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1 This content has been downloaded fom IOPscience Please scoll down to see the full text Download details: IP Addess: 4858 This content was downloaded on 9/6/8 at : Please note that tems and conditions apply You may also be inteested in: Linea and nonlinea sloshing in a cicula conical tank I P Gavilyuk, I A Lukovsky and A N Timokha Femion-dyon bound states and femion numbe factionisation M Ravendanadhan and M Sabi Electon enegy distibutions in Townsend dischages in hydogen T E Kenny and J D Caggs Influence of esidual achomatic abeation on the isochonicity in the FAIR collecto ing S Litvinov, A Dolinskii, I Koop et al Mathematical modeling is also physics intedisciplinay teaching between mathematics and physics in Danish uppe seconday education Claus Michelsen Influence of the phase font paametes on the path of a beam of ays Nail R Sadykov Dischage time of a cylindical leaking bucket M Blasone, F Dell Anno, R De Luca et al Popagation of singulaities along chaacteistics of Maxwell's equations Elisabetta Baletta and Soin Dagomi Wind enegy: an application of Benoulli's theoem genealied to isentopic flow of ideal gases R De Luca and P Desidei

2 IOP Concise Physics Electomagnetism Poblems and solutions Caolina C Ilie and Zachaiah S Schecengost Chapte Mathematical techniques Thee ae a vaiety of mathematical techniques equied to solve poblems in electomagnetism The aim of this chapte is to povide poblems that will build confidence in these techniques Concepts fom vecto calculus and cuvilinea coodinate systems ae the pimay focus Theoy Dot and coss poduct Given vectos A = Ax x ˆ + Ay yˆ + A ˆ and B = Bx xˆ + By yˆ + B ˆ A B = A B + A B + A B = AB cos θ A B = x x y y xˆ yˆ ˆ A A A B B B x y x y with A B = AB sin θ whee A = A = Ax + Ay + A, B = B = Bx + By + B, and θ is the angle between A and B Sepaation vecto This notation is outlined by David J Giffiths in his book Intoduction to Electodynamics (999, ) Given a souce point and field point, the sepaation vecto points fom to and is given by = = ( x x ) xˆ + ( y y ) yˆ + ( ) ˆ doi:88/ ch - ª Mogan & Claypool Publishes 6

3 and the unit vecto pointing fom to is ˆ = = = ( x x ) xˆ + ( y y ) yˆ + ( ) ˆ ( x x ) + ( y y ) + ( ) As explained by Giffiths, this notation geatly simplifies late equations Tansfomation matix Given vecto A = Ax x ˆ + Ay yˆ + A ˆ in coodinate system K, the components of A in coodinate system K ae detemined by otational matix R given by R xx Rxy Rx R = Ryx Ryy Ry Rx Ry R with Ax A x = A R A y y A A 4 Gadient Given a scala functiont, the gadients fo vaious coodinate systems ae given below Catesian T = T ˆ + ˆ + x x T y y T ˆ Cylindical Spheical T = T ˆ + ϕ ϕ ˆ + s s T T s ˆ = T ˆ + θ θˆ T + θ ϕ ϕ ˆ T T sin 5 Divegence Given vecto function v, the divegences fo vaious coodinate systems ae given below Catesian = vx + vy v + v x -

4 Cylindical v = s s + vϕ ( svs ) s ϕ + v Spheical v = + ( θ θ) + v sin v sin θ θ vϕ sin θ ϕ 6 Cul Given vecto function v, the culs fo vaious coodinate systems ae given below Catesian = v ˆ + ˆ + vy v ˆ x vx v x y vy vx x Cylindical v = ϕ ˆ + ϕˆ + v v ϕ ˆ s ϕ s vs v s s s sv vs ϕ Spheical = θ v θ v v ( sin vϕ ) ˆ + sin θ θ ϕ sin θ ϕ + θ ϕ ˆ θ v v ( θˆ ϕ v ) 7 Laplacian Given a scala function T, the Laplacians fo vaious coodinate systems ae given below Catesian = + + T T T T x Cylindical T = + + s s s T T T s s ϕ Spheical T = + θ + T T T sin sin θ θ θ sin θ ϕ -

5 8 Line integal Given vecto function v and path P, a line integal is given by b v d l, a P whee a and b ae the end points, and dl is the infinitesimal displacement vecto along P In Catesian coodinates dl = dxxˆ + dyyˆ + d ˆ 9 Suface integal Given vecto function v and suface S, a suface integal is given by v d a, S whee da is the infinitesimal aea vecto that has diection nomal to the suface Note that da always depends on the suface involved Volume integal Given scala function T and volume V, a volume integal is given by V T d, τ whee dτ is the infinitesimal volume element In Catesian coodinates dτ = dx dy d Fundamental theoem fo gadients b ( T) dl = T b T a a P Fundamental theoem fo divegences (Gauss s theoem, Geen s theoem, divegence theoem) V v dτ = v da Fundamental theoem fo culs (Stoke s theoem, cul theoem) S 4 Cylindical pola coodinates Hee ou infinitesimal quantities ae v da = v dl S dl = dssˆ + sdϕϕˆ + dˆ P and dτ = s ds dϕ d -4

6 5 Spheical pola coodinates Hee ou infinitesimal quantities ae dl = dˆ + dθθˆ + sinθdϕϕˆ and dτ = sin θ d dθ d ϕ 6 One-dimensional Diac delta function The one-dimensional Diac delta function is given by x a δ( x a) = x = a and has the following popeties δ( x a)dx = f( x) δ( x a)d x = f( a) δ( kx) = δ ( x ) k 7 Theoy of vecto fields If the cul of a vecto field F vanishes eveywhee, then gadient of a scala potential V : F F = V F can be witten as the If the divegence of a vecto vanishes eveywhee, then F can be expessed as the cul of a vecto potential A: Poblems and solutions F = F = A Poblem Given vectos A = xˆ + 9yˆ + 5ˆ and B = xˆ 7yˆ + 4 ˆ, calculate A B and A B using vecto components and find the angle between A and B using both poducts -5

7 Solution A B = (xˆ + 9yˆ + 5 ˆ) ( xˆ 7yˆ + 4 ˆ) = ()() + (9)( 7) + (5)(4) = 6 + A B = 4 xˆ yˆ ˆ A B = = [(9)(4) ( 7)(5)] xˆ + [()(5) ()(4)] yˆ + [()( 7) ()(9)] ˆ Aˆ Bˆ = 7xˆ 7yˆ ˆ To find the angle θ between A and Bˆ we must fist calculate A and B: A = = 5 B = + ( 7) + 4 = 66 Using the dot poduct, we have = θ θ = A B AB cos cos θ = 7 Using the coss poduct, we have A B = AB sin θ 7 + ( 7) + ( ) = 5 66 sin θ θ = 67 Note, howeve, that we can see that the angle between A and B is geate than 9 Fo any agument γ, 9 sin ( γ) 9 Since the angle between A and B is geate than 9, we must adjust fo this by subtacting ou angle fom 8 Theefoe, θ = 8 67 = 7 as expected -6

8 Poblem The scala tiple poduct states A ( B C ) = B ( C A ) Pove this by expessing each side in tems of its components Solution Stating with the left-hand side, the coss poduct is B C = xˆ yˆ ˆ B B B C C C x y x y ( y y) x x ( x y y x) = BC BC xˆ + BC BC yˆ + BC BC ˆ Now, dotting A with ( B C ) A B C = A BC BC + A BC BC + A BC BC x y y y x x x y y x = ABC ABC + ABC ABC + ABC ABC x y x y y x y x x y y x = B C A C A + B C A C A + B C A C A x y y y x x x y y x A B C = B CA ˆ + ˆ + ˆ y CA y x CA x CA x y CA x y CA y x Note the tem in backets is pecisely C A, theefoe A ( B C ) = B ( C A ) as desied This pocedue can easily be applied again to pove the final pat of the tiple poduct, A ( B C ) = B ( C A ) = C ( A B ) Poblem Given souce vecto = cos θ xˆ + sin θ yˆ and field vecto =, ˆ find the sepaation vecto and the unit vecto ˆ Solution We have = = ˆ cos θ xˆ + sin θ yˆ = cos θ xˆ sin θ yˆ + ˆ To detemine the unit vecto ˆ, we must fist find the magnitude of, = ( cos θ) + ( sin θ) + = cos θ + sin θ + = + -7

9 So θ θ ˆ = = ˆ ˆ + ˆ + x y cos sin Poblem 4 Given A in coodinate system K, find the otational matix to give the components in system K Solution Fom the figues, we have = = = A A A A A A,, x y y x We want to find the otational matix R that satisfies = A A A R A A A x y x y Fom ou equations above = A A A A A A x y x y Electomagnetism -8

10 Theefoe, R = Poblem 5 Find the gadient of the following functions: a) 4 T = x + y + b) T = x ln y c) T = x y + Solutions a) T = x + y + 4 b) T = x ln y c) T = x y + T = T ˆ + ˆ + x x T y y T ˆ= 4x xˆ + yyˆ + ˆ T = T ˆ + ˆ + x x T y y T ˆ= x y xˆ + x y yˆ + x ln ln y ˆ T = T ˆ + ˆ + x x T y y T ˆ= xyxˆ + x yˆ + ˆ Poblem 6 Find the divegence of the following functions: a) v = xyxˆ y yˆ + ˆ b) v = ( x + y) xˆ + ( y + ) yˆ + ( + x) ˆ Solutions a) v = xyxˆ y yˆ + ˆ = v + v + v v = y 4y + x x y b) v = ( x + y) xˆ + ( y + ) yˆ + ( + x) ˆ = vx + vy v + v x = + + = Poblem 7 Find the cul of the following functions: a) v = xyxˆ y yˆ + ˆ b) v = ( x + y) xˆ + ( y + ) yˆ + ( + x) ˆ c) v = sin xxˆ + cos yyˆ -9

11 Solutions a) v = xyxˆ y yˆ + ˆ = v ˆ + ˆ + vy v ˆ x vx v x y vy vx x = ˆ + ( ) y ( xy) x x + y ( xy) ˆ x = + y xˆ + ( ) yˆ + ( x) ˆ v = y xˆ xˆ ˆ y b) v = ( x + y) xˆ + ( y + ) yˆ + ( + x) ˆ = v ˆ + ˆ + vy v ˆ x vx v x y vy vx x = ( + x) ( y + ) ˆ + ( x + y) ( + x) x ˆ y y x + ( y + ) ( x + y) ˆ x v = ˆ x yˆ ˆ c) v = sin xxˆ + cos yyˆ = v ˆ + ˆ + vy v ˆ x vx v x y vy vx x = () (cos y) ˆ + (sin x) () x ˆ y x y + (cos y) (sin x) ˆ= x -

12 Poblem 8 Pove ( T ) = Solution T = xˆ yˆ ˆ x T x T T = T T ˆ + T T x yˆ x x + T T ˆ x x ( T ) = Poblem 9 Find the Laplacian of the following functions: a) T = x + y + x + b) x T = e + siny cos( ) c) T = sin x cos y d) = ˆ v xyx + yˆ ˆ Solutions a) T = x + y + x + = T T x + T + T = + + = b) x T = e + siny cos( ) = T T x + T + T x = e sin y cos( ) 4 sin y cos( ) x = e 5siny cos( ) c) T = sin x cos y = T T x + T + T = sin x cos y sin x cos y = sinx cos y -

13 d) = ˆ v xyx + yˆ ˆ = v v x + + ˆ + v v v x x v v v ˆ x + + v v yˆ x x x y y y v = ( + + ) xˆ + ( + + ) yˆ+ ( + + ) ˆ = yˆ Poblem Test the divegence theoem with = ˆ + ˆ v xyx y y + ( x y) ˆ and the volume below Solution The divegence theoem states V v dτ = v d a Stating with the left-hand side, we have the divegence v = y + y + x = y + + x S We must split the volume into two pieces, (a) y and (b) y (a) τ= vd y + + x dy dx d = 5 -

14 (b) 4 y τ= vd y + + ( ) x dy dx d = 76 5 So, 5 76 v dτ= + = V Now we solve v d a, which must be evaluated ove the six sides S (i) We must split this egion into two sections (a) and (b), and da = dy d ˆ with x = In (a), y, v da = () y dy d = 4 In (b), y and 4 y 4 y v da = () y d dy = 6 -

15 (ii) Hee, da = dy d xˆ and x =, sov da = () y( dy d x) xˆ = (iii) Hee, da = dx dy ˆ and = = v d a x () y dy dx = (iv) Hee, da = dx dy ˆ and = = v d a x () y ( dx d y) = 8 (v) Hee da = dx d yˆ and y =, sov da = ( dx d ) = (vi) Hee, we have da = dx d nˆ whee nˆ = A = ˆ y + ˆ and B = xˆ n n We can find (the edges of the volume): n by cossing vectos n = A B = xˆ yˆ ˆ = 4yˆ + ˆ So n = 4 + = 5 and We can also obtain d by consideing 5 5 nˆ = yˆ + ˆ 5 5-4

16 so d = 5 d Now 5 da = x yˆ + ˆ = yˆ + ˆ x d d 5 5 d d 5 5 and = 4 y y = So v da = + y ( x y) dx d Theefoe = + = x dx d 4 5 v d a = = S as expected Poblem Test the cul theoem with = ˆ v xy x + y yˆ 5 + 4x ˆ and the suface below -5

17 Solution The cul theoem states S ( v ) da = v d l Stating with the left-hand side, the cul is given by P v = xˆ yˆ ˆ x 5xy y 4x = yxˆ 8xyˆ xyˆ We also have da = dy d xˆ with ( y ) + 4So S 4 ( y ) + 4 ( v ) da = y d dy = Now to solve v dl ove the two paths (i) and (ii): P 4 5 (i) Hee we have x =, =, and dl = d yyˆ Sov d l = y( )dy = (ii) Hee we have dl = dyyˆ + d ˆ, x =, and = ( y ) + 4 l = + = + 4 v d y dy 4( ) d y ( y ) 4 dy = P P 5 4-6

18 So, as expected l = + 4 v d = 4 P 5 5 Poblem Test the gadient theoem with T = x y and path = y and = y fom (,, ) (,, ) Solution The gadient theoem states b T d l = T b T a a Stating with the ight side, we have T(,,) T(,,) = () () = b Now to solve T dl, the gadient of T is given by a T = ( x y ) xˆ + ( x y ) yˆ + ( x y ) ˆ x = ˆ ˆ T x y y + 6 x y ˆ -7

19 Hee, dl = dyyˆ + dˆ with x = So l = + T d y dy 6()( ) y d = y dy y d Fo path (i), we have = y d = y dyso and T dl = y y d y y (y d y) = 4y dy b T dl = 4y dy = a as expected Fo path (ii), we have = y d = y dyso and 4 T dl = y y dy y y dy = 5y dy also as expected b 4 T dl = 5y dy = a Poblem Veify the following integation by pats given f = = ˆ + ˆ A x 4xyy xˆ and the suface below, f A da = + l A f da fa d S S P xy and -8

20 Solution Stating with the left-hand side A = xˆ yˆ ˆ x 4xy x = [ ( x)] yˆ + 4yˆ = ( x + ) yˆ + 4 yˆ Now f A = xy x + yˆ + 4 xy ˆ Hee we have da = dx d nˆ whee nˆ = xˆ + yˆ and n = so nˆ = xˆ + yˆ Also fom we have Now Theefoe, S dx = dx withy = x da = dx d xˆ + yˆ = dx d ( xˆ + yˆ ) = + ˆ f A da xy ( x ) y + 4 xy ˆ ( xˆ + yˆ )dx d = x( x) ( x + )dx d = Next, we will solve the fa dl tem fo the fou segments P 7 9-9

21 Segment (i) = f = xy () = Segment (ii) x = f = () y = Segment (iii) Segment (iv) dl = dxxˆ + d yyˆ, =, and y= x dy= d x y = f = x( ) = So l f A d = xy dx + 4xy d y = x( x) 4 x( x) dx and P l = fa d x( x) 4 x( x) dx = Now to solve the [ A f] da tem Fist, we have S = ˆ + ˆ f y x xyy + xy ˆ 6 -

22 So xˆ yˆ ˆ A ( f ) = 4xy x y xy xy = + ˆ 4xy xy x+ xy xy yˆ + xy 4 xyˆ As befoe, da = dx d ( xˆ + yˆ )So S [ A ( f)] da = 4 x ( x) + x ( x) x ( x) x( x) dx d So [ A ( f )] d a = S S 8 [ A ( f)] da + fa dl = + = P as expected Poblem 4 Find the divegence and cul of the following functions: a) v = ˆ + cos θ sin ϕθˆ + sin θ cos ϕϕˆ b) v = s cos ϕ s ˆ + cos ϕ sin ϕϕˆ+ sin ϕ ˆ Solutions a) v = ˆ + cos θ sin ϕθˆ + sin θ cos ϕϕˆ v = + ( θ θ) + v sin v sin θ θ vϕ sin θ ϕ = + θ θ ϕ + 4 sin cos sin sin θ θ sin θ ϕ (sin θ cos ϕ) ϕ = 4 sin + sin θ + cos θ + ϕ ( sin ) sin θ -

23 sin ϕ = 4 + sin θ sin θ sin ϕ sin ϕ v = 4 + csc θ sin θ v = θ v θ ( sin vϕ) ˆ sin θ θ ϕ v + sin θ ϕ θˆ + θ ϕ ˆ ϕ θ v v v = θ ϕ ( sin cos ) (cos θ sin ϕ) ˆ sin θ θ ϕ + sin θ ϕ ( ) ( θ ϕ θˆ sin cos ) + θ ϕ ( cos sin ) ϕ sin θ cos ϕ = ( sin θ cos θ cos ϕ cos θ cos ϕ) ˆ θ ˆ sin θ cos θ sin ϕ + ϕ ˆ cos θ cos ϕ sin θ cos ϕ θ ϕ v = θ ˆ θ ˆ cos sin ( csc ) + ϕ ˆ b) v = s cos ϕ s ˆ + cos ϕ sin ϕ ϕˆ + sin ϕ ˆ ϕˆ v = + ϕ s s sv v ( s) s ϕ + v = ( ϕ) + ϕ ϕ + ϕ s s s cos (cos sin ) ( s ϕ sin ) = cosϕ + sin ϕ + cos ϕ + sin ϕ s v = cosϕ + sinϕ + cos ϕ sin ϕ s -

24 v = ϕ ˆ + ϕˆ + v v ϕ ˆ s ϕ s vs v s s s sv vs ϕ = ϕ ϕ ϕ ˆ + ϕ ( sin ) (cos sin ) s ( ϕ ϕˆ s ϕ s cos ) ( s sin ) + ϕ ϕ ( ϕ ˆ s s s cos sin ) ( ϕ s cos ) = cos ϕ sˆ + (cos ϕ sin ϕ + s sin ϕ ) ˆ s s sin ϕ v = cos ϕ sˆ + (cos ϕ + s ) ˆ s s Poblem 5 Find the gadient and Laplacian of: a) T = (cos θ sin ϕ + sin θ cos ϕ) b) T = sin ϕ s cos ϕ Solutions a) T = (cos θ sin ϕ + sin θ cos ϕ) = T ˆ + θ θ ˆ T + θ ϕ ϕ ˆ T T sin = θ ϕ + θ ϕ ˆ + θ ϕ + θ ϕ θˆ (cos sin sin cos ) ( sin sin cos cos ) + (cos θ cos ϕ sin θ sin ϕ) ϕˆ sin θ = (cos θ sin ϕ + sin θ cos ϕ) ˆ + (cos θ cos ϕ sin θ sin ϕ) θˆ + (cos θ cos ϕ sin θ sin ϕ) ϕˆ sin θ = θ + ϕ ˆ + θ + ϕ θˆ T sin cos + θ + ϕ ϕˆ sin θ cos Note we could have witten T as T = sin( θ + ϕ) and then computed the gadient T = + θ + T T T sin sin θ θ θ sin θ ϕ -

25 = θ + ϕ sin + sin θ θ + cos( θ + ϕ) sin θ ϕ sin θ cos( θ + ϕ) = 6sin( θ + ϕ) + cosθ cos( θ + ϕ) sinθ sin( θ + ϕ) sin θ + ( sin( θ + ϕ) ) sin θ [ ] cos θ T = 5sin( θ + ϕ) + cos( θ + ϕ) sin θ sin( θ + ϕ) sin θ b) T = sin ϕ s cos ϕ T = T ˆ + ϕ ϕ ˆ + s s T T s ˆ = ( ϕ ϕ) ˆ + ( ϕ ϕ) ϕˆ s s s sin cos sin s cos s ϕ + ( ϕ ϕ) ˆ sin s cos = cos ϕ sˆ + ϕ ϕ ϕ ϕ ˆ + ϕ ˆ s cos s cos ( sin ) sin cos ϕ = ϕ ˆ T cos s + + s sin ϕϕˆ + sin ϕˆ s T = + s s s T T s s ϕ + T T s T ϕ T = ϕ = ϕ s T s = ϕ s s s T cos cos cos s = ϕ + ϕ ϕ T cos s cos sin = sin ϕ + s sin ϕ + cos ϕ ϕ = ϕ T sin = sin ϕ cos ϕ T = + sin ϕ + cos ϕ sin ϕ + sin ϕ s s s -4

26 ϕ cos T = sin ϕ + sin ϕ s s s Poblem 6 Test the divegence theoem with v = cos ϕ ˆ + cos θ sin θ θˆ + sin ϕϕˆ and the volume below (the uppe half of the sphee of adius R with a cone R of adius a = cut out) Solution The divegence theoem states V v dτ = v d a S Stating with the left-hand side, the divegence is v = + ( θ θ) + v sin v sin θ θ vϕ sin θ ϕ = ( ϕ) + ( θ θ) + cos sin cos sin θ θ ( sin ϕ) sin θ ϕ = cosϕ + sin θ cos θ sin θ + sin θ cos ϕ sin θ v = cosϕ + cos θ sin θ + Fo the volume, cos ϕ sin θ a π π π R, tan = tan = θ ϕ π R 6 6, -5

27 So V R π ϕ τ = cos ϕ + θ θ + ( v d cos cos sin sin θ ) dϕ dθ d sin θ π 6 π v dτ = πr V Now fo the ight-hand side, we have thee sufaces: the bottom (i), the oute shell (ii), and the inne pat whee the cone is cut out (iii) We have v = cos ϕ ˆ + sin ϕ ϕˆ + cos θ sin θ θˆ Fo (i), we have da = d dϕθˆ and θ = π So π π v da = cos sin d dϕ = and v da = () i Fo (ii), we have = R and da = sinθ dθ dϕ ˆ = R sinθ dθ dϕ ˆ So and v da = R cosϕsinθ dθ dϕ π π v a R ϕ θ θ ϕ d = cos sin d d = ( ii) Fo (iii), we have θ = π and da = sin θ d d ϕ θ ˆ ϕ θ 6 = d d ˆSo and π 6 π π v da = cos sin = π v a R 8 d = dϕ d = π ( iii) R -6

28 Theefoe, as expected S v da = πr Poblem 7 Test the cul theoem with v = ss ˆ + sin ϕ cos ϕ ϕˆ + scos ϕ ˆ and half of a cylindical shell with adius R and height h Solution The cul theoem states S v da = v d l P Stating with the left-handed side, we have Since we ae dotting So da = s dϕ d sˆ = R dϕ d sˆ da with v, we only need the ŝ component of the cul: v = ϕ ˆ = ϕ v v ϕ ϕ ˆ s ϕ s [ ] s ( s cos ) (sin cos ) s s ϕ = sin ϕ sˆ v da = R sinϕ dϕ d -7

29 We have ϕ π and h so ϕ ϕ S h π v da = R sin d d = h R Fo the left-hand side, we have fou cuves with v = ssˆ + sin ϕcos ϕ ϕˆ + cos ϕ ˆ Fo cuve (i), dl = d ϕϕˆ, =, and s = RSo v dl = sinϕcosϕ dϕ and π sin ϕcos ϕ dϕ = Fo cuve (ii), dl = d ˆ, ϕ = π, and s = RSo and v dl = s cos ϕ d = R cos π d = R d h R d = h R -8

30 Fo cuve (iii), dl = d ϕϕˆ, = h, and s = RSo and v dl = sinϕcosϕ dϕ sin ϕcos ϕ dϕ = Fo cuve (iv), dl = d ˆ, ϕ =, and s = RSo and So, as expected π v dl = s cos ϕ d = R cos()d = R d P R d = h R h v dl = h R h R = h R Poblem 8 Test the gadient theoem using T = s sin ϕ and the half helix path (adius R, height h) -9

31 Solution The gadient theoem states T d l = T b T a P Stating with the ight-hand side π π π π T = = ( b) T( a) T R,, h T R, Rh R = h R, sin () sin Now, the gadient is = T T ˆ + ϕ ϕ ˆ + ˆ= ϕ ˆ + ϕ ϕ ˆ + ϕ ˆ s s T T s s sin cos s sin We also have s = R and l = s d ϕϕˆ + d ˆ = R d ϕϕˆ + d ˆ So T dl = R cosϕ dϕ + R sinϕ d We need a way to elate and ϕ Note that as ϕ inceases, inceases linealy So, using the equation of line = γϕ ( ϕ), when = and ϕ = π, when = h and ϕ = π, so and Using ou expessions fo and So b π = γ ϕ +, γ π π h h = + γ =, π = h π ϕ h = h d d π ϕ d, we have T l = π ϕ + ϕ + π ϕ + ϕ ϕ R h h d cos R h h h sin π d π l = π ϕ + ϕ + π ϕ + ϕ ϕ = π T R h h d cos R h h h hr sin d a as expected π -

32 Poblem 9 Evaluate the following integals: a) (x x + 4) δ( x )dx b) ( x + 4) δ( x )dx 6 x c) sin δ( x π)dx d) (x + ) δ(4 x)dx e) x δ (x + )dx a f) δ( x b)dx Solutions a) x x + 4 δ( x )d x Since (, ) and f( x) = x x + 4, we have x x + 4 δ( x )d x = f() = () + 4 = b) c) Since (, ), we have 6 x + 4 δ( x )d x x + 4 δ( x )dx = x sin δ( x π)d x -

33 x Since π (, 6) and f( x) = sin 6, we have x π sin δ( x π)d x = f( π) = sin = d) x + δ(4 x)d x Since (, ) and f( x) = x +, we have x + δ(4 x)dx = f = + = 4 () 4 () 4 e) This can be ewitten as x δ (x + )d x x δ x + x = x δ x + x = x δx + x d d d Since (, ) and f( x) = x, we have x δ x + x = f = = d 8 f) Hee we have a δ( x b)dx = a δ( x b)d x { if < b < a othewise Poblem Suppose we have two vecto fields F = y ˆ and F = xxˆ + yyˆ + ˆ Calculate the divegence and cul of each Which can be witten as the gadient of a scala and which can be witten as the cul of a vecto? Find a scala and a vecto potential -

34 Solution Fo F, we have = ˆ + ˆ + ˆ F ( ˆ ) = x x y y y y = and F = xˆ yˆ ˆ x y = yxˆ Fo and F, we have = ˆ + ˆ + ˆ F x x y y ( xx ˆ + yy ˆ + ˆ ) = + + = F = xˆ yˆ ˆ x x y = ( ) xˆ + ( ) yˆ + ( ) ˆ = Since F =, F can be expessed as F = A We can find A by consideing By inspection: A = A This is satisfied by A = A xˆ yˆ ˆ x y A ˆ + A x A ˆ + x y A x A ˆ y x y x A = A A = A,, = y x x y x y x A = yxy, ˆ which is just one example Since F =, F can be expessed as F = VWe can find V by consideing -

35 = V ˆ + ˆ + ˆ F x x V y y V By inspection: This is satisfied by = V = V = V x, y, x x y V = + + which is again just one example Bibliogaphy Byon F W and Fulle R W 99 Mathematics of Classical and Quantum Physics (New Yok: Dove) Giffiths D J 999 Intoduction to Electodynamics d edn (Englewood Cliffs, NJ: Pentice Hall) Giffiths D J Intoduction to Electodynamics 4th edn (New Yok: Peason) Halliday D, Resnick R and Walke J Fundamentals of Physics 9th edn (New Yok: Wiley) Halliday D, Resnick R and Walke J Fundamentals of Physics th edn (New Yok: Wiley) Pucell E M and Moin D J Electicity and Magnetism d edn (Cambidge: Cambidge Univesity Pess) Rogawski J Calculus: Ealy Tanscendentals nd edn (San Fancisco, CA: Feeman) -4