Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
Why does a TV loop have one turn?

Why does a TV loop have one turn?

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
See similar textbooks
icon
Related questions
Question
Why does a TV loop have one turn?
Problem 6.5 A circular-loop TV antenna with 0.02-m² area is in the presence of a uniform-amplitude 300-MHz signal. When oriented for maximum response, the loop I develops an emf with a peak value of 30 (mV). What is the peak magnitude of B of the incident wave? Solution: TV loop antennas have one turn. At maximum orientation, Eq. (6.5) evaluates to D = [B· ds with magnitude B we can express B as B = Bo cos (oI + ao) with @ = arbitrary reference phase. From Eq. (6.6), = +BA for a loop of area A and a uniform magnetic field B|. Since we know the frequency of the field is f= 300 MHz, 2n x 300 × 10° rad/s and ao an Vemf = –N =-A–[Bo cos(mt + a)] = AB90 sin(ot + o). di dt Vemf is maximum when sin(wt + ao) = 1. Hence, 30 x 10 = AB0@=0.02 × Bo × 6r x 10*, which yields Bo = 0.8 (nA/m).
Transcribed Image Text:Problem 6.5 A circular-loop TV antenna with 0.02-m² area is in the presence of a uniform-amplitude 300-MHz signal. When oriented for maximum response, the loop I develops an emf with a peak value of 30 (mV). What is the peak magnitude of B of the incident wave? Solution: TV loop antennas have one turn. At maximum orientation, Eq. (6.5) evaluates to D = [B· ds with magnitude B we can express B as B = Bo cos (oI + ao) with @ = arbitrary reference phase. From Eq. (6.6), = +BA for a loop of area A and a uniform magnetic field B|. Since we know the frequency of the field is f= 300 MHz, 2n x 300 × 10° rad/s and ao an Vemf = –N =-A–[Bo cos(mt + a)] = AB90 sin(ot + o). di dt Vemf is maximum when sin(wt + ao) = 1. Hence, 30 x 10 = AB0@=0.02 × Bo × 6r x 10*, which yields Bo = 0.8 (nA/m).
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Transmission Line Parameter (T-parameter)
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS