Find the radiation resistance of the wire joining the two ends of the dipole. (This is the resistance that would give the same average power loss—to heat—as the oscillating dipole in fact puts out in the form of radiation.) Show that$\mathit{R}\mathbf{=}\mathbf{790}{\mathbf{\left(}\frac{\mathbf{d}}{\mathbf{\lambda }}\mathbf{\right)}}^{\mathbf{2}}\mathit{\Omega }$ , where$\mathit{\lambda }$ is the wavelength of the radiation. For the wires in an ordinary radio (say, d = 5 cm ), should you worry about the radiative contribution to the total resistance?

Write the expression for the power dissipated across a wire at time t.

$\mathit{P}\mathbf{=}\mathit{I}{\left(t\right)}^{\mathbf{2}}\mathit{R}$ …… (1)

Here, I is the current through the wire, and R is the resistance.

Write the expression for the average power radiated by an oscillating electric dipole.

$<P>\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{p}}_{\mathbf{0}}^{\mathbf{2}}{\mathbf{\omega }}^{\mathbf{4}}}{\mathbf{12}\mathbf{\Pi }\mathbf{c}}$ …… (2)

Here,${\mu }_0$ is the magnetic constant,$p_0$ is the maximum dipole moment of the dipole,$\omega$ is the angular frequency, and c is the speed of light.

Write the expression for the current through a wire at time t.

$I\left(t\right)=\frac{dq\left(t\right)}{dt}$ …… (3)

Here, $q\left(t\right)$ is the charge at time t, which is given as:

$q\left(t\right)=q_0\cos \omega t$

Substitute $q\left(t\right)=q_0\cos \omega t$in equation (3).

$$\begin{gathered} I\left(t\right)=\frac{d}{dt}\left(q_0\cos \omega t\right) \\ I\left(t\right)=-q_0\omega \sin \omega t \end{gathered}$$

Substitute $I\left(t\right)=-q_0\omega \sin \omega t$in equation (1).

$$\begin{gathered} P={\left(-q_0\omega \sin \omega t\right)}^{2}R \\ P=q_0^2{\omega }^{2}R{\sin }^2\omega t \end{gathered}$$

The average of ${\sin }^2\omega t$is given as $\frac{1}{2}$.

Hence, the above equation becomes,

$$\begin{gathered} P=q_0^2{\omega }^{2}R\left(\frac{1}{2}\right) \\ P=\frac{q_0^2{\omega }^{2}R}{2}........\left(4\right) \end{gathered}$$

Write the expression for the maximum dipole moment of the dipole.

$p_0=q_{0}d$

Substitute $p_0=q_{0}d$in equation (2).

Equate equations (4) and (5).

$$\begin{gathered} \frac{q_0^2{\omega }^{2}R}{2}=\frac{{\mu }_0{q}_0^2{d}^2{\omega }^4}{12\Pi c} \\ R=\frac{{\mu }_0{d}^2{\omega }^2}{6\pi c}.......\left(6\right) \end{gathered}$$

Write the relation between angular frequency and wavelength.

$\omega =\frac{2\pi c}{\lambda }$

Substitute $\omega =\frac{2\pi c}{\lambda }$in equation (6).

$$\begin{gathered} R=\frac{{\mu }_0{d}^2{\left({\displaystyle \frac{2\pi c}{\lambda }}\right)}^2}{6\pi c} \\ R=\frac{{\mu }_0{d}^{2}4{\pi }^2{c}^2}{6\pi c{\lambda }^2} \\ R=\frac{2{\mu }_0\pi c}{3}{\left(\frac{d}{\lambda }\right)}^2........\left(7\right) \end{gathered}$$

Substitute ${\mu }_0=4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m}$and $c=3\times {10}^8\mathrm{m}/\mathrm{s}$in equation (7).

role="math" $$\begin{gathered} R=\frac{2\left(4\pi \times {10}^{-7}\mathrm{H}/m\right)\pi \times \left(3\times {10}^{-8}\mathrm{m}/\mathrm{s}\right)}{3}{\left(\frac{d}{\lambda }\right)}^2 \\ R=790{\left(\frac{d}{\lambda }\right)}^2\Omega .........\left(7\right) \end{gathered}$$

Write the expression for the wavelength in terms of speed of light and frequency.

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

Here, f is the frequency of the FM wave$\left(f=100\mathrm{MHz}\right)$ .

Substitute$c=3\times {10}^8\mathrm{m}/\mathrm{s}$ and$f=100\mathrm{MHz}$ in the above expression.

$$\begin{gathered} \lambda =\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{\left(100\mathrm{MHz}\times {\displaystyle \frac{{10}^6\mathrm{Hz}}{1\mathrm{MHz}}}\right)} \\ \lambda =3\mathrm{m} \end{gathered}$$

Substitute $d=5\mathrm{cm}$and $\lambda =3\mathrm{m}$in equation (7).

$$\begin{gathered} R=790{\left(\frac{5\mathrm{cm}\times {\displaystyle \frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}}}{3\mathrm{m}}\right)}^2\Omega \\ R=790{\left(\frac{0.05}{3}\right)}^2\Omega \\ R=0.219\Omega \end{gathered}$$

The resistance is small, althought it is larger than the ohmic resistance of the wire present in the radio. This suggest that the raditiona effects the final resistance.

Therefore, the radiation resistance of the wire joining the two ends of the dipole is $R=790{\left(\frac{d}{\lambda }\right)}^2\Omega$, and as the resistance is larger than the ohmic resistance of the wire in the radio, we do not need to worry about the radiative contribution to the total resistance.