(2) An electromagnetic harmonic wave is traveling in a medium having refractive index n = 1.44. If the wave is traveling in a direction given by û = (+2)/√2, and its electric field component is specified by the following function: E = (2x −9+2) × 10³ cos(k r-5.196 x 10¹5 t) T Determine: (a) (b) (c) The speed of the wave. The direction and magnitude of the magnetic field vector. The wave vector.

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
icon
Related questions
Question

Please Answer Question (2)

(a)
(b)
Attempt all questions:
1) Two sections of a long thick cable are fed with a uniform
current given by 10.5 cos(100zt) 2 A/m². The radius
of the cable is equal to a 5 cm and the separation
between the two sections w= 2 mm << a. Assume that
the current flows in such a way that the surface charges
on the two sides of the gap are uniformly distributed and
the fringing of the field lines outside the gap is negligible.
a) Determine the electric field in the gap between the
ends of the two sections, as a function of t.
b) Find the displacement current through a circle of
radius p < a in the region midway between the ends
of the two sections.
Determine:
(c)
(d)
c) Determine the magnetic field vector at points in the
gap between the ends of the two sections at p < a
from the axis.
(2) An electromagnetic harmonic wave is traveling in a medium having refractive index
n = 1.44. If the wave is traveling in a direction given by û = (+2)/√2, and its
electric field component is specified by the following function:
E = (2x -ŷ + 2) × 103 cos(k r-5.196 x 10¹5 t) T
(e)
d) Calculate Poynting vector on the edge of the
gap at p a for 0≤z≤w. Integrate the
Poynting vector over the side area of the gap
to show that the electromagnetic power through the
side of the gap equals the electric power P = V(t)1(t),
where V(t) is the voltage between the ends of the two
sections of the wire.
x
The speed of the wave.
The direction and magnitude of the magnetic field vector.
The wave vector.
W
The wavelength and frequency of the wave.
The average intensity of the wave.
Transcribed Image Text:(a) (b) Attempt all questions: 1) Two sections of a long thick cable are fed with a uniform current given by 10.5 cos(100zt) 2 A/m². The radius of the cable is equal to a 5 cm and the separation between the two sections w= 2 mm << a. Assume that the current flows in such a way that the surface charges on the two sides of the gap are uniformly distributed and the fringing of the field lines outside the gap is negligible. a) Determine the electric field in the gap between the ends of the two sections, as a function of t. b) Find the displacement current through a circle of radius p < a in the region midway between the ends of the two sections. Determine: (c) (d) c) Determine the magnetic field vector at points in the gap between the ends of the two sections at p < a from the axis. (2) An electromagnetic harmonic wave is traveling in a medium having refractive index n = 1.44. If the wave is traveling in a direction given by û = (+2)/√2, and its electric field component is specified by the following function: E = (2x -ŷ + 2) × 103 cos(k r-5.196 x 10¹5 t) T (e) d) Calculate Poynting vector on the edge of the gap at p a for 0≤z≤w. Integrate the Poynting vector over the side area of the gap to show that the electromagnetic power through the side of the gap equals the electric power P = V(t)1(t), where V(t) is the voltage between the ends of the two sections of the wire. x The speed of the wave. The direction and magnitude of the magnetic field vector. The wave vector. W The wavelength and frequency of the wave. The average intensity of the wave.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

Blurred answer
Knowledge Booster
Maxwell Equation
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON