Chapter(23+(HW(Solutions((Please(refer(to(textbook(for(questions+(numbers(indicated(here)((1.##Picture(the(Problem:#The#ima ge #s h o w s#a#ring#of#radius#3.1#cm#oriented#at#an#angle#of#θ#=#16º#from#a#B#=#0.055#T#magnetic#field.###Strategy:#Solve#equation#23A1#for#the#flux.##Solution:#Calculate#the#flux:(# Φ = BAcosθ= 0.055 T( )π0.031 m( )2cos16°Φ = 1.6 × 10−4 Wb##Insight:#The#maximum#flux#through#this#coil,#1.66×10−#4#Wb,#occurs#when#the#angle#θ#is#zer o.#(3.##Picture(the(Problem:#The#ima ge #s h o w s #a #re c ta n g u la r#lo o p #o rie n te d #4 2 #d e g re e s#fr o m#a#magnetic#field.###Strategy:##Solve#equation#23A1#for#the#magnetic#field.##Solution:#Calculate##the#magnetic#field:## B =ΦA cosθ=4.8 × 10−5 Tm20.051 m( )0.068 m( )cos 47°= 0.020 T##Insight:#The#minimum#magnetic#field#that#would#produce#this#flux#would#occur#when#the#rectangle#is#parallel#to#the#magnetic#field.##(4.##Picture(the(Problem:#A#house#h a s #a #flo o r#o f#d imensions #2 2 #m#by#18#m .##T h e #lo c al #magnetic#fie ld #d u e #to #E a r th #h a s#a #horizontal#component#2.6×10A5#T#and#a#downward#vertical#component#4.2×10A5#T.##Strategy:#The#horizontal#component#of#the#magnetic#field#is#parallel#to#the#floor,#so#it#does#not#contribute#to#the#flux.##Use#equation#23A1#to#calculate#the#flux#using#the#vertical#component.###Solution:#Calculate#the#magnetic#flux:(# Φ = BAcosθ= B⊥A = 4.2 × 10−5T( )22 m( )18 m( )= 1.7 × 10−2 Wb##Insight:#The#flux#through#the#vertical#walls#of#the#house#is#determined#by#the#horizontal#component#of#the#magnetic#field#instead#o f#th e#v ert ica l#co m p o n en t.#(10.##Picture(the(Problem:#The#ma gn e t ic#fl u x#t h ro u g h #a #co il#o s cill at es #in#t im e #as#indicated#by#the#graph#at#right.###Strategy: The magnitude of the flux is greatest when the flux is at a maximum or a minimum on the graph. The magnitude of the emf is greatest when the flux has the greatest positive or negative slope. #Solution: 1. (a) The magnetic flux has its greatest magnitude at t = 0 s, 0.2 s, 0.4 s, and 0.6 s . #2. (b) The magnitude of the induced emf is greatest at t = 0.1 s, 0.3 s, and 0.5 s . #Insight:##Note#that#the#magnitude#of#the#induced#emf#is#zero#when#the#magnitude#of#the#flux#is#a#maximum#and#the#magnitude#of#the#induced#emf#is#a#maximum#when#the#flux#is#zero.#((20.##Picture(the(Problem:#A#metal#ring#is#dropped#into#a#localized#region#of#constant#magnetic#field,#as#indicat ed #in #the #figu r e#a t#the #rig h t.#Th e#magnetic#field #is#ze ro #ab o v e#a nd #b elo w #th e #indicated#r eg io n .###Strategy:#Use#Lenz’s#Law#to#determine#the#direction#that#the#induced#current#must#flow#in#order#to#oppose#the#change#in#the#magnetic#flux#through#the#ring.##Solution:#1.((a)#At#location#1#the#magnetic#flux#through#the#ring#is#increasing#in#the#outAofAtheApage#direction.##The#induced#c u rre n t#will#flow#cloc k w is e#in #o rd er #to #pr o du c e#in toAtheApage#flux#to#oppose#this#change.##At#location#2#the#flux#through#the#ring#is#not#chang ing#so#tha t#the#indu ced #curren t#is#zero.##At#location#3#the #out AofAtheApage#flux#through#the#ring#is#decreasing,#so#the#induced#current#will#flow#countercloc kw ise#to#oppo se#this#change#by #produ cing#ou t A ofA th eApage#flux.##In#summary:#Location#1,#clockw ise;#loca tion #2,#zero;#location#3,#counterclockw ise.##2.(The#best#explanation#is#I.#Clockwise#at#1#to#oppose# th e# fie ld ;#zero#at#2#because#the#field# is # uniform;#counterclockw ise#at#3#to#try#to#maintain#the#field.##Statement#II#has#the#current#directions#reversed,#and#statement#III#does#not#properly#recognize#the#rate#of#change#of#the#magnetic#flux #th roug h#the #loop.##Insight:#Statement#II#would#be#correct#if#the#magnetic#field#pointed#into#the#page.#((21.##Picture(the(Problem:#A#metal#ring#is#dropped#into#a#localized#region#of#constant#magnetic#field,#as#indicat ed #in #the #figu r e#a t#the #rig h t.#Th e#magnetic#field#is#zero#above#and#below#the#indicated#r eg io n .###Strategy:#Use#Lenz’s#Law#to#determine#the#direction#that#the#induced#current#must#flow#in#order#to#oppose#the#change#in#the#magnetic#flux#through#the#ring.##Then#use#the#RightAHand#Rule#for#magnetic#forces#to#determine#the#force#on#the#ring.##Solution:#1.((a)#From#the#previous#problem#we#know#that#at#location#1#the#induced#current#will#flow#clockwise.##There#is#no#mag n etic #fo rc e#o n #th e#to p #o f#th e#rin g #where#the#field #is#ze ro ,#b u t#a t#th e#b o ttom#of#the#ring#the#force#on#a#current#flowing#fro m #rig h t#to #left#w ill#b e #in#th e #up ward#directio n.##T h e#fo rce s#o n #the #sid es #of#th e#rin g #ca n ce l,#so#th e #ne t#fo rce #on #the#ring#is#upward.##At#location#2#no#cu rrent#flow s#(see #previou s#prob lem )#so#th ere#is#no#m agnetic#force.##At#location#3#the#current#is#counterclockw ise#(se e#prev ious#pro blem )#an d#the re#is#no#force#on #the#bo ttom #of#the#ring#b ecau se#it#is#in#the#fieldAfree#regio n .##At #the #to p#o f#th e#rin g #th e#magnetic#for ce#o n #a#cu r re nt #flowing#from#rig h tAtoAle ft#is#in#the#upw a rd #direction.#In#summary:#Location#1,#upward;#location #2,#zero;#location#3,#upward.##2.(The#best#explanation#is#III.#Upward#at#1#to#oppose#entering#the#field ;# ze ro#at#2#because#the#field#is# uniform;#upward#at#3#to#oppose#lea vin g #th e# fie ld .##Statement#I#has#the#force#direction#at#location#3#reversed#and#statement#II#does#not#recognize#that#the#induced#current#(and #therefore#the#ma gnetic#force)#is#zero#at#location#2.# ##Insight:#The#action#of#the#magnetic#force#will#be#to#reduce#the#downward#acceleration#of#the#ring#both#w he n #it#ent ers #and#when#it#leaves#the#field#region.##At#all#other#times#the#acceleration#of#the#ring#is#9.81#m/s2#downward.#(26.##Picture(the(Problem:#The#ima ge #s h o w s #a #lo o p #of #w ir e #d ro p p in g #b et ween#the#p o le s #of#a#magnet.###Strategy: Use Lenz’s Law to determine the direction of the induced current. #Solution: 1. (a) When the loop is above the magnet, the magnetic field is increasing and directed out of the page. The current in the loop will oppose the increasing field by flowing clockwise. #2. (b) When the loop is below the magnet, the magnetic field is decreasing and is directed out of the page. The current in the loop will oppose the decreasing field by flowing counterclockwise. #Insight: When the loop is directly between the two poles the flux is a maximum, and therefore momentarily not changing. At this point the induced current is zero.