Homework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is:

Similar documents
This immediately suggests an inverse-square law for a "piece" of current along the line.

U>, and is negative. Electric Potential Energy

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface

STD: XI MATHEMATICS Total Marks: 90. I Choose the correct answer: ( 20 x 1 = 20 ) a) x = 1 b) x =2 c) x = 3 d) x = 0

SPA7010U/SPA7010P: THE GALAXY. Solutions for Coursework 1. Questions distributed on: 25 January 2018.

Answers to test yourself questions

Physics 1502: Lecture 2 Today s Agenda

1 Using Integration to Find Arc Lengths and Surface Areas

Radial geodesics in Schwarzschild spacetime

Chapter 28 Sources of Magnetic Field

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

Chapter 25 Electric Potential

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

Physics 11b Lecture #11

Physics Courseware Physics II Electric Field and Force

Electric Field F E. q Q R Q. ˆ 4 r r - - Electric field intensity depends on the medium! origin

CHAPTER 18: ELECTRIC CHARGE AND ELECTRIC FIELD

Optimization. x = 22 corresponds to local maximum by second derivative test

Homework: Study 6.2 #1, 3, 5, 7, 11, 15, 55, 57

Mark Scheme (Results) January 2008

Section 35 SHM and Circular Motion

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by

CHAPTER 7 Applications of Integration

Electric Potential. and Equipotentials

Physics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems.

ELECTRO - MAGNETIC INDUCTION

Solutions to Midterm Physics 201

Ch 26 - Capacitance! What s Next! Review! Lab this week!

(a) Counter-Clockwise (b) Clockwise ()N (c) No rotation (d) Not enough information

10 Statistical Distributions Solutions

The Area of a Triangle

MATHEMATICS IV 2 MARKS. 5 2 = e 3, 4

Previously. Extensions to backstepping controller designs. Tracking using backstepping Suppose we consider the general system

Physics 604 Problem Set 1 Due Sept 16, 2010

(A) 6.32 (B) 9.49 (C) (D) (E) 18.97

1. The sphere P travels in a straight line with speed

Fluids & Bernoulli s Equation. Group Problems 9

AQA Maths M2. Topic Questions from Papers. Circular Motion. Answers

Chapter 2: Electric Field

Friedmannien equations

EECE 260 Electrical Circuits Prof. Mark Fowler

ELECTROSTATICS. 4πε0. E dr. The electric field is along the direction where the potential decreases at the maximum rate. 5. Electric Potential Energy:

9.4 The response of equilibrium to temperature (continued)

FI 2201 Electromagnetism

Topics for Review for Final Exam in Calculus 16A

Continuous Charge Distributions

Week 8. Topic 2 Properties of Logarithms

Prerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) ,

u(r, θ) = 1 + 3a r n=1

$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4!" or. r ˆ = points from source q to observer

Solutions to Problems : Chapter 19 Problems appeared on the end of chapter 19 of the Textbook

10 m, so the distance from the Sun to the Moon during a solar eclipse is. The mass of the Sun, Earth, and Moon are = =

PX3008 Problem Sheet 1

SOLUTIONS TO CONCEPTS CHAPTER 11

r = (0.250 m) + (0.250 m) r = m = = ( N m / C )

Lecture 4. Electric Potential

PH126 Exam I Solutions

CHAPTER 29 ELECTRIC FIELD AND POTENTIAL EXERCISES

Lecture 10. Solution of Nonlinear Equations - II

Electric Potential and Gauss s Law, Configuration Energy Challenge Problem Solutions

r a + r b a + ( r b + r c)

π,π is the angle FROM a! TO b

dx was area under f ( x ) if ( ) 0

Chapter 21: Electric Charge and Electric Field

SSC Mains Mock Test 226 [Answer with Solution]

3.1 Magnetic Fields. Oersted and Ampere

CHAPTER? 29 ELECTRIC FIELD AND POTENTIAL EXERCISES = 2, N = (5.6) 1 = = = = = Newton

( )( )( ) ( ) + ( ) ( ) ( )

Homework Assignment 5 Solution Set

SSC [PRE+MAINS] Mock Test 131 [Answer with Solution]

Two dimensional polar coordinate system in airy stress functions

(1) It increases the break down potential of the surrounding medium so that more potential can be applied and hence more charge can be stored.

A LEVEL TOPIC REVIEW. factor and remainder theorems

Problem Set 3 SOLUTIONS

Course Updates. Reminders: 1) Assignment #8 available. 2) Chapter 28 this week.

On the Eötvös effect

Classical Electrodynamics

Fourier-Bessel Expansions with Arbitrary Radial Boundaries

Conservation of Linear Momentum using RTT

Homework Set 3 Physics 319 Classical Mechanics

Lecture 11: Potential Gradient and Capacitor Review:

Quality control. Final exam: 2012/1/12 (Thur), 9:00-12:00 Q1 Q2 Q3 Q4 Q5 YOUR NAME

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016

General Relativity Homework 5

Prof. Anchordoqui Problems set # 12 Physics 169 May 12, 2015

Chapter 6 Thermoelasticity

Physical Security Countermeasures. This entire sheet. I m going to put a heptadecagon into game.

Flux. Area Vector. Flux of Electric Field. Gauss s Law

Chapter 4 Two-Dimensional Motion

Winter 2004 OSU Sources of Magnetic Fields 1 Chapter 32

CHAPTER 2 ELECTROSTATIC POTENTIAL

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Physics 111. Uniform circular motion. Ch 6. v = constant. v constant. Wednesday, 8-9 pm in NSC 128/119 Sunday, 6:30-8 pm in CCLIR 468

GRAVITATION. Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18 PG 1

( ) ( ) ( ) ( ) ( ) # B x ( ˆ i ) ( ) # B y ( ˆ j ) ( ) # B y ("ˆ ( ) ( ) ( (( ) # ("ˆ ( ) ( ) ( ) # B ˆ z ( k )

Chapter I Vector Analysis

Energy Dissipation Gravitational Potential Energy Power

u( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 2-18, pp 44-48): Determine the equation of the following graph.

Example

Transcription:

. Homewok 3 MAE 8C Poblems, 5, 7, 0, 4, 5, 8, 3, 30, 3 fom Chpte 5, msh & Btt Point souces emit nuetons/sec t points,,, n 3 fin the flux cuent hlf wy between one sie of the tingle (blck ot). The flux fo point souce is: ϕ fom poblem. 4 The flux is then the sum of the point souces fluxes. ϕ 4 4 The fist tem is the flux fom points n which e equl the secon tem is the flux fom point 3. The cuent fom points n cncle so you only nee to clculte the cuent fom point 3. J D ϕ D 3 The cuent hs iection which is in the iection of the vecto fom point 3 to the point you e clculting the flux. 5. This cse in n infinite moeto so using the flux eqution 5.33. At the cente of the sque the cuent is zeo ue to symety. The flux is simply the sum of the 4 point souces. ϕ i e D

The flux t point ii is gin the sum of the fluxes. e e ϕ ii D D The cuent fom ech point is just, J D ϕ e which is given equtions befoe 5.33 4 The cuents fom point n cncel. ooking t the igm, the hoizonl potion of the cuents fom poitns 3 n 4 cncel. The sum of the veticl cuents with thei iection being own in the igm. Distnce fom point 3 to ii sme fo 4 to ii. J 3 e J 4 4 cuent fom point 3 n 4 t ii J e sin tn ( ) in the ownw iection 4 Whee the sine n tngent tems come fom foming ight tingle with points ii,, 3 n then using the tio of the sies to fin the ngle t point ii n then using tht ngle to fin the potion of the cuent in the ownw iection.

7. Point souce in n infinite moeto, ϕ e 4 D J D ϕ 4 e ) To fin the numbe of neutons pssing though sufce, fin the cuent t the sufce then integte ove the sufce e. J e 4 whee is the ius of the sphee which hs e 4 totl numbe of neutons e b) The numbe bsobe pe secon within the sphee is equl to Σ ϕ V in ou cse the volume integl is ove the volume of the sphee e Σ Σ ϕ V 4 e e 4 D D 0 c) Veify the continuity eqution, the numbe bobe the numbe leke equls the numbe pouce. e Σ e e Σ D Σ

0. Cuent equls zeo by symety, n the flux cn hve no spcil epenence ue to symety. Diffusion eqution ϕ ϕ D Eq (5.9) flux must be constnt ϕ C ϕ Σ plug into iffusion eqution, C C D D Σ 4. Agin using the iffusion eqution, ϕ ϕ D Thee is now non-unifomity in so the solutions cn vy in the iection. A genel solution to the iffusion eqution is, ϕ A sinh cosh B C Now using bouny conitions, the flux nees to be finite t 0 so B 0. Note tht the fist tem oes NOT go to infinity s goes to zeo, this cn be seen by expning sinh s n infinite seies. Agin we hve, C D C D Σ ϕ A sinh Σ Now using, ϕ( R ) 0 0 A sinh R ( R ) Σ A ( R ) Σ R sinh ϕ ( R ) Σ R sinh sinh Σ Σ sinh ( R ) R sinh

using, sinh( x) sinh( x) ϕ Σ sinh ( R ) sinh R b) J D ϕ D Σ sinh ( R ) sinh R ( R ) J D sinh R cosh sinh c) How mny neutons lek fom the sphee? JR ( ) 4 R ) D4 R ( R ) sinh R sinh R R cosh R R Avege pobbility tht souce neuton will leve the sphee is the numbe tht lek ivie by the totl numbe pouce D4 R ( R ) sinh R sinh R R cosh R R D4 R ( R ) sinh R sinh R R cosh R R

5. J D ϕ sin ϕ i A i R J DA i cos R R sin R R : 50cm ekge.cm30 5 cos( ) sin( ) 4R cms.606 0 7 : neutons pe secon s R R.7cm0 6 cos( ) sin( ) 4R cms.34 0 8 : neutons pe secon s R R 3.05cm0 6 cos( ) sin( ) 4R cms 4.45 0 7 : neutons pe secon s R R 8. J D ϕ cos y ϕ T Acos x cos z Cn't figue out how to mke it til so just clling it. J D ϕ T x x y ϕ T y z ϕ T z whee x is the unit vecto in the x iection ) J DA cos x cos y cos z sin x cos x sin y z sin x y cos y cos z z b) Evlute J otte into x ht t x / n then integte fom y -/ to / n z -/ to /. J, y, z DA cos cos y cos z sin cos sin y z sin x y cos y cos z z

J, y, z x DA eking neutons cos y cos z DA x DA4 cos y y pe sie c) By symety, the totl numbe is just the nswe fom pt b) times 6. DA 4 3. ) b : 9400 4 cm N A.600 4 : mol b) N A b cm s 6.05 gm.97 0 9 mol gmy.80 9 0 4 7.5 0 5 ecys pe secon y s 7.50 5 3.70 0.953 0 5 cuies 30. N : 00 i : 0.. N n : 0.. N x : i T :.85cm Σ :.097 cm 8i cm N D :.6cm : 8cm : 0 8 cm s

ϕ : T D sinh ( x) T cosh T ϕ 0 3 80 60 40 0 whee x is in units of metes n the flux is in units of neutons/cm^*s. 3. 0 0 0.0 0.04 0.06 0.08 x Using Eq. 5.64 to clculte the chnge in the theml iffusion pmete ( ) T ( ρ, T) T ρo, T o ρ o ρ T T o m ssuming To T T : T ρ Tb : T ρ b Tc : T ρ c.838cm.849cm.85cm 000 0.435 ρ :.004 00 000.435 ρ b :.00043 00 000 0..435 ρ c :.00004 000. sinh ( x) T T ϕ : ϕb : D cosh T Tb D sinh ( x) Tb cosh Tb ϕc : T D sinh ( x) Tc cosh Tc This clucltion tkes into ccount the chnge in ensity ue to ing boic ci to wte. The smll concenttions esult in vey smll chnge in the flux.

0 3 80 ϕ ϕb ϕc 60 40 0 0 0 0.0 0.04 0.06 0.08 x Clely this oesn't look like enough of n effect to mke this poblem wothwhile, I think the coect wy to o this poblem involves using the bsoption coss-section fo boic ci. I ws unble to fin this numbe.