# For a short dipole antenna:

1. The radiation intensity is maximum along the normal to the dipole axis
2. The current distribution along its length is uniform irrespective of the length
3. The effective length equals its physical length
4. The input impedance is independent of the location of the feed – point

Option 1 : The radiation intensity is maximum along the normal to the dipole axis

## Electric and Magnetic Field MCQ Question 1 Detailed Solution

A dipole antenna is a linear antenna, usually fed in the center and producing maximum of radiation in the plane normal to the axis.

It is said to be short dipole when length is less than đťť€⁄đťź’ and current distribution is sinusoidal.

Radiation intensity is maximum along the normal to the dipole axis.

# For a short dipole antenna:

1. The radiation intensity is maximum along the normal to the dipole axis
2. The current distribution along its length is uniform irrespective of the length
3. The effective length equals its physical length
4. The input impedance is independent of the location of the feed – point

Option 1 : The radiation intensity is maximum along the normal to the dipole axis

## Electric and Magnetic Field MCQ Question 2 Detailed Solution

A dipole antenna is a linear antenna, usually fed in the center and producing maximum of radiation in the plane normal to the axis.

It is said to be short dipole when length is less than đťť€⁄đťź’ and current distribution is sinusoidal.

Radiation intensity is maximum along the normal to the dipole axis.

# If the radiation electric field at the point (r, θ, Ď•) due to a short dipole antenna located at the origin is E, then the field at the point (2r, θ, Ď•) is

1. E/8
2. E/4
3. E/2
4. E

Option 3 : E/2

## Electric and Magnetic Field MCQ Question 3 Detailed Solution

The electric field of a short dipole or Hertzian dipole is given by

$$\begin{array}{l} {E_r} = \frac{{\eta {I_o}dl}}{{2\pi }}\cos \theta \left[ {\frac{1}{{{r^2}}} - \frac{j}{{\beta {r^3}}}} \right]{e^{ - j\beta r}}\\ {I_\theta } = \frac{{\eta {I_o}dl}}{{4\pi }}\cos \theta \left[ {\frac{{j\beta }}{r} + \frac{1}{{{r^2}}} - \frac{j}{{\beta {r^3}}}} \right]{e^{ - j\beta r}} \end{array}$$

Here, $$\frac{1}{{{r^3}}}$$ term corresponds to electrostatic field of an electric dipole.

The, $$\frac{1}{{{r^2}}}$$ term corresponds to near field or inductive field.

The, $$\frac{1}{r}$$ term corresponds to far field or radiation field.

$$\begin{array}{l} E\;\alpha \frac{1}{r}\\ \therefore Field\;at\;point\;\left( {2r,\theta ,\phi } \right)is\;E = \frac{E}{2} \end{array}$$

# A Hertzian dipole radiates an input power of $$15\ {\rm{kW}}$$ with $$70{\rm{\% }}$$ efficiency. The maximum rms electric field intensity at $$10\ {\rm{km}}$$ isGiven, directivity of Hertzian dipole is $$1.5$$ and intrinsic impedance of free space, $${{\rm{\eta }}_{\rm{o}}} = 120{\rm{\pi }}$$

1. $$94.5\ {\rm{mV}}/{\rm{m}}$$
2. $$68.7\ {\rm{mV}}/{\rm{m}}$$
3. $$90.5\ {\rm{V}}/{\rm{m}}$$
4. $$97.2\ {\rm{mV}}/{\rm{m}}$$

Option 2 : $$68.7\ {\rm{mV}}/{\rm{m}}$$

## Electric and Magnetic Field MCQ Question 4 Detailed Solution

By definition, directive gain $${{\rm{G}}_{\rm{d}}}\left( {{\rm{\theta }},\phi } \right)$$ is given by, $${{\rm{G}}_{\rm{d}}}\left( {{\rm{\theta }},\phi } \right) = \frac{{4{\rm{\pi }}{{\rm{r}}^2}{{\rm{P}}_{{\rm{avg}}}}}}{{{{\rm{P}}_{\rm{r}}}}}$$

$${{\rm{P}}_{{\rm{avg}}}}$$ is maximum in the direction where $${{\rm{G}}_{\rm{d}}}\left( {{\rm{\theta }},\phi } \right)$$ is maximum.

Now, $${\left. {{{\rm{G}}_{\rm{d}}}\left( {{\rm{\theta }},\phi } \right)} \right|_{{\rm{max}}}} = {\rm{D}} = 1.5$$

Thus, $${\left. {{{\rm{P}}_{{\rm{avg}}}}} \right|_{{\rm{max}}}} = \frac{{{\rm{D}} \cdot {{\rm{P}}_{\rm{r}}}}}{{4{\rm{\pi }}{{\rm{r}}^2}}}$$

Let the electric field at this point in $${{\rm{E}}_0}$$

Then, $${\left. {{{\rm{P}}_{{\rm{avg}}}}} \right|_{{\rm{max}}}} = \frac{{{\rm{E}}_0^2}}{{2{{\rm{\eta }}_0}}}$$

Thus, $$\frac{{{\rm{E}}_0^2}}{{2{{\rm{\eta }}_0}}} = \frac{{{\rm{D}} \cdot {{\rm{P}}_{\rm{r}}}}}{{4{\rm{\pi }}{{\rm{r}}^2}}}$$

$$\Rightarrow {\rm{E}}_0^2 = \frac{{{\rm{D}} \cdot {{\rm{P}}_{\rm{r}}} \cdot {{\rm{\eta }}_0}}}{{2{\rm{\pi }}{{\rm{r}}^2}}}$$

Now, radiated power $${{\rm{P}}_{\rm{r}}} = 0.7 \times 15 \times {10^3} = 10.5\ {\rm{kW}}$$

$$\begin{array}{l} \Rightarrow {{\rm{E}}_0} = {\left[ {\frac{{1.5 \times 10.5 \times {{10}^3} \times 120{\rm{\pi }}}}{{2{\rm{\pi }} \times {{\left( {10 \times {{10}^3}} \right)}^2}}}} \right]^{\frac{1}{2}}}= 97.2\ {\rm{mV}}/{\rm{m}} \end{array}$$

Now, rms value

$$\rm \\ \Rightarrow {{\rm{E}}_{\rm{rms}}} =E_0/\sqrt{2}=68.73\ mV/m$$
The text passage contains multiple-choice questions and detailed solutions related to the properties and behavior of dipole antennas, specifically focusing on radiation intensity, electric field, and power calculations.
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# For a short dipole antenna:

1. The radiation intensity is maximum along the normal to the dipole axis
2. The current distribution along its length is uniform irrespective of the length
3. The effective length equals its physical length
4. The input impedance is independent of the location of the feed – point

Option 1 : The radiation intensity is maximum along the normal to the dipole axis

## Electric and Magnetic Field MCQ Question 1 Detailed Solution

A dipole antenna is a linear antenna, usually fed in the center and producing maximum of radiation in the plane normal to the axis.

It is said to be short dipole when length is less than đťť€⁄đťź’ and current distribution is sinusoidal.

Radiation intensity is maximum along the normal to the dipole axis.

# For a short dipole antenna:

1. The radiation intensity is maximum along the normal to the dipole axis
2. The current distribution along its length is uniform irrespective of the length
3. The effective length equals its physical length
4. The input impedance is independent of the location of the feed – point

Option 1 : The radiation intensity is maximum along the normal to the dipole axis

## Electric and Magnetic Field MCQ Question 2 Detailed Solution

A dipole antenna is a linear antenna, usually fed in the center and producing maximum of radiation in the plane normal to the axis.

It is said to be short dipole when length is less than đťť€⁄đťź’ and current distribution is sinusoidal.

Radiation intensity is maximum along the normal to the dipole axis.

# If the radiation electric field at the point (r, θ, Ď•) due to a short dipole antenna located at the origin is E, then the field at the point (2r, θ, Ď•) is

1. E/8
2. E/4
3. E/2
4. E

Option 3 : E/2

## Electric and Magnetic Field MCQ Question 3 Detailed Solution

The electric field of a short dipole or Hertzian dipole is given by

$$\begin{array}{l} {E_r} = \frac{{\eta {I_o}dl}}{{2\pi }}\cos \theta \left[ {\frac{1}{{{r^2}}} - \frac{j}{{\beta {r^3}}}} \right]{e^{ - j\beta r}}\\ {I_\theta } = \frac{{\eta {I_o}dl}}{{4\pi }}\cos \theta \left[ {\frac{{j\beta }}{r} + \frac{1}{{{r^2}}} - \frac{j}{{\beta {r^3}}}} \right]{e^{ - j\beta r}} \end{array}$$

Here, $$\frac{1}{{{r^3}}}$$ term corresponds to electrostatic field of an electric dipole.

The, $$\frac{1}{{{r^2}}}$$ term corresponds to near field or inductive field.

The, $$\frac{1}{r}$$ term corresponds to far field or radiation field.

$$\begin{array}{l} E\;\alpha \frac{1}{r}\\ \therefore Field\;at\;point\;\left( {2r,\theta ,\phi } \right)is\;E = \frac{E}{2} \end{array}$$

# A Hertzian dipole radiates an input power of $$15\ {\rm{kW}}$$ with $$70{\rm{\% }}$$ efficiency. The maximum rms electric field intensity at $$10\ {\rm{km}}$$ isGiven, directivity of Hertzian dipole is $$1.5$$ and intrinsic impedance of free space, $${{\rm{\eta }}_{\rm{o}}} = 120{\rm{\pi }}$$

1. $$94.5\ {\rm{mV}}/{\rm{m}}$$
2. $$68.7\ {\rm{mV}}/{\rm{m}}$$
3. $$90.5\ {\rm{V}}/{\rm{m}}$$
4. $$97.2\ {\rm{mV}}/{\rm{m}}$$

Option 2 : $$68.7\ {\rm{mV}}/{\rm{m}}$$

## Electric and Magnetic Field MCQ Question 4 Detailed Solution

By definition, directive gain $${{\rm{G}}_{\rm{d}}}\left( {{\rm{\theta }},\phi } \right)$$ is given by, $${{\rm{G}}_{\rm{d}}}\left( {{\rm{\theta }},\phi } \right) = \frac{{4{\rm{\pi }}{{\rm{r}}^2}{{\rm{P}}_{{\rm{avg}}}}}}{{{{\rm{P}}_{\rm{r}}}}}$$

$${{\rm{P}}_{{\rm{avg}}}}$$ is maximum in the direction where $${{\rm{G}}_{\rm{d}}}\left( {{\rm{\theta }},\phi } \right)$$ is maximum.

Now, $${\left. {{{\rm{G}}_{\rm{d}}}\left( {{\rm{\theta }},\phi } \right)} \right|_{{\rm{max}}}} = {\rm{D}} = 1.5$$

Thus, $${\left. {{{\rm{P}}_{{\rm{avg}}}}} \right|_{{\rm{max}}}} = \frac{{{\rm{D}} \cdot {{\rm{P}}_{\rm{r}}}}}{{4{\rm{\pi }}{{\rm{r}}^2}}}$$

Let the electric field at this point in $${{\rm{E}}_0}$$

Then, $${\left. {{{\rm{P}}_{{\rm{avg}}}}} \right|_{{\rm{max}}}} = \frac{{{\rm{E}}_0^2}}{{2{{\rm{\eta }}_0}}}$$

Thus, $$\frac{{{\rm{E}}_0^2}}{{2{{\rm{\eta }}_0}}} = \frac{{{\rm{D}} \cdot {{\rm{P}}_{\rm{r}}}}}{{4{\rm{\pi }}{{\rm{r}}^2}}}$$

$$\Rightarrow {\rm{E}}_0^2 = \frac{{{\rm{D}} \cdot {{\rm{P}}_{\rm{r}}} \cdot {{\rm{\eta }}_0}}}{{2{\rm{\pi }}{{\rm{r}}^2}}}$$

Now, radiated power $${{\rm{P}}_{\rm{r}}} = 0.7 \times 15 \times {10^3} = 10.5\ {\rm{kW}}$$

$$\begin{array}{l} \Rightarrow {{\rm{E}}_0} = {\left[ {\frac{{1.5 \times 10.5 \times {{10}^3} \times 120{\rm{\pi }}}}{{2{\rm{\pi }} \times {{\left( {10 \times {{10}^3}} \right)}^2}}}} \right]^{\frac{1}{2}}}= 97.2\ {\rm{mV}}/{\rm{m}} \end{array}$$

Now, rms value

$$\rm \\ \Rightarrow {{\rm{E}}_{\rm{rms}}} =E_0/\sqrt{2}=68.73\ mV/m$$
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The text passage contains multiple-choice questions and detailed solutions related to the properties and behavior of dipole antennas, specifically focusing on radiation intensity, electric field, and power calculations.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS