\pi \times {\text{10}}^{15}=2\pi f\\ f=\frac{\text{2}\text{.40}\pi \times {\text{10}}^{15}}{2\pi }\\ f=1.200\times {10}^{14}\text{Hz}\end{array}$

To determine

(d)

The period of the electromagnetic wave.

Answer to Problem 36QAP

The period of the electromagnetic wave is, $T=8.33\times {10}^{-15}\text{s}$

Explanation of Solution

Given:

$B(x,t)=(0.7\mu \text{T})\sin [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t]$

Frequency, $f=1.200\times {10}^{14}\text{Hz}$

Formula used:

The period of electromagnetic wave is,

$T=\frac{1}{f}$

Where,

$T$ =Period

$f$ =Frequency

Calculation:

The period of electromagnetic wave is,

$\begin{array}{l}T=\frac{1}{f}\\ T=\frac{1}{1.200\times {10}^{14}}\\ T=8.33\times {10}^{-15}\text{s}\end{array}$

To determine

(e)

The wavelength of the electromagnetic wave.

Answer to Problem 36QAP

The wavelength of the electromagnetic wave is, $\lambda =0.25\times {10}^{-6}\text{m}$

Explanation of Solution

Given:

$B(x,t)=(0.7\mu \text{T})\sin [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t]$

Formula used:

The wave number of electromagnetic waves is,

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

Where,

$k$ =Wave-number

$\lambda$ =Wavelength

Calculation:

The magnetic field is given,

$B(x,t)=B_o\cos [kx-\omega t+\varphi ]\mathrm{...}(1)$

$\begin{array}{l}B(x,t)=(0.7\mu \text{T})\sin [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t]\\ B(x,t)=(0.7\mu \text{T})\cos [(8\pi \times {10}^6{\text{m}}^{-1})x-(2.40\pi \times {10}^{15}{\text{s}}^{-1})t+\varphi ]\mathrm{...}(2)\end{array}$

Comparing equation (1) and (2) we get,

$\begin{array}{l}{B}_o=0.7\text{μT}\\ \omega =\text{2}\text{.40}\pi \times {\text{10}}^{15}{\text{s}}^{-1}\\ k\text{=8}\pi \times {\text{10}}^6{\text{m}}^{-1}\end{array}$

The wave number of electromagnetic waves is,

$\begin{array}{l}k=\frac{2\pi }{\lambda }\\ \text{8}\pi \times {\text{10}}^6=\frac{2\pi }{\lambda }\\ \lambda =\frac{\text{2}\pi }{\text{8}\pi \times {\text{10}}^6}\\ \lambda =0.25\times {10}^{-6}\text{m}\end{array}$