Chapter 22, Problem 61P

(a)

To determine

The wavelength of radio wave in air.

Answer to Problem 61P

The wavelength of radio wave in air is $\underset{\_}{526\text{m}}$.

Explanation of Solution

Write the expression for wavelength.

$\lambda =\frac{c}{f}$ (I)

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

Conclusion:

Substitute $3\times {10}^8\text{m/s}$ for $c$, and $570\times {10}^3\text{Hz}$ for $f$ in equation (I).

$\begin{array}{c}\lambda =\frac{3\times {10}^8\text{m/s}}{570\times {10}^3\text{Hz}}\\ =526\text{m}\end{array}$

Therefore, the wavelength of radio wave in air is $\underset{\_}{526\text{m}}$.

(b)

To determine

The capacitance in the tuning circuit.

Answer to Problem 61P

The capacitance in the tuning circuit is $\underset{\_}{390\text{pF}}$.

Explanation of Solution

Write the expression for the frequency in an LC circuit.

$f_0=\frac{1}{2\pi \sqrt{LC}}$ (II)

Here, $L$ is the inductance, and $C$ is the capacitance.

Rearrange equation (I) to obtain an expression for $C$.

$C=\frac{1}{4{\pi }^{2}L{f}_0^2}$ (III)

Conclusion:

Substitute, $0.2\times {10}^{-3}\text{H}$ for $L$, $570\times {10}^3\text{Hz}$ for $f_0$ in equation (III).

$\begin{array}{c}C=\frac{1}{4{\left(3.14\right)}^2\left(0.2\times {10}^{-3}\text{H}\right)\left(570\times {10}^3\text{Hz}\right)}\\ =3.90\times {10}^{-11}\text{F}\\ \text{=390}\text{pF}\end{array}$

Therefore, the capacitance in the tuning circuit is $\underset{\_}{390\text{pF}}$.

(b)

To determine

The maximum induced emf in the antenna.

Answer to Problem 61P

The maximum induced emf in the antenna is $\underset{\_}{380\text{μV}}$.

Explanation of Solution

Write the expression for induced emf in the coil.

$\left|\epsilon \right|=N\frac{\Delta {\Phi }_B}{\Delta t}$ (IV)

Here, $N$ is the number of turns, $\Delta {\Phi }_B$ is the change in magnetic flux, and $\Delta t$ is the change in time.

Write the expression for the maximum magnetic flux.

$\begin{array}{c}{\Phi }_B=BA\\ =B\pi r^2\end{array}$ (V)

Here, $B$ is the magnetic field, and $r$ is the radius of the coil.

Replace $B$ by $B_{\text{m}}\sin \left(kz+\omega t\right)$ in equation and find the value of $\frac{\Delta B}{\Delta t}$.

$\frac{\Delta B}{\Delta t}=\omega B_{\text{m}}\cos \left(kz+\omega t\right)$

The maximum value of $\frac{\Delta B}{\Delta t}$ is obtained when $kz+\omega t$ become equal to 0, replace $\omega$ by $2\pi f$ in the expression of $\frac{\Delta B}{\Delta t}$.

$\begin{array}{c}{\left(\frac{\Delta B}{\Delta t}\right)}_{\text{max}}=\omega B_{\text{m}}\\ =2\pi f{B}_{\text{m}}\end{array}$ (VI)

Write the expression for the relationship between electric field and magnetic field.

$\begin{array}{c}{E}_{\text{m}}=c{B}_{\text{m}}\\ B_{\text{m}}=\frac{E_{\text{m}}}{c}\end{array}$ (VII)

Rewrite equation (IV).

$\begin{array}{c}\left|\epsilon \right|=N\pi r^2\frac{\Delta B}{\Delta t}\\ =N\pi r^2{\left(\frac{\Delta \left(B_{\text{m}}\sin \left(kz+\omega t\right)\right)}{\Delta t}\right)}_{\text{max}}\\ =N\pi r^2\left(2\pi f\right)B_{\text{m}}\\ =N\pi r^2\left(2\pi f\right)\frac{E_{\text{m}}}{c}\end{array}$ (VIII)

Conclusions:

Substitute, $50$ for $N$, $0.016\text{m}$ for $r$, $570000\text{Hz}$ for $f$, $0.80\text{V/m}$ for $E_m$, and $3\times {10}^8\text{m/s}$ for $c$ in equation (VIII).

$\begin{array}{c}\left|\epsilon \right|=50\times 3.14\times {\left(0.016\text{m}\right)}^2\times 2\times 3.14\times 570000\text{Hz}\frac{0.80\text{V/m}}{3\times {10}^8\text{m/s}}\\ =380\text{μV}\end{array}$

Therefore, the maximum induced emf in the antenna is $\underset{\_}{380\text{μV}}$.