( ) Make-up Tests. From Last Time. Electric Field Flux. o The Electric Field Flux through a bit of area is

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1 Mon., 3/23 Wed., 3/25 Thus., 3/26 Fi., 3/27 Mon., 3/30 Tues., 3/ Using Gauss s & nto to Ampee s Maxwell s, Gauss s, and Ampee s Quiz Ch 21, Lab 9 Ampee s Law (wite up) ,10 nto to Faaday s & Lenz s RE24 RE ,.7 Faaday & Emf & nductance RE27 Lab 9 Wite-up HW21:RQ.12, 15,17; P.20, 23, 26 RE26 Make-up Tests Fom Last Time Electic Field Flux o The Electic Field Flux though a bit of aea is Φ E A E o The Electic Field Flux out though a closed suface Φ E da E (whee A points out of the enclosed volume) Gauss s Law o Φ E = Q E da = ε enclosed o This Time Using Gauss s Law to find the Electic Field: Fo some distibutions of chage, Gauss s law to detemine the magnitude of the electic field. n paticula, distibutions with vey simple field geometies. 1. Use a symmety agument to detemine the diection of the electic field 2. Daw a Gaussian suface with each patch eithe pependicula o paallel to the electic field 3. Apply Gauss s law to find the magnitude of the electic field Example 1: Spheical shell with adius R and unifomly distibuted chage Q By symmety, the electic field must point adially and its magnitude can only depend on the distance fom the cente of the shell. A sphee of adius is a good Gaussian suface since it will be pependicula to the electic field eveywhee. The electic flux is: E n ˆ A = E 4π 2 ( )

2 ( ), so E = 0 ( q inside = Q), so < R: no chage inside q inside = 0 > R: all chage is inside E n ˆ A = E( 4π 2 )= E = Q 4πε 0 2 q inside ε 0 = Q ε 0 We ve been using these esults fo quite a while now (the second pat was just stated)! Note: Gauss s law only helps get the magnitude of the electic field, not the diection. Example 2: Lage unifomly chaged plate with chage pe aea of Q/A By symmety, the electic field nea the cente of the plate must point pependiculaly away fom the plate. ts magnitude could depend on the distance fom the plate. A box that extends on each side of the plate is a good Gaussian suface since the electic field will be paallel o pependicula to each side. A box f the sides pependicula to the plate each have an aea A box, the electic flux is: E n ˆ A = 2EA box The amount of chage inside the Gaussian suface is E n ˆ A = 2EA box = E = Q A 2ε 0 q inside ε 0 q inside = Q A A box, so = 1 ε 0 Q A A box Note: Need the Symmety. Gauss s law cannot be simply used to detemine the magnitude of the electic field fo chage distibutions that don t have the ight kind of symmety. Fo example, a cube with unifomly distibuted chage. Step 1 fails in this case. Monday, Mach. 23,

3 Chages on Metals: We can make some quick, qualitative obsevations about fields and chages in metals using Gauss s Law. Use what we know about the electic field to detemine popeties of chage. (1) Static equilibium : E = 0 inside metal The electic flux though any shape of closed suface is zeo, so thee is no net chage inside. E = 0 thus Q encl = 0 (2) Steady state: n a unifom, cuent-caying wie with constant aea, E has a constant magnitude and points along the wie. Daw a Gaussian suface just inside the wie (diagam below). The electic flux on the suface is zeo, because the electic field points in one end and out of the othe. Theefoe, thee can be no net chage inside the wie. E unifom thus Q encl = Tansition between two mateials, Suppose thee is a tansition between wies with the same aea A and mobile electon density n, but diffeent mobilities u 1 > u 2. The electon cuents must be the same (Node Rule), so E 1 < E 2. Daw a Gaussian suface just inside the wie that coss the tansition (diagam below). The electic flux on the suface is positive, so thee must be a net positive chage inside the suface. t is on the inteface (see Fig on p. 642). E not unifom thus Q encl not= E 1 E 2 Monday, Mach. 23,

4 Whiteboad Activity: Applying Gauss s Law A. Unifomly-Chaged Rod A thin od of length L has a positive chage Q distibuted unifomly along its length. Field Geomety. Use a symmety agument to detemine the diection of the electic field nea the cente of the od. Choosing the Gaussian bubble. What shape of Gaussian suface can you daw so that each pat is eithe pependicula o paallel to the electic field? Doing the Math. Use Gauss s law to find the magnitude of the electic field at a adial distance fom the od nea its cente. B. Solid Unifomly-Chaged Sphee A solid sphee of adius R has a positive chage Q distibuted unifomly thoughout its volume. Field Geomety. Use a symmety agument to detemine the diection of the electic field. Monday, Mach. 23,

5 Choosing a Gaussian Bubble. What shape of Gaussian suface can you daw so that each pat is eithe pependicula o paallel to the electic field? Doing the Math. Use Gauss s law to find the magnitude of the electic field at a distance fom the cente of the sphee fo: > R < R Monday, Mach. 23,

6 Gauss s Law fo Magnetism: Flux is a vey geneal mathematical idea. The flux of anything (wate fo example) though a closed suface depends on the souces / sinks of that something enclosed by the suface. f neithe (o equal souces and sinks) ae enclosed, then thee s no net flux. Applied to Magnetism. Anothe way to say that thee ae no magnetic dipoles is that thee is no (zeo) magnetic flux though any closed suface: B n ˆ da = 0. n othe wods, thee ae no points that the magnetic field adiates away fom. That s a boing, and not so useful elation. Howeve, thee is a elations, simila to Gauss s Law, that is similaly useful fo magnetism. Ampee s Law: You can detemine what is inside a loop by knowing the magnetic field aound the loop. Sketch some examples: B-field tangent to a loop & counteclockwise cuent out of page B-field tangent to a loop & clockwise cuent into page We want to show that the path integal of the magnetic field aound a closed loop is popotional to the amount of cuent piecing though the loop: B d l = µ 0 inside path. 1. Find the popotionality constant use a single, staight wie caying cuent in the cente of a cicle of adius. ntegate counteclockwise (diection of d l ). The magnetic field is the same size eveywhee on the cicle: B wie µ 0 2 4π, so the path integal is appoximately: B d l = B wie 2π = µ 0 Agument fo Geneality. Though not poven hee, it tuns out that this is bette than an appoximation, it s an equality. You may accept that the appoximation becomes an equality in the limit that goes to zeo; then, having established that it is tue fo the diffeentially thin loop aound a diffeentially small cuent, you can ague that it must scale up to be tue fo a big loop aound the same diffeentially small cuent, and then, the suppe position pinciple can be applied to add up the effects of not so diffeentially small. Monday, Mach. 23,

7 This is what we want. t is tue egadless of the adius of the loop (B vaies as 1/ and the cicumfeence vaies as ). 2. Loop Geomety doesn t matte. Show that the shape of the loop doesn t matte fo a single wie. Any shape of loop can be made with adial segments and acs centeed on the wie. d l 3. Only depends on en-looped cuent. The path integal of the magnetic field aound a closed loop is zeo if a wie is outside. Once again, any shape of loop can be made with adial segments and acs centeed on the wie. Howeve, fo evey segment with a positive path integal, thee will be one that is negative and the same size. Theefoe, the total path integal is zeo. d l 4. Multiple Cuents. Ampee s law holds fo multiple cuents Ex: thee cuents one into the loop, one out of the loop, and one outside of the loop 2 dl 1 Add these equations togethe: B 1 d l = µ 0 1 B 2 d l = µ 0 2 B 3 d l = 0 3 Monday, Mach. 23,

8 ( B 1 + B 2 + B 3 ) d l = µ B d l = µ 0 inside path ( ) The diection of the cuents inside the loop must be taken into account. f you cul the finges of you ight hand in the diection of the path aound the loop, you thumb will point in the diection that is positive fo cuent. Cuents pointing in the opposite diection ae negative. What Ampee s Law Does not say (applies fo whole loop, not segments) Note that Ampee s law says it is the complete loop integal that depends exclusively on the piecing cuent it does not say that the field o even B dl ove a given segment must depend only on the piecing cuent. f you imagine thee paallel wies and daw an Ampeian loop aound just one of them, cetainly the field at evey point on the loop is influenced by all thee cuents; howeve, Ampee s law does say that, when you integate ove the whole loop the contibutions of extenal cuents cancel out of the sum. We noted something simila fo Gauss s Law last time you get the simple esult only when you sum flux though the whole suface. Also, in this fom, it only speaks fo continuous cuents that is, it cannot handle a single moving point chage. Using Ampee s Law to find the Magnetic Field: Like Gauss s Law, Ampee s Law is of paticula use when the field geometies ae simple. Fo some distibutions of chage, Ampee s law to detemine the magnitude of the magnetic field. 1. Field Geomety. Use a symmety agument to detemine the diection of the magnetic field 2. Ampeian Loop. Daw a Ampeean loop that is eithe pependicula o paallel to the magnetic field 3. Math. Apply Ampee s law to find the magnitude of the magnetic field Monday, Mach. 23,

9 C. Thick Cuent-Caying Wie A long, thick wie of adius R caies a cuent. Field Geomety. Use a symmety agument to detemine the diection of the magnetic field nea the cente of the wie s length. o We have cylindical symmety in the cuent flow, so we must have cylindical symmety in the magnetic field. That necessitates that, at any point a distance fom the cente of the wie, the field must have the same stength and the same diection (in tems of and θ), so that if the wie is otated, the field, just like the cuent, looks unchanged. Geneally, this says the field looks like: θ nˆ No Radial Component by Gauss s Law o Note: as illustated, symmety aguments alone don t foce the field to be tangential to the suface. That comes fom Gauss s Law fo Magnetism: B n ˆ da = 0 o magine fo a moment that we enwap the wie in a Gaussian shell, ou Ampeian ing is just a coss-section of that. Applying symmety, we still have that B is constant and of constant oientation elative to the aea eveywhee on the suface, so Bsin θ da = Bsin θlπ = 0. The only way to make this equal zeo as we know it must is fo sinθ =0, o θ = 90. Ampeian Loop. What shape of Ampeean loop can you daw so that the magnetic field is tangent o pependicula to each segment? Math. Use Ampee s law to find the magnitude of the magnetic field at a adial distance fom the cente of the wie fo: > R Now, applying Ampee s Law gives Monday, Mach. 23,

10 B d l = µ B2π = µ µ B = π insidepath insidepath insidepath Exactly what we have fo an infinitesimally thin wie. n fact, notice that the only assumption that we made about the cuent density was that it was cylindically symmetic: that coves a line cuent (no width), a hollow shell of cuent, and eveything in between say, a cuent that dops of acoss the adius of the wie. As long as it s adially symmetic, the field looks the same. < R if we assume unifom cuent density, /A, then Hollow wie. Look at the case of a hollow shell of cuent: R The math woks out just the same: B d l = µ B2π = µ µ B = insidepath 2π insidepath insidepath What about inside this hollow tube of cuent? fo >R Coaxial Cable The symmety and Gaussian aguments ae the same, but inside is 0, so inside B = 0. Now, what about a Coaxial cable? That s got a thin wie in the middle, caying cuent one way, and a hollow tube encicling it and caying equal and opposite cuent. R Monday, Mach. 23,

11 µ 0 wie The field inside the tube is puely due to the inne wie, B =, <R 2π While the field outside is due to both of the wie and tube of cuent, µ 0 wie µ 0tube B = +, >R, 2π 2π but if they have equal and opposite cuents, then the two tems cancel so B = 0, >R. That, in fact, is one of the appeals of coaxial wies. \ Solenoid. Now fo something a little moe complicated: the solenoid. Symmety: demands that the field is cylindically symmetic. f we appoximate the solenoid as infinitely long, i.e., we want an appoximate answe good nea the middle of its length, we can ague fo linea symmety too that is, anywhee we look along the solenoid s length, the field must be pointing in the same diection. Consideing the fields due two o thee consecutive ings, we can also ague that they cancel each othe s adial components of the field (o a Gaussian suface could be employed to ule this out.) So we e left with only axial components. Monday, Mach. 23,

12 Example 3: solenoid with N/L waps pe length caying cuent The magnetic field inside the solenoid points in the diection shown below (RHR). Daw a ectangula loop that has a side paallel to the magnetic field. The loop can extend vey fa away fom the solenoid so that the contibution of the top end must be zeo. d The path integal is Bd and the amount of cuent in the loop is d( N L), so: B Bd = µ 0 [ d( N L) ] B = µ 0 N L This is much moe difficult to do using the Biot-Savat Law (see Ch. 17). Note: it is not tivial to eason that the field is puely axial. Applying Ampee s Law in the plane of a cuent loop gives 0 (the cuent loop doesn t piece the Ampeian loop) so thee s no angula component; applying Gauss s law tells us thee s no adial component. Apply to a Tous. Applying Ampee s Law Fiday: Pactice with Gauss s law and Ampee s law Monday, Mach. 23,

Ch 30 - Sources of Magnetic Field! The Biot-Savart Law! = k m. r 2. Example 1! Example 2!

Ch 30 - Sources of Magnetic Field! The Biot-Savart Law! = k m. r 2. Example 1! Example 2! Ch 30 - Souces of Magnetic Field 1.) Example 1 Detemine the magnitude and diection of the magnetic field at the point O in the diagam. (Cuent flows fom top to bottom, adius of cuvatue.) Fo staight segments,