Magnetic fields (origins) CHAPTER 27 SOURCES OF MAGNETIC FIELD. Permanent magnets. Electric currents. Magnetic field due to a moving charge.

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1 Magnetic fields (oigins) CHAPTER 27 SOURCES OF MAGNETC FELD Magnetic field due to a moving chage. Electic cuents Pemanent magnets Magnetic field due to electic cuents Staight wies Cicula coil Solenoid Foce between two wies Gauss s Law and Ampèe s Law Magnetic field inside a wie Tooidal solenoid Stationay electic Stationay electic chage field Moving electic Moving electic field chage + (cuent) Magnetic field Much eseach in the 19th centuy: magnetism and electicity unified by James Clek Maxwell ( ). Magnetic poles appea in pais (N & S), always out in Like poles epel Unlike poles attact

2 Magnetic field of a single moving chage: d = 0 ˆ + v The magnetic field is cicula aound the diection of motion of the chage. The magnetic field at a point (P), due to a chage q moving with velocity v is: = k q v ˆ 2, whee k = µ " 4π with µ " = 4π 10 7 T.m/A and ˆ = pemeability of fee space φ P d = 0. Question 27.0: A point chage, q = 3.2 µc, is moving along the x-axis with velocity v = 2.50 m/s. What is the magnetic field poduced by this chage at the point ( x,y) = ( 2.00 m,1.50 m) when it passes the point x = 0.50 m?

3 We ll solve this poblem in two ways. z ˆ k y ˆ j P (2.50,1.50) x + φ q v ˆ i and ˆ = = (a) The field is given by = µ " q v ˆ 4π 2, whee v = (2.50 m/s)ˆ i (2.00ˆ i +1.50ˆ j )m (2.00m) 2 + (1.50m) 2 ˆ = (0.80ˆ i ˆ j )m. = µ " ( C)(2.50 m)ˆ i (0.80ˆ i ˆ j )m 4π (2.00m) 2 + (1.50m) 2 ( ) k ˆ = (b) We have v ˆ = v ˆ sin φ k ˆ, whee sin φ = = ( T) ˆ k (2.00) 2 + (1.50) 2 = z ˆ k y ˆ j P (2.50,1.50) x + φ q v ˆ i = µ " q v ˆ sin φ 4π 2 = ( ) ( C)(2.50 m/s)(1)(0.60) k ˆ 6.25 = ( T) ˆ k. k ˆ

4 Magnetic field due to a cuent (iot-savat Law): P d = 0 d l The magnetic field at a point, due to a small element of wie d l caying a cuent is: d l ˆ d = µ # 4π Fo a complete cicuit: = µ # 4π ˆ φ 2. d l ˆ 2. d = 0 The diection of, with espect to the cuent, is given by the ight hand ule: o

5 Magnetic field of a staight wie: The magnetic field line contous aound d l ae cicula. The magnetic field shown above is fo a cuent out of the page. Remembe, the chages also poduce an electic field, but the electic field lines ae adial. θ 2 P θ d (out) θ 1 R ˆ j The iot-savat law is d = µ " d l ˆ 4π 2. Since the contibutions to the total magnetic field at P ae all paallel (outwads towads us) we need only calculate the magnitude of the field. Then, we have d P = µ " dy 4π 2 sin φ = µ " dy 4π 2 cosθ, whee φ is the angle between d l and ˆ. ut y = R tanθ, so diffeentiating we find φ d l = dyˆ j y dy = Rsec 2 θ.dθ = R 2 R 2 dθ = 2 R dθ.

6 Then, substituting fo dy we get d P = µ 2 dθ 4π 2 R cosθ = µ 4πR cosθ.dθ P = µ 4πR θ 2 cosθ.dθ = µ 4πR [ sin θ] θ 2 θ 1 θ 1 P = µ ( 4πR sin θ 2 + sin θ 1 ). Magnetic field " Cuent P θ 2 θ 1 R R = 0.25L Question 27.1: What is the magnetic field at the point P on the axis of the wie? R = 0.5L θ 1 = 0 θ 2 = 0

7 Magnetic field of an infinitely long wie: P " ˆ d l Fom the iot-savat Law: = µ " d l ˆ 4π 2, ut d l l ˆ fo all elements d l. d l ˆ = 0 = 0. P θ 2 θ 1 Fo a staight wie (R) = µ 4π (sin θ 1 + sin θ 2 ). As the wie becomes infinitely long: R θ 1 90 and θ sin θ 1 + sin θ 2 2, so (R) = µ 2π

Explanation of Oested s expeiments (1819): = 0 Diection of the Eath s magnetic field Magnetic field lines fo a single cuent caying wie, indicated by ion filings Looking down the wie = µ 2π

8 Explanation of Oested s expeiments (1819): = 0 Diection of the Eath s magnetic field Magnetic field lines fo a single cuent caying wie, indicated by ion filings Looking down the wie = µ 2π Magnetic fields, like electic fields, add vectoially. The compass needles point in the diection of the esultant of the Eath s magnetic field and the magnetic field poduced by the cuent in the wie.

9 Magnetic field due to two paallel wies: ˆ j P ˆ i 4.00 cm 5.00 cm L R Magnetic field lines add vectoially: shown hee is the combined field fo two paallel wies caying a cuent in opposite diections. Question 27.2: Two, vey long, paallel wies, L and R, each cay cuents = 3.00 A in opposite diections, as shown above. f the wies ae spaced 5.00 cm apat, what is the magnetic field vecto at a point P a distance 4.00 cm fom R, if LRP = 90? What would be the magnetic field midway between the two wies if the cuents ae in the same diection?

10 R 1 P θ 2 d L R L Let the magnetic field at P due to the left hand wie (L) be L and the magnetic field at P due to the ight hand wie (R) be R. We have ( ) = θ = tan Note that, " L is pependicula to LP and " R is paallel to LR. The esultant magnetic field in the ˆ i diection is i = ( L sin(90 θ) ) R, and the esultant magnetic field in the ˆ j diection is j = L cos(90 θ). So, the esultant magnetic field is: " = iˆ i + jˆ j. ˆ j ˆ i L " R 90 θ d P θ 1 2 R 90 θ " L Fom ealie, and using the given values, L = µ µ = 2π 1 2π d and and = 4π π = T R = µ 2πd = 4π = T. 2π 0.04 i = ( L sin(90 θ) ) R = T j = L cos(90 θ) = T. " = ( ˆ i ˆ j ) T, " and = T. CHECK: You will find the esultant field at P is paallel to PL. That is a geneal esult fo all points P, when RP is pependicula to LR.

11 Question 27.3: The cuent in the wie shown below is 8.0 A. Find the magnetic field stength at P. 2 cm 8 A 1 cm P a 8 A Label the ends of the staight sections a-f. Sections a-b and e-f do not contibute to the magnetic field at P, but b-c, c-d and d-e all poduce magnetic fields into the page (so we can simply add them). c b 2 cm P d Field due to b-c: 1 = µ 4π (sin θ 1 + sin θ 2 ) = 4π A 4π 0.01 m (sin 45 + sin 0) = T. ut this is the same as the field due to d-e. Field due to c-d: 2 = µ 4π (sin θ 1 + sin θ 2 ) = 4π A 4π 0.01 m (sin 45 + sin 45 ) = T. Theefoe total field is: 1 cm e f ( ) T = T (into the page).

12 Magnetic field on the axis of a cicula coil (of adius R): z The cuent is the same at evey point aound the coil, so the components of the magnetic field pependicula to the x-diection (i.e., in the y z plane) cancel, i.e., the field esticted to the x-diection: d l y R d d x d x = dsin θ. ut, d = d = µ " d l ˆ 4π 2 = µ " dl 4π (x 2 + R 2 ), whee we have substituted d l ˆ = dl since d l ˆ. x θ d y θ d y d x d x = µ dl 4π (x 2 + R 2 ) sin θ = µ dl 4π (x 2 + R 2 ) = µ Rdl 4π (x 2 + R 2 ) 3 2. R x 2 + R 2 To obtain the total field we integate aound the loop. NOTE: x and R ae both constant, = µ 4π x = d x = µ 4π R (x 2 + R 2 ) 3 2 2πR = µ 4π = µ R 2 2(x 2 + R 2 ) 3 2, R (x 2 + R 2 ) 3 2 dl 2πR 2 (x 2 + R 2 ) 3 2 i.e., # µ = xˆ i = R 2 ˆ 2(x 2 + R 2 ) 3 2 i. When x = 0: x = µ 2R.

x max = µ " 2R x = µ " R 2 2(x 2 + R 2 ) 3

13 x max = µ " 2R x = µ " R 2 2(x 2 + R 2 ) 3 2 x 0 x Vaiation of the magnetic field along the axis of a cicula coil. The diection of the field is given by the ight hand ule. y R z x x x Magnetic field lines in the vicinity of a cicula coil caying a cuent.

14 P Question 27.4: A closed cicuit consists of two semicicles of adii 40.0 cm and 20.0 cm that ae connected by staight segments. A cuent of 3.00 A flows in the cicuit in a clockwise diection. Find the magnetic field stength at the point P. P The staight sections do not contibute to the field at P. The net field is due to the two semi-cicula acs. Put R 1 = 0.40 m and R 2 = 0.20 m. = = 1 µ " (inwads) and 2 = 1 µ " (inwads) 2 2R 1 2 2R 2 = = 1 µ " + 1 µ " = µ " R 1 2 2R 2 4 R 1 R 2 = 4π A m m = T (inwads).

5: The magnetic field lines poduced by a cuent caying coil ae vey simila to the magnetic field lines of a

15 left hand face ight hand face Above, magnetic field lines poduced by a cuent caying coil. elow, magnetic field lines of a pemanent magnet. Question 27.5: The magnetic field lines poduced by a cuent caying coil ae vey simila to the magnetic field lines of a pemanent magnet. Theefoe, one face of the coil acts like a NORTH pole and the othe acts like a SOUTH pole. Which face of the coil shown hee is the noth pole? Notice, they ae essentially identical.

equivalent magnets N S N S ATTRACTON S N equivalent

16 equivalent magnets N S N S ATTRACTON S N equivalent magnets N S S N y definition, magnetic field lines tavel outwad fom a NORTH pole. Theefoe, the ight hand face of this coil acts like a noth pole. REPULSON Thus, a cuent caying coil has the popeties and chaacteistics of a pemanent magnet. Cuent CCW. McDonald ule Cuent CW

17 N S N S Cuent CW Question 27.6: A cuent caying wie and the noth end of a magnet ae placed close to each othe, as shown above. Will thee be a mutual ataction o mutual epulsion between the coil and the magnet? Fom the N-pole end of the magnet, the cuent in the coil appeas clockwise. y the McDonald ule, a cw cuent means that face of the coil is a S-pole. So, thee is a mutual attaction between the coil and magnet

A SOLENOD is a long, tightly wound coil with a constant adius.

solenoid ae paallel to the axis and the magnetic field stength at the cente is: = µ n whee

18 A SOLENOD is a long, tightly wound coil with a constant adius. Magnetic field of a solenoid: Total of N tuns n = N l l The magnetic field lines within a solenoid ae paallel to the axis and the magnetic field stength at the cente is: = µ n whee n is the numbe of tuns pe unit length (n = N l ), N is the total numbe of tuns and l is the length. z ( l >> adius) µ " n ~ 50% x We can incease the stength of the magnetic field by inseting a coe... then µ " K m µ " µ, i.e, = µn. Values of K m vay fom 1 to ~ l 2 l 2 l ~ 50% x y S N S N x

19 Foce between two staight wies (Ampèe 1820): The magnetic field due to wie 1 at position P is: P = µ " 1 ˆ j. 2πR Since wie 2 caies a cuent, 2, thee is a foce: F = 2 l P. acting on the segment l. Since l is in the k ˆ diection, F is in the ˆ i diection, i.e., to the left and attactive. Also, as P and l ae pependicula, the magnitude of the foce is 2 1 P F P l 1 2 R F = 2 l P = 2 l P. ˆ j ˆ k ˆ i 2 1 P F Substituting fo P we find P l 1 2 R F = 2 l P = 2 l µ # 1 2πR ˆ j = µ # 1 2 l 2πR, to the left so the foce pe unit length is F l = µ # A simila analysis indicates that the magnetic field ˆ k 1 2 2πR. poduced by the cuent in wie 2 esults in a foce of equal magnitude towads the ight on wie 1, i.e., in the +ˆ i diection. Hence, two wies caying cuents in the same diection ae attacted to each othe. Coollay: two wies caying cuents in opposing diections epel each othe. ˆ i

20 Definition of the AMPÈRE: 1.00 m f two paallel wies cay the same cuent, i.e., 1 = 2 =, the foce pe unit length between the wies is: F 2 l = µ " 2πR. Since µ " = 4π 10 7 T m/a, a cuent of one AMPÈRE can be defined as: 10.0 cm Question 27.7: Thee vey long staight paallel wies pass though the vetices of an equilateal tiangle with sides of length 10.0 cm. f the cuent in each wie is 15.0 A in the diections shown,... the cuent in two paallel wies of infinite length placed 1.00 m apat that poduces a foce on each wie of N pe mete of length. i.e., if 1 = 2 = 1 A and R = 1.00 m. F l = 4π π N/m (a) What is the magnetic field at the location of the uppe wie due to the cuents in the two lowe wies? (b) What foce pe unit length does the uppe wie expeience?

21 (a) Note that 1 is pependicula to the line joining #1 and the uppe wie and 2 is pependicula to the line y (ˆ j ) 2 30 " 30 " x (ˆ i ) 60 " 60 " 60 " 1 #1 # 2 joining #2 and the uppe wie. Using the ight hand ule thei diections ae as shown. Fom the geomety of the poblem, the magnetic field vectos ae at 30 " to the x- diection, so the esultant magnetic field is = 1 cos30 " + 2 cos30 ". ut 1 = 2 =, whee = µ " 2π = 4π A 2π 0.10 m = T. = 2( cos30 " )ˆ i = 2( cos30 " )ˆ i = 2( T 0.866)ˆ i = T ˆ i. y (ˆ j ) 2 30 " 30 " x (ˆ i ) 60 " 60 " 60 " 1 #1 # 2 (b) The foce the lowe wies exet on the uppe wie is F = l = lˆ k ˆ i = lˆ j, i.e., in the y-diection. F l = ˆ j = ( 15.0 A T)ˆ j = ( N)ˆ j.

22 Question 27.8: dentify all of the foces acting on the windings of a cuent caying solenoid... The foces acting on the windings of a solenoid ae etween each tun the cuents ae paallel so thee ae attactive foces that tend to educe the length of the coil. 2. Acoss a tun the cuents ae anti-paallel so thee ae epulsive foces that tend to incease the adius of the coil.

23 Gauss s Law in magnetism: Unlike electical chages, single, isolated magnetic poles - monopoles - do not exist... magnetic poles only occu in pais. So, the net magnetic flux though any closed suface is zeo, always. Look... A sot of equivalent to Gauss s Law in magnetism is... Ampèe s Law: The line integal of the magnetic field aound a closed path is popotional to the algebaic sum of the enclosed cuents, + Electic chage with electic field lines; lines stat fom a +ve chage and end on a ve chage. Magnetic poles with magnetic field lines; magnetic field lines ae continuous so if a line exits a Gaussian suface it must also ente. So, Gauss s Law tells us... the net magnetic flux though any closed suface is zeo, always. i.e., c d l = µ # enc. Howeve, Ampèe s Law is only useful if... the cuent is constant, and the system has high symmety. So it is not so geneally applicable as Gauss s Law. Let s look at some examples.

24 Example of the use of Ampèe s Law: Ampee s Law: c d l = µ # enc. We choose the closed path to be a cicle pependicula to the wie and centeed on the wie. Then is constant on the path (by symmety). d l at all points on the path. c d l = d c l i.e., = µ # 2π, = ( 2π) = µ # which is the same esult we obtained befoe fo a long staight wie d l Ampeian closed path (a cicle centeed on the wie) d l Magnetic field inside and outside a cuent caying wie: s thee a magnetic field inside a conducting wie? R 1: nside the wie ( < R). Take a cicula Ampeian loop 1 centeed on the wie. y symmety, 1 is constant (and paallel to d l ) d l = 1 dl 1 = 1 ( 2π). The faction of the cuent enclosed by 1 is enc = Aea enclosed by Total aea = π2 πr 2 = 2 R ( 2π) = µ # enc = µ # R 2 i.e., 1 = µ # 2πR 2 1 ( < R). d 1 l R 1

25 2: Outside the wie ( > R). 2 d l R R 2 1 = µ 2πR 2 max = µ 2πR 2 = µ 2π y symmety 2 is constant (and paallel to d l ) d l = 2 2 dl = 2 2dl = 2 ( 2π) = µ # enc. The cuent enclosed by the Ampeian loop 2 is. 2 ( 2π) = µ # i.e., 2 = µ # ( > R). 2π This is just like the field fom an infinitely long wie. Note: it is independent of the adius R. nside the wie, the field inceases linealy fom the cente to a maximum at the suface of the wie. Outside the wie, the field is the same as the field poduced by an infinite wie. 0 R 2R 3R 4R 5R The magnetic field inside ( < R) and outside ( > R) a cuent caying wie of adius R. Hee s what the field looks like...

26 R 1 R 2 R 2 R 1 Tube caying cuent Question 27.9: s thee a magnetic field inside a cuent caying hollow tube, i.e., fo < R 1? f so, in what diection is it? Daw a cicula Ampeian loop adius (centeed on the wie with < R 1 ). Using Ampèe s Law, d l = µ # enc. ut thee is no enclosed cuent so, d l = 0. cannot be cicula, i.e., l to d l. Could be pependicula to d l, i.e., paallel to the axis of the tube? Think of the tube as being made up of individual, small elements each caying a faction of the cuent δi: Tube caying cuent δi

27 δi i i Each element acts like a wie caying pat of the total cuent. t poduces a magnetic field that is concentic, i.e., not down the tube. ut, cancellation fom all the diametically opposite elements poduces zeo net field inside So, thee is no magnetic field inside the tube at all. Note that addition of the individual fields outside the tube poduce a cicula, clockwise set of field lines. = µ 2π ( > R 2). Question 27.10: A coaxial cable is connected so that the instantaneous cuent to a load flows though the inne cental conducto and etuns though the oute coppe baid, as shown. f you could see magnetic field lines, what would they look like outside the cable if the instantaneous cuent in the cental conducto is diected towads you? A: Cicula, clockwise. : Cicula, counte-clockwise. C: Radial, outwads. D: Radial, inwads. E: Thee is no field.

28 i i C i i Daw a closed, cicula Ampeian loop (C) aound the coaxial cable. We know d C l = µ # enc. ut, the net cuent enclosed enc = 0, so d l = 0, i.e., cannot be cicula. Could be adial?... that would make d l = 0. No, because magnetic field lines must be continuous, i.e., loop aound. Theefoe, the only conclusion is that thee is no field outside the coaxial cable. souce i i load Connecting coaxial cables this way ensues that thee is no extenal magnetic field even if the cuent i changes. So, the answe is E. Question 27.11: A coaxial cable is connected so that the instantaneous cuent to a load flows though the cental conducto and etuns though the coppe baid as shown. f you could see magnetic field lines, what would they look like between the conductos when the instantaneous cuent in the cental conducto is diected towads you? A: Cicula, clockwise. : Cicula, counte-clockwise. C: Radial, outwads. D: Radial, inwads. E: Thee is no field.

29 i Magnetic field of a tooidal solenoid: d l i C Loop 1 Daw a closed, cicula loop (C) with adius, aound the cente wie and inside the insulating mateial. Applying Ampèe s Law we get d l = µ # enc. C ( 2π) = µ # i i.e., () = µ #i 2π. This is the same as fo a cuent caying wie, and the diection of the field is given by the ight hand ule, i.e., counte-clockwise. So, the answe is. Note: the oute conducto has absolutely no effect Take thee closed (cicula) loops... Loop 1: Net cuent inside Loop 1 = 0 so = 0 eveywhee inside the ing. Loop 2: y symmety, is a tangent to the path and l to d l. Note that enc = N. d l = 2π = µ enc = µn, 2 whee µ = K m µ # and K m is the elative pemeability of the coe. N tuns of wie Loop 2 i.e., = µn 2π. Loop 3

30 Loop 3 Loop 3: Net cuent inside Loop 3 = 0 (because of cancellation), so = 0 eveywhee outside the ing. Theefoe, in an ideal tooidal solenoid, the field is esticted to the coe of the solenoid and, at adius, is given by: = µn 2π. Note that although the field lines ae concentic, the field vaies acoss the coe; it is lagest whee is smallest. Question 27.12: A tightly wound 1000-tun tooidal solenoid has an inne adius of 1.00 cm and an oute adius of 2.00 cm, and caies a cuent of 1.50 A. The coe of the solenoid is made of soft ion with a elative pemeability K m = (a) What is the magnetic field stength at a distance of 1.10 cm fom the cente of the tooid? (b) What is the magnetic field stength at a distance of 1.50 cm fom the cente of the tooid?

50cm) = 5500 (4π 10 7 )(1000)(1.50 A) 2π(1.50 10 2 m) = 110T. Since magnetic field lines ae continuous we do not lose much flux (lines/unit aea) in a naow gap.

31 How can we get to the field inside the coe?? in gap in The magnetic field in the coe of the solenoid at adius, is given by: = µn 2π, whee µ = 5500µ. (a) Substituting the numeical values (1.10cm) = 5500 (4π 10 7 )(1000)(1.50 A) 2π( m) = 150T. (b) Substituting the numeical values (1.50cm) = 5500 (4π 10 7 )(1000)(1.50 A) 2π( m) = 110T. Since magnetic field lines ae continuous we do not lose much flux (lines/unit aea) in a naow gap. f we wind the coil on a medium with pemeability µ, the field in the gap gap in µn 2π and since fo some mateials µ K m µ " 10 5 µ " we can obtain vey lage magnetic field stengths in the gap. ** This is the basis of an electomagnet **

32 Typical laboatoy electomagnet Cuent caying coils S N Continuous magnetic field cicuit High pemeability mateial

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,