Sources of the Magnetic Field. Physics 231 Lecture 8-1

Transcription

1 Souces of the Magnetic Field Physics 31 Lectue 8-1

2 Magnetic Field of a Point Chage Given a point chage, q, we know that it geneates an electic field egadless of whethe it is moving o not f the chage is in fact moving then the chage also geneates a magnetic field The magnetic field geneated by the chage howeve does not behave in the same way the electic field of the chage does Physics 31 Lectue 8-

3 Magnetic Field of a Point Chage t is found that the magnetic field is pependicula to both the velocity of the chage and the unit vecto fom the chage to the point in question The magnitude of the field is given by µ q v sinφ 4π whee φ is the angle between the velocity and unit vecto The magnetic field lines fom concentic cicles about the velocity vecto Physics 31 Lectue 8-3

4 Magnetic Field of a Point Chage The vecto equation fo this is The diection of the magnetic field is given by one of the ight-hand ules Point you thumb, of you ight hand, in the diection of the velocity and cul you finges. The diection of the field is then the diection you finges cul. The magnetic field fo a negative chage will be in the opposite diection. µ 4π qv ˆ Physics 31 Lectue 8-4

5 Magnetic Field of a Cuent Element f instead of a single chage, we have a cuent what is the magnetic field then given by We stat with the pinciple of supeposition we add the magnetic fields of the individual chages that make up the cuent Given n moving chages pe unit volume each caying a chage q, then the total chage in this element is dq nq Adl whee A is the coss sectional aea Physics 31 Lectue 8-5

6 Magnetic Field of a Cuent Element We then have fo d d dq vd sinφ µ n q vd Adl sin µ 4π 4π ut fom ou wok on cuents we also have that Theefoe O in vecto fom µ dl sinφ d 4π µ dl ˆ d 4π φ n q v d A Physics 31 Lectue 8-6

7 iot-savat Law Fo a complete cicuit we then use the integal fom of the pevious equation µ dl ˆ 4π This can be a vey difficult integal to evaluate The level of difficulty depends upon the vaiable chosen to integate ove and the pesence of any symmeties that can be eploited We will develop anothe, much simple method, shotly Physics 31 Lectue 8-7

8 Field of a Staight Wie We set up the poblem as shown The integal becomes µ 4π a a dy ( + y ) 3/ The esult of this integal is µ π a 4 + a n the limit that a >> µ π Physics 31 Lectue 8-8

9 Magnetic Foce etween Two Cuent Caying Wies We have two wies caying cuents and sepaated by a distance The fist wie caying cuent sets up a magnetic field at the second wie given by µ π Physics 31 Lectue 8-9

10 Magnetic Foce etween Two Cuent Caying Wies Fom befoe we know that a wie caying a cuent that is placed in a magnetic field will epeience a foce pe unit length that is given by F L ' µ π This is in fact what the second wie will epeience And fom F ' L This foce is diected towads the othe wie the foce is attactive Physics 31 Lectue 8-1

11 Magnetic Foce etween Two Cuent Caying Wies f we had stated with the wie caying the cuent, we would have ended up with the same esult f the wies wee caying cuents in opposite diections, then the foce between the wies would be epulsive Physics 31 Lectue 8-11

12 Cicula Cuent Loop Anothe cicuit fo which it is wothwhile to calculate the magnetic field is the cicula cuent loop We stat by utilizing the iot-savat law d µ 4π dl ˆ with + a a being the adius of the loop Physics 31 Lectue 8-1

13 Cicula Cuent Loop A cuent element at the top will yield the magnetic field as shown A cuent element at the bottom will yield a magnetic field of the same magnitude, but diected downwad The vetical components will cancel and all that will be left is a hoizontal magnetic field of incemental value µ dl µ a dl 4 a d cosθ π + a 4π + ( ) 3/ Physics 31 Lectue 8-13

14 Cicula Cuent Loop The integal is ove just the cicumfeence of the loop as all othe vaiables ae constant µ a µ a dl 4π + a ( ) 3/ ( + a ) 3/ f thee ae N loops, tuns, then the esult is just N times above esults µ N + a a ( ) 3/ Physics 31 Lectue 8-14

15 Cicula Cuent Loop An impotant point fo the field of a cuent loop is at the cente of the loop whee µ N a A cuent loop consisting of N tuns is often efeed to as a Solenoid A final caution: These esults fo a cuent loop only apply on the ais of the loop Physics 31 Lectue 8-15

16 Ampee s Law Just as we developed an easy way to calculate the electic field in symmetic situations, Gauss Law, thee is a simila technique we can use fo calculating magnetic fields This is known as Ampee s Law dl µ whee the integal is aound a closed path and is the total cuent passing though the aea bounded by this path Physics 31 Lectue 8-16

17 Ampee s Law Thee is one big diffeence to emembe between the two methods Gauss Law is an integal ove a closed suface Q E da ε Wheeas Ampee s Law is an integal ove a closed path dl µ Physics 31 Lectue 8-17

18 Ampee s Law Given a path aound which the integation is to be done, thee is a sign convention fo the cuent Detemine the diection of the nomal, as defined by the sense of movement aound the integation path f the cuent is going in the same diection as the nomal, the cuent is consideed to be positive othewise it is negative Physics 31 Lectue 8-18

19 Physics 31 Lectue 8-19 Long Staight Wie Cuent in same diection as nomal of integation path dl dl π µ µ π µ µ This is the same as we deived fom the iot-savat Law

20 Long Staight Wie Cuent in opposite diection to nomal of integation path dl π µ π µ dl µ µ Again this is the same as we deived fom the iot- Savat Law Physics 31 Lectue 8-

21 Caution Ampee s Law gives you only the magnetic field due to any cuents that cut though the aea bounded by the integation path Fo the indicated integation path, the esult of Ampee s Law is zeo Physics 31 Lectue 8-1

22 Eample Given thee cuents and thei associated loops A C 1) Fo which loop is dl the smallest? A C Same The cuents that ae used in Ampee s Law ae the cuents that ae within the loop that is used fo integation Loop A has no cuent going though it wheeas the othe two loops do have cuents going though them, the integal fo Loop A is zeo Physics 31 Lectue 8-

23 Eample Given thee cuents and thei associated loops A C ) Now compae loops and C. Fo which loop is dl the geatest? C Same The ight hand side of Ampee s law is the same fo both loops, theefoe the integal fo both loops is the same Physics 31 Lectue 8-3

24 Eample A cuent flows in an infinite staight wie in the +z diection as shown. A concentic infinite cylinde of adius R caies cuent in the -z diection. 1) What is the magnetic field (a) at point a, just outside the cylinde as shown? (a) (a) < (b) (a) (c) (a) > y a b The loop used fo application of Ampee s Law will be outside the cylinde The total cuent going though this loop is the sum of the two cuent yielding a cuent of. This gives a magnetic that points in the positive diection at a Physics 31 Lectue 8-4

25 Eample A cuent flows in an infinite staight wie in the +z diection as shown. A concentic infinite cylinde of adius R caies cuent in the -z diection. ) What is the magnetic field (b) at point b, just inside the cylinde as shown? (a) (b) < (b) (b) (c) (b) > y a b This time, the Ampee loop only encloses cuent which is in the +z diection the loop is inside the cylinde! The cuent in the cylindical shell does not contibute to at point b. Physics 31 Lectue 8-5

26 Long Cylindical Conducto The poblem can be divided into two egions Outside the conducto nside the conducto Physics 31 Lectue 8-6

27 Long Cylindical Conducto Then using Ampee s Law we then have o µ π Outside the conducto We stat with an integation path aound the conducto at an abitay distance fom the cental ais of the conducto dl µ dl π µ Physics 31 Lectue 8-7

28 Long Cylindical Conducto We define a cuent density nside the conducto Since we have not been told othewise, we assume that the cuent is unifomly distibuted acoss the coss section of the conducto J A π R We now apply Ampee s Law aound an integation path centeed on the cylindical ais and adius < R dl µ with being the cuent going though the aea defined by the integation path Physics 31 Lectue 8-8

29 Long Cylindical Conducto dl π nside the conducto π Jπ π R and So afte equating, we then have µ µ µ R R π R The magnetic field inceases linealy with the distance fom the ais of the cylinde Physics 31 Lectue 8-9

30 Long Cylindical Conducto nside the wie: ( < a) µ π R a Outside the wie: ( > a ) µ π Physics 31 Lectue 8-3

31 Solenoid We choose a path of integation as shown with one pat of the path inside the solenoid and one pat of the path outside We assume that the path on the outside, cd, is sufficiently fa fom the solenoid, compaed to the diamete of the solenoid, that the field is appoimately equal to zeo Physics 31 Lectue 8-31

32 Solenoid We use Ampee s Law dl µ Fist the Left-Hand-Side dl a b: Using the ight hand ule fo cuents, the field is in the same diection as the path and it is constant, we then have + b a d c dl dl b + + c b a d dl a dl dl L Physics 31 Lectue 8-3

33 Solenoid The same is tue fo d a Physics 31 Lectue 8-33 b c: Along the pat of the path that is inside the solenoid, is pependicula to dl, so that contibution is zeo, while fo the pat of the path that is outside of the solenoid, ~, so we then have c a d dl c d: Hee is zeo so we again have b dl d c dl

34 Solenoid Now fo the Right-Hand-Side The total cuent passing though the aea defined by the integation path is N whee N is the numbe of tuns and is the cuent in the loop, so the ight-hand-side is µ N Now equating the left and ight-hand-sides we have O solving fo L µ µ N L N Physics 31 Lectue 8-34

35 Solenoid The magnetic field of a solenoid looks like the following with the eact shape of the field lines being dependent upon the the numbe of tuns, the length, and the diamete of the solenoid Physics 31 Lectue 8-35