(10.1) Find the electric field for an infinitesimal dipole radiator.

Electric Field of an Infinitesimal Dipole Radiator

This problem is most naturally expressed in spherical coordinates. We know that the vector potential is:

\[ A_r = \frac{\mu_0 I_0 d e^{-ikr}}{4 \pi r} \cos\theta, \quad A_\theta = -\frac{\mu_0 I_0 d e^{-ikr}}{4 \pi r} \sin\theta, \quad A_\phi = 0 \]

The electric field is given by:

\[ E = \frac{1}{i\omega \mu_0 \epsilon_0} \nabla (\nabla \cdot A) - i \omega A \]

First, we compute \( \nabla \cdot A \):

\[ \nabla \cdot A = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 A_r) + \frac{1}{r \sin\theta} \frac{\partial}{\partial \theta} (A_\theta \sin\theta) \]

Substituting the given potentials and simplifying:

\[ \nabla \cdot A = \frac{\mu_0 I_0 d}{4 \pi} e^{-ikr} \cos\theta \left( \frac{1}{r^2} - i \frac{k}{r} \right) \]

Next, we compute the gradient:

\[ \nabla f = \frac{\partial f}{\partial r} \hat{r} + \frac{1}{r} \frac{\partial f}{\partial \theta} \hat{\theta} + \frac{1}{r \sin\theta} \frac{\partial f}{\partial \phi} \hat{\phi} \]

Calculating \( \nabla (\nabla \cdot A) \):

\[ [\nabla (\nabla \cdot A)]_r = \frac{\partial}{\partial r} (\nabla \cdot A) \]

\[ = \frac{\mu_0 I_0 d}{4 \pi} e^{-ikr} \cos\theta \left( \frac{2}{r^3} + i \frac{2k}{r^2} - k^2 \frac{1}{r} \right) \]

Similarly, computing the \( \theta \) and \( \phi \) components:

\[ [\nabla (\nabla \cdot A)]_\theta = \frac{1}{r} \frac{\partial}{\partial \theta} (\nabla \cdot A) = 0 \]

Thus, substituting into the field equation and recalling that \( k = \sqrt{\omega \mu_0 \epsilon_0} \), we obtain:

\[ E_r = \frac{I_0 d}{4 \pi} e^{-ikr} \cos\theta \left( \frac{2i \omega \epsilon_0}{r^3} + \frac{2 k \omega \epsilon_0}{r^2} - \frac{k^2 i \omega \epsilon_0}{r} \right) \]

\[ E_\theta = \frac{I_0 d}{4 \pi} e^{-ikr} \sin\theta \left( \frac{i}{\omega \epsilon_0 r^3} + \frac{k}{\omega \epsilon_0 r^2} \right) \]

\[ E_\phi = 0 \]

At large distances \( (r \gg \lambda) \), the dominant term is the radiating wave proportional to \( 1/r \), confirming the dipole's radiation pattern.

(10.2) What is the magnitude of the Poynting vector at a distance of 1 km from an antenna radiating 1 kW of power, assuming that it is an isotropic radiator with a wavelength much less than 1 km? What is the peak electric field strength at that distance?

Poynting Vector and Electric Field Strength

Given an isotropic radiator emitting power \( P_{\text{total}} = 1 \) kW at a distance of \( r = 1 \) km, we calculate the magnitude of the Poynting vector and the peak electric field strength.

For an isotropic source, the power spreads uniformly over a sphere of radius \( r \):

\[ \langle |\mathbf{P}| \rangle = \frac{P_{\text{total}}}{4 \pi r^2} \]

Substituting values:

\[ \langle |\mathbf{P}| \rangle = \frac{10^3}{4 \pi \times 10^6} = 8 \times 10^{-5} \text{ W/m}^2 \]

For a plane electromagnetic wave:

\[ \langle |\mathbf{P}| \rangle = \langle |E||H| \rangle \]

Since the intrinsic impedance of free space is:

\[ \eta = \sqrt{\frac{\mu_0}{\epsilon_0}} \approx 377 \Omega, \]

we use:

\[ |H| = \frac{1}{\eta} |E| \]

Thus:

\[ \langle |\mathbf{P}| \rangle = \frac{1}{2 \eta} E_{\max}^2 \]

Rearrange to solve for \( E_{\max} \):

\[ E_{\max} = \sqrt{2 \eta \langle |\mathbf{P}| \rangle} \]

Substituting values:

\[ E_{\max} = \sqrt{2 \times 377 \times 8 \times 10^{-5}} \]

\[ = \sqrt{0.06032} = 0.24 \text{ V/m} \]

Final Answer

Magnitude of the Poynting vector at \( r = 1 \) km:

\[ \langle |\mathbf{P}| \rangle = 8 \times 10^{-5} \text{ W/m}^2 \]

Peak electric field strength at \( r = 1 \) km:

\[ E_{\max} = 0.24 \text{ V/m} \]

(10.3) For what value of Rload is the maximum power delivered to the load in Figure 9.3?

Maximum Power Transfer

For what value of \( R_{ ext{load}} \) is the maximum power delivered to the load?

By Ohm’s Law:

\[ V = I (R_r + R_l) \]

where \( R_r \) is the radiation resistance, and \( R_l \) is the load resistance. Thus, solving for \( I \):

\[ I = \frac{V}{R_r + R_l} \]

The power dissipated in the load resistance is:

\[ P = I^2 R_l = \frac{V^2 R_l}{(R_r + R_l)^2} \]

To maximize this, we take the derivative:

\[ \frac{dP}{dR_l} = V^2 \left( (R_r + R_l)^{-2} - 2 R_l (R_r + R_l)^{-3} \right) \]

Setting this derivative to zero gives:

\[ R_r + R_l = 2 R_l \]

or solving for \( R_l \):

\[ R_l = R_r \]

Thus, the maximum power is delivered to the load when the load resistance equals the radiation resistance.

(10.4) For an infinitesimal dipole antenna, what are the gain and the area, and what is their ratio?

Antenna Gain and Area

For an infinitesimal dipole antenna, we determine the gain, the effective area, and their ratio.

From the text, the time-averaged Poynting vector is given by:

\[ \langle |\mathbf{P}| \rangle = \frac{I_0^2 k^2 d^2}{32 \pi^2 r^2} \sqrt{\frac{\mu_0}{\epsilon_0}} \sin^2\theta \]

The total radiated power is:

\[ W = I_0^2 \pi^3 \sqrt{\mu_0 \epsilon_0} \left( \frac{d}{\lambda} \right)^2 \]

Thus, the antenna gain is:

\[ G = \frac{\max_{\theta, \phi} \langle |P(r=1,\theta,\phi)| \rangle}{W / 4\pi} = \frac{4\pi \frac{I_0^2 k^2 d^2}{32 \pi^2} \sqrt{\frac{\mu_0}{\epsilon_0}}}{3 I_0^2 \pi \sqrt{\frac{\epsilon_0}{\mu_0}} \left( \frac{\lambda d}{\lambda^2} \right)} \]

Substituting \( k = \frac{2\pi}{\lambda} \), we obtain:

\[ G = \frac{(2\pi \lambda^{-1})^2 8 \pi^2}{3 \lambda^2} = \frac{3}{2} \]

To find the effective area, recall that the maximum power received is:

\[ \frac{V^2}{8R} = \langle |P| \rangle A = \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_{\max}^2 A \]

Solving for \( A \):

\[ A = \frac{V^2}{4R} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{1}{E_{\max}^2} \]

Using the relation:

\[ R = 2 \pi^3 \sqrt{\mu_0 \epsilon_0} \left( \frac{d}{\lambda} \right)^2 \]

and noting that the voltage relates to the maximum field as \( V = E_{\max} d \), we get:

\[ A = \frac{3}{8\pi} \lambda^2 \]

Finally, the ratio of gain to area is:

\[ \frac{A G}{\lambda^2} = \frac{3}{8\pi} \lambda^2 \times \frac{3}{2} = \frac{1}{4\pi} \lambda^2 \]