Chapter 22, Problem 45GP

(a)

To determine

The length of an AM antenna would have to be if it were $\frac{1}{2}\lambda$ .

Answer to Problem 45GP

  $150m$

Explanation of Solution

Given info:

Frequency, $f=1MHz={10}^{6}Hz$

Speed of light, $c=3.00\times {10}^{8}m/s$

Wavelength, $\frac{1}{2}\lambda$

Formula used:

The relation between frequency and wavelength: $f=\frac{c}{\lambda }$

Where,

  $f$ is frequency

  $\lambda$ is wavelength

  $c$ is speed of light

Calculation:

Rearrange the relation between frequency and wavelength: $f=\frac{c}{\lambda }$ . So, $\lambda =\frac{c}{f}$

According to the question,

  $\lambda =\frac{c}{f}=\frac{3.0\times {10}^{8}m/s}{{10}^{6}Hz}=300m$

  $\frac{1}{2}\lambda =\frac{1}{2}\left(300m\right)=150m$

(b)

To determine

The length of an AM antenna would have to be if it were $\frac{1}{4}\lambda$ .

Answer to Problem 45GP

  $75m$

Explanation of Solution

Given info:

Frequency, $f=1MHz={10}^{6}Hz$

Speed of light, $c=3.00\times {10}^{8}m/s$

Wavelength, $\frac{1}{4}\lambda$

Formula used:

The relation between frequency and wavelength: $f=\frac{c}{\lambda }$

Where,

  $f$ is frequency

  $\lambda$ is wavelength

  $c$ is speed of light

Calculation:

Rearrange the relation between frequency and wavelength: $f=\frac{c}{\lambda }$ . So, $\lambda =\frac{c}{f}$

According to the question,

  $\lambda =\frac{c}{f}=\frac{3.0\times {10}^{8}m/s}{{10}^{6}Hz}=300m$

  $\frac{1}{4}\lambda =\frac{1}{4}\left(300m\right)=75m$