Practice Homework 20 2010

Practice Homework 20 2010 - choe (dec865) – Practice HW...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: choe (dec865) – Practice HW 20 Solutions – Weathers – (22202) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The electric field between the plates of an air capacitor of plate area 0 . 09 m 2 is changing at a rate of 23500 V / m / s. The permitivity of free space is 8 . 85419 × 10- 12 C 2 / N / m 2 . What is Maxwell’s displacement current? Correct answer: 1 . 87266 × 10- 8 A. Explanation: Let : ǫ = 8 . 85419 × 10- 12 C 2 / N / m 2 , A = 0 . 09 m 2 , and d E dt = 23500 V / m / s , where A is the area and d E dt is the changing rate of the electric field. The displacement current is I D = ǫ A d E dt = ( 8 . 85419 × 10- 12 C 2 / N / m 2 ) × ( . 09 m 2 ) (23500 V / m / s) = 1 . 87266 × 10- 8 A . 002 10.0 points The electric field in an electromagnetic wave (in vacuum) is described by E = E max sin( k x − ω t ) , where E max = 20 N / C and k = 9 . 3 × 10 8 m- 1 . The speed of light is 2 . 99792 × 10 8 m / s. Find the amplitude of the corresponding magnetic wave. Correct answer: 6 . 67128 × 10- 8 T. Explanation: Let : c = 2 . 99792 × 10 8 m / s and E max = 20 N / C . The relation between the magnitude of the electric field and that of the magnetic field for an electromagnetic wave is given by B = E c . Thus, the magnitude of the magnetic wave is B max = E max c = 20 N / C 2 . 99792 × 10 8 m / s = 6 . 67128 × 10- 8 T . 003 (part 1 of 2) 10.0 points What is the wavelength corresponding to a frequency of 2 × 10 19 Hz of electromagnetic wave in free space? The speed of light is 2 . 99792 × 10 8 m / s. Correct answer: 1 . 49896 × 10- 11 m. Explanation: Let : f = 2 × 10 19 Hz and c = 2 . 99792 × 10 8 m / s . The relation between frequency and wave- length is given by λ = c f = 2 . 99792 × 10 8 m / s 2 × 10 19 Hz = 1 . 49896 × 10- 11 m ....
View Full Document

This document was uploaded on 09/23/2010.

Page1 / 4

Practice Homework 20 2010 - choe (dec865) – Practice HW...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online