Find the radiation resistance (Prob. 11.3) for the oscillating magnetic dipole in Fig. 11.8. Express your answer in terms of$\mathit{\lambda }$and b , and compare the radiation resistance of the electric dipole. [ Answer: $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}{\left(\frac{b}{\lambda }\right)}^{\mathbf{4}}\mathit{\Omega }$]

Write the expression for the power loss in an oscillatory magnetic dipole.

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

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

Write the expression for the total radiated power.

$<P>\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{m}}_{\mathbf{0}}^{\mathbf{2}}{\mathbf{\omega }}^{\mathbf{4}}}{\mathbf{12}\mathbf{\pi }{\mathbf{c}}^{\mathbf{3}}}$ …… (2)

Here,${\mu }_0$ is the magnetic constant,$m_0$ is the magnetic dipole moment,$\omega$ is the angular frequency, and c is the speed of light.

Write the expression for the current through a wire loop.

$I\left(t\right)=I_0\cos \omega t$

Substitute $I\left(t\right)=I_0\cos \omega t$in equation (1).

$$\begin{gathered} P={\left(I_0\cos \omega t\right)}^{2}R \\ P=I_0^{2}R{\cos }^2\omega t \end{gathered}$$

The average value of${\cos }^2\omega t$is $\frac{1}{2}$.

Hence, the equation for P becomes,

$P=\frac{1}{2}{I}_0^{2}R$ …… (3)

Write the model for an oscillating magnetic dipole.

$$\begin{gathered} m\left(t\right)=\pi b^{2}I\left(t\right)\hat{z} \\ m\left(t\right)=m_0\cos \omega t\hat{z} \end{gathered}$$

Here, $m_0=\pi b^2{I}_0$

Substitute $m_0=\pi b^2{I}_0$in equation (2).

…… (4)

Equate equations (3) and (4).

Write the relation between angular frequency and wavelength.

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

Substitute $\omega =\frac{2\pi c}{\lambda }$in equation (5).

$$\begin{gathered} R=\frac{{\mu }_0{\pi }^2{b}^4{\left({\displaystyle \frac{2\pi c}{\lambda }}\right)}^4}{6\pi c^3} \\ R=\frac{{\mu }_0{\pi }^2{b}^4}{3\pi c^3}{\left(\frac{2\pi c}{\lambda }\right)}^4 \\ R=\frac{8}{3}{\pi }^5{\mu }_{0}c{\left(\frac{b}{\lambda }\right)}^4........\left(6\right) \end{gathered}$$

Substitute${\mu }_0=4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m}$ and $c=3\times {10}^8\mathrm{m}/\mathrm{s}$in equation (6).

$$\begin{gathered} R=\frac{8}{3}{\pi }^5\left(4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right){\left(\frac{b}{\lambda }\right)}^4 \\ R=3.076\times {10}^5{\left(\frac{b}{\lambda }\right)}^4\Omega \approx 3\times {10}^5\left(\frac{b}{\lambda }\right)\Omega \\ R=3\times {10}^5{\left(\frac{b}{\lambda }\right)}^4\Omega \end{gathered}$$

Therefore, the radiation resistance for the oscillating magnetic dipole is

$3\times {10}^5{\left(\frac{b}{\lambda }\right)}^4\Omega$.