text: # 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
  
  $$\rm \\ \Rightarrow {{\rm{E}}_{\rm{rms}}} =E_0/\sqrt{2}=68.73\ mV/m$$
explanation: The provided text extract is a series of multiple-choice questions and their detailed solutions related to electric and magnetic fields, specifically focusing on dipole antennas. Each section contains a question followed by a detailed solution explaining the correct answer.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # 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
  
  $$\rm \\ \Rightarrow {{\rm{E}}_{\rm{rms}}} =E_0/\sqrt{2}=68.73\ mV/m$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text extract is a series of multiple-choice questions and their detailed solutions related to electric and magnetic fields, specifically focusing on dipole antennas. Each section contains a question followed by a detailed solution explaining the correct answer. 
  
  Label: multiple-question-and-answer