14. Magnetic Field III

Transcription

1 University of Rhode sland PHY 204: Elementary Physics Physics Course Materials Magnetic Field Gerhard Müller University of Rhode sland, Creative Commons License This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License. Follow this and additional works at: Abstract Part fourteen of course materials for Elementary Physics (PHY 204), taught by Gerhard Müller at the University of Rhode sland. Documents will be updated periodically as more entries become presentable. Some of the slides contain figures from the textbook, Paul A. Tipler and Gene Mosca. Physics for Scientists and Engineers, 5 th /6 th editions. The copyright to these figures is owned by W.H. Freeman. We acknowledge permission from W.H. Freeman to use them on this course web page. The textbook figures are not to be used or copied for any purpose outside this class without direct permission from W.H. Freeman. Recommended Citation Müller, Gerhard, "14. Magnetic Field " (2015). PHY 204: Elementary Physics. Paper This Course Material is brought to you for free and open access by the Physics Course Materials at DigitalCommons@UR. t has been accepted for inclusion in PHY 204: Elementary Physics by an authorized administrator of DigitalCommons@UR. For more information, please contact digitalcommons@etal.uri.edu.

2 Electric Field of a Point Charge electric charge generates electric field exerts force locally electric charge exerts force over distance (1) Electric field E generated by point charge q: E = k q r 2 ˆr (2) Force F 1 exerted by field E on point charge q 1 : F1 = q 1 E (1+2) Force F 1 exerted by charge q on charge q 1 : F 1 = k qq 1 r 2 ˆr (static conditions) ǫ 0 = C 2 N 1 m 2 k = 1 4πǫ 0 = Nm 2 C 2 q + source point r r field point ^r field point E ^r E 27/10/2015 [tsl14 1/33]

3 Magnetic Field of a Moving Point Charge moving electric charge generates magnetic field exerts force locally moving electric charge exerts force over distance (1) Magnetic field generated by point charge q: = µ 0 4π q v ˆr (2) Force F 1 exerted by field on point charge q 1 : F1 = q 1 v 1 (1+2) There is a time delay between causally related events over distance. r 2 field point Permeability constant µ 0 = 4π 10 7 Tm/A q r^ v + source point r 27/10/2015 [tsl210 2/33]

4 Magnetic Field Application (1) A particle with charge q = 4.5nC is moving with velocity v = m/sî. Find the magnetic field generated at the origin of the coordinate system. Position of field point relative to particle: r = 4mî 3mĵ Distance between Particle and field point: r = p (4m) 2 + (3m) 2 = 5m Magnetic field: = µ 0 4π = µ 0 4π = µ 0 4π q v ˆr r 2 = µ 0 4π q v r r 3 q( m/sî) (4mî 3mĵ) (5m) 3 q( m/sî) (3mĵ) (5m) 3 = Tˆk. 27/10/2015 [tsl212 3/33]

5 Law of iot and Savart Current element: d s = dq v [1Am = 1Cm/s] Magnetic field of current element: d = µ 0 dqv sin θ 4π r 2 = µ 0 ds sin θ 4π r 2 Vector relation: d = µ 0 d s ˆr 4π r 2 Magnetic field generated by current of arbitrary shape: = µ 0 4π Z d s ˆr r 2 (Law of iot and Savart) d r r^ θ ds 27/10/2015 [tsl213 4/33]

6 Magnetic Field of Circular Current Law of iot and Savart: d = µ 0 4π d z = d sin θ = d d z = µ 0 4π z = µ 0 4π z = µ 0 2 R z 2 + R 2 Rdl (z 2 + R 2 ) 3/2 R (z 2 + R 2 ) 3/2 R 2 Z 2πR 0 (z 2 + R 2 ) 3/2 dl dl z 2 + R 2 Field at center of ring (z = 0): z = µ 0 2R Magnetic moment: µ = πr 2 Field at large distance (z R) : z µ 0 2π µ z 3 27/10/2015 [tsl214 5/33]

7 Magnetic Field Application (11) The electric field E x along the axis of a charged ring and the magnetic field x along the axis of a circular current loop are E x = Q 4πǫ 0 x (x 2 + R 2 ) 3/2, x = µ 0 2 R 2 (x 2 + R 2 ) 3/2 (a) Simplify both expressions for x = 0. (b) Simplify both expressions for x R. (c) Sketch graphs of E x (x) and x (x). E x x x x 27/10/2015 [tsl228 6/33]

8 Magnetic Field on the Axis of a Solenoid Number of turns per unit length: n = N/L Current circulating in ring of width dx : ndx Magnetic field on axis of ring: d x = µ 0(ndx ) 2 Magnetic field on axis of solenoid: x = µ Z x2 0n dx 2 R2 x 1 [(x x ) 2 + R 2 ] 3/2 = µ 0n 2 R 2 [(x x ) 2 + R 2 ] 3/2! x x 1 p (x x1 ) 2 + R x x 2 p 2 (x x2 ) 2 + R 2 27/10/2015 [tsl215 7/33]

9 Magnetic Field Generated by Current in Straight Wire (1) Consider a field point P that is a distance R from the axis of the wire. d = µ 0 4π dx r 2 sin φ = µ 0 4π x = R tan θ dx dθ = dx r 2 cos θ R cos 2 θ = R R 2 /r 2 = r2 R d = µ 0 r 2 dθ 4π r 2 R cos θ = µ 0 cos θdθ 4π R = µ 0 4π = µ 0 4π Z θ2 cos θdθ R θ 1 R (sin θ 2 sin θ 1 ) Length of wire: L = R(tan θ 2 tan θ 1 ) Wire of infinite length: θ 1 = 90, θ 2 = 90 = µ 0 2πR 27/10/2015 [tsl216 8/33]

10 Magnetic Field Generated by Current in Straight Wire (2) Consider a current in a straight wire of infinite length. The magnetic field lines are concentric circles in planes prependicular to the wire. The magnitude of the magnetic field at distance R from the center of the wire is = µ 0 2πR. The magnetic field strength is proportional to the current and inversely proportional to the distance R from the center of the wire. The magnetic field vector is tangential to the circular field lines and directed according to the right-hand rule. 27/10/2015 [tsl217 9/33]

11 Magnetic Field Generated by Current in Straight Wire (3) Consider the magnetic field in the limit R 0. = µ 0 4π sin θ 1 = sin θ 2 = µ 0 4π R (sin θ 2 sin θ 1 ) a a 2 + R = 1 q R2 a 2 2a 4a 2 + R = 1 q 2 R = µ 0 4π R 2 4a R2 4a 2 1 R 2 a 2 «R 2 a R 2 4a 2 3R R 0 8a 2 0 a R a θ 2 θ 1 a 27/10/2015 [tsl380 10/33]

12 Magnetic Field Application (2) The currents 1, 2 in two long straight wires have equal magnitude and generate a magnetic field as shown at three points in space. Find the directions ( J, N ) for 1, 2 in configurations (a) and (b). (a) (b) /10/2015 [tsl218 11/33]

13 Magnetic Field at Center of Square-Shaped Wire Consider a current-carrying wire bent into the shape of a square with side 2a. Find direction and magnitude of the magnetic field generated at the center of the square. a Ι a a a Ι Ι a 45 o 45 o a a Ι a = 4 µ 0 4π h a i sin(45 ) sin( 45 ) = 2µ0 πa. 27/10/2015 [tsl518 12/33]

14 Magnetic Field Application (5) f the current in (a) generates a magnetic field 0 = 1T pointing out of the plane find magnitude and direction of the fields 1, 2, 3 generated by in (b), find magnitude and direction of the fields 4, 5, 6 generated by in (c). (a) (b) (c) 0 =1T /10/2015 [tsl221 13/33]

15 Magnetic Field Application (6) A current-carrying wire is bent into two semi-infinite straight segments at right angles. (a) Find the direction ( J, N ) of the magnetic fields 1,..., 6. (b) Name the strongest and the weakest fields among them. (c) Name all pairs of fields that have equal strength L 6 L /10/2015 [tsl222 14/33]

16 Magnetic Field Application (15) A current-carrying wire is bent into two straight segments of length L at right angles. (a) Find the direction ( J, N ) of the magnetic fields 1,..., 6. (b) Name the strongest and the weakest fields among them. (c) Name all pairs of fields that have equal strength L 6 L /10/2015 [tsl245 15/33]

17 Magnetic Field Next to Current-Carrying Ribbon Consider a very long ribbon of width w carrying a current in the direction shown. The current density is assumed to be uniform. Find the magnetic field generated a distance d from the ribbon as shown. 0 d dx w x Divide the ribbon into thin strips of width dx. Treat each strip as a wire with current d = dx/w. Sum up the field contributions from parallel wires. = µ 0 2πw d = µ 0 2π Z d+w d d x = µ 0 2πw dx x dx x = µ 0 1 2πw ln + w d 27/10/2015 [tsl519 16/33]

18 Force etween Parallel Lines of Electric Charge Electric charge densities: λ a, λ b Electric field generated by line a: E a = 1 λ a 2πǫ 0 d Electric force on segment of line b: F ab = λ b LE a Electric force per unit length (repulsive): F ab L = 1 λ a λ b 2πǫ 0 d d L F ab E a λ a > 0 λ b > 0 27/10/2015 [tsl231 17/33]

19 Force etween Parallel Lines of Electric Current Electric currents: a, b Magnetic field generated by line a: a = µ 0 a 2π d Magnetic force on segment of line b: F ab = b L a Magnetic force per unit length (attractive): F ab L = µ 0 2π a b d d a b F ab a L 27/10/2015 [tsl232 18/33]

20 Force etween Perpendicular Lines of Electric Current Electric currents: a, b Magnetic field generated by line a: a = µ 0 a 2π r Magnetic force on segment dr of line b: df ab = b a dr Magnetic force on line b: F ab = µ 0 2π a b Z r2 r 1 dr r = µ 0 2π a b ln r 2 r 1 r 1 r 2 dr a F b ab a 27/10/2015 [tsl233 19/33]

21 s There Absolute Motion? Forces between two long, parallel, charged rods λ > 0 1 λ > 0 1 = v λ 1 1 d v F λ > λ > 0 2 = v λ 2 at rest F E in uniform motion F E F E L = 1 λ 1 λ 2 2πǫ 0 d F E F L c = = 1 λ 1 λ 2 2πǫ 0 d F E 1 ǫ0 µ 0 = ms 1 (left), L = 1 λ 1 λ 2 2πǫ 0 d, F L = µ 0 2π «1 v2 c 2 = 1 λ 1 λ 2 2πǫ 0 d (speed of light) 1 2 d, (right) λ 1 = λ 1 p 1 v 2 /c 2, λ 2 = λ 2 p 1 v 2 /c 2 (due to length contraction) 27/10/2015 [tsl211 20/33]

22 Magnetic Field Application (4) An electric current flows through the wire as indicated by arrows. (a) Find the direction ( J, N ) of the magnetic field generated by the current at the points 1,..., 5. (b) At which points do we observe the strongest and weakest magnetic fields? /10/2015 [tsl220 21/33]

23 Magnetic Field Application (12) Consider two infinitely long straight currents 1 and 2 as shown. Find the components x and y of the magnetic field at the origin of the coordinate system. y 2 = 4A 3m 1 = 2A x 4m 27/10/2015 [tsl234 22/33]

24 Magnetic Field Application (13) Two straight electric currents 1 and 2 of infinite length directed perpendicular to the xy-plane generate a magnetic field of magnitude = T in the direction shown. Find the magnitude and direction (, ) of each current. y 2 3m 55 o 1 x 4m 27/10/2015 [tsl235 23/33]

25 Magnetic Field Application (3) Two semi-infinite straight wires are connected to a segment of circular wire in three different ways. A current flows in the direction indicated. (a) Find the direction ( J, N ) of the magnetic fields 1, 1, 3. (b) Rank the magnetic fields according to strength. L L L L 1 L 2 L 3 27/10/2015 [tsl219 24/33]

26 Magnetic Field Application (8) Three squares with equal clockwise currents are placed in the magnetic field of a straight wire with a current flowing to the right. Find the direction (,, zero) of the magnetic force acting on each square. (1) (2) (3) 27/10/2015 [tsl225 25/33]

27 Magnetic Field Application (10) Consider two currents of equal magnitude in straight wires flowing perpendicular to the plane. n configurations (a) and (b), find the direction (,,, ) of the magnetic field generated by the two currents at points P, Q, R, S P (a) P (b) R S R S Q Q 27/10/2015 [tsl227 26/33]

28 Magnetic Field Application (9) Two wires of infinite length contain concentric semicircular segments of radii 1m and 2m, respectively. f one of the wires carries a 6A current in the direction indicated, what must be the direction (, ) and magnitude of the current in the other wire such that the magnetic field at the center of the semicircles vanishes? (a) (b) 6A 6A 27/10/2015 [tsl226 27/33]

29 Magnetic Field Application (14) Consider two pairs of rectangular electric currents flowing in the directions indicated. (a) What is the direction (, ) of the magnetic force experienced by the black rectangle in each case? (b) Which black rectangle experiences the stronger magnetic force? (1) (2) 27/10/2015 [tsl238 28/33]

30 ntermediate Exam : Problem #1 (Spring 05) An infinitely long straight current of magnitude = 6A is directed into the plane ( ) and located a distance d = 0.4m from the coordinate origin (somewhere on the dashed circle). The magnetic field generated by this current is in the negative y-direction as shown. (a) Find the magnitude of the magnetic field. (b) Mark the location of the position of the current on the dashed circle. y 0.4m x 27/10/2015 [tsl342 29/33]

31 ntermediate Exam : Problem #1 (Spring 05) An infinitely long straight current of magnitude = 6A is directed into the plane ( ) and located a distance d = 0.4m from the coordinate origin (somewhere on the dashed circle). The magnetic field generated by this current is in the negative y-direction as shown. (a) Find the magnitude of the magnetic field. (b) Mark the location of the position of the current on the dashed circle. y Solution: (a) = µ 0 2π d = 3µT. 0.4m x 27/10/2015 [tsl342 29/33]

32 ntermediate Exam : Problem #1 (Spring 05) An infinitely long straight current of magnitude = 6A is directed into the plane ( ) and located a distance d = 0.4m from the coordinate origin (somewhere on the dashed circle). The magnetic field generated by this current is in the negative y-direction as shown. (a) Find the magnitude of the magnetic field. (b) Mark the location of the position of the current on the dashed circle. y Solution: (a) = µ 0 2π d = 3µT. 0.4m x (b) Position of current is at y = 0, x = 0.4m. 27/10/2015 [tsl342 29/33]

33 ntermediate Exam : Problem #1 (Spring 06) Consider two infinitely long, straight wires with currents of equal magnitude 1 = 2 = 5A in the directions shown. Find the direction (in/out) and the magnitude of the magnetic fields 1 and 2 at the points marked in the graph. a 1 2m 2m 2 2m 2m b 27/10/2015 [tsl355 30/33]

34 ntermediate Exam : Problem #1 (Spring 06) Consider two infinitely long, straight wires with currents of equal magnitude 1 = 2 = 5A in the directions shown. Find the direction (in/out) and the magnitude of the magnetic fields 1 and 2 at the points marked in the graph. a 1 2m 2m Solution: 2 1 = µ 0 2π 5A 4m 5A «4m = 0 (no direction). 2m 2m b 27/10/2015 [tsl355 30/33]

35 ntermediate Exam : Problem #1 (Spring 06) Consider two infinitely long, straight wires with currents of equal magnitude 1 = 2 = 5A in the directions shown. Find the direction (in/out) and the magnitude of the magnetic fields 1 and 2 at the points marked in the graph. a 1 2m Solution: 2 1 = µ 0 2π 2 = µ 0 2π 5A 4m 5A «4m 5A 2m 5A «4m = 0 (no direction). = 0.25µT (out of plane). 2m 2m 2m b 27/10/2015 [tsl355 30/33]

36 ntermediate Exam : Problem #2 (Spring 07) Consider two very long, straight wires with currents 1 = 6A at x = 1m and 3 = 3A at x = 3m in the directions shown. Find magnitude and direction (up/down) of the magnetic field (a) 0 at x = 0, = 6A 1 = 3A 3 (b) 2 at x = 2m, (c) 4 at x = 4m x [m] 27/10/2015 [tsl366 31/33]

37 ntermediate Exam : Problem #2 (Spring 07) Consider two very long, straight wires with currents 1 = 6A at x = 1m and 3 = 3A at x = 3m in the directions shown. Find magnitude and direction (up/down) of the magnetic field (a) 0 at x = 0, = 6A 1 = 3A 3 (b) 2 at x = 2m, (c) 4 at x = 4m. Solution: x [m] (a) 0 = µ 0(6A) 2π(1m) + µ 0(3A) 2π(3m) = 1.2µT + 0.2µT = 1.0µT (down), 27/10/2015 [tsl366 31/33]

38 ntermediate Exam : Problem #2 (Spring 07) Consider two very long, straight wires with currents 1 = 6A at x = 1m and 3 = 3A at x = 3m in the directions shown. Find magnitude and direction (up/down) of the magnetic field (a) 0 at x = 0, = 6A 1 = 3A 3 (b) 2 at x = 2m, (c) 4 at x = 4m. Solution: x [m] (a) 0 = µ 0(6A) 2π(1m) + µ 0(3A) 2π(3m) (b) 2 = µ 0(6A) 2π(1m) + µ 0(3A) 2π(1m) = 1.2µT + 0.2µT = 1.0µT (down), = 1.2µT + 0.6µT = 1.8µT (up), 27/10/2015 [tsl366 31/33]

39 ntermediate Exam : Problem #2 (Spring 07) Consider two very long, straight wires with currents 1 = 6A at x = 1m and 3 = 3A at x = 3m in the directions shown. Find magnitude and direction (up/down) of the magnetic field (a) 0 at x = 0, = 6A 1 = 3A 3 (b) 2 at x = 2m, (c) 4 at x = 4m. Solution: x [m] (a) 0 = µ 0(6A) 2π(1m) + µ 0(3A) 2π(3m) (b) 2 = µ 0(6A) 2π(1m) + µ 0(3A) 2π(1m) (c) 4 = µ 0(6A) 2π(3m) µ 0(3A) 2π(1m) = 1.2µT + 0.2µT = 1.0µT (down), = 1.2µT + 0.6µT = 1.8µT (up), = 0.4µT 0.6µT = 0.2µT (down). 27/10/2015 [tsl366 31/33]

40 Unit Exam : Problem #1 (Spring 08) Consider two circular currents 1 = 3A at radius r 1 = 2m and 2 = 5A at radius r 2 = 4m in the directions shown. (a) Find magnitude and direction (, ) of the resultant magnetic field at the center. (b) Find magnitude µ and direction (, ) of the magnetic dipole moment generated by the two currents. 2 1 r 2 r 1 27/10/2015 [tsl381 32/33]

41 Unit Exam : Problem #1 (Spring 08) Consider two circular currents 1 = 3A at radius r 1 = 2m and 2 = 5A at radius r 2 = 4m in the directions shown. (a) Find magnitude and direction (, ) of the resultant magnetic field at the center. (b) Find magnitude µ and direction (, ) of the magnetic dipole moment generated by the two currents. Solution: (a) = µ 0(3A) 2(2m) µ 0(5A) 2(4m) = ( ) 10 7 T = T r 2 r /10/2015 [tsl381 32/33]

42 Unit Exam : Problem #1 (Spring 08) Consider two circular currents 1 = 3A at radius r 1 = 2m and 2 = 5A at radius r 2 = 4m in the directions shown. (a) Find magnitude and direction (, ) of the resultant magnetic field at the center. (b) Find magnitude µ and direction (, ) of the magnetic dipole moment generated by the two currents. Solution: (a) = µ 0(3A) 2(2m) µ 0(5A) 2(4m) = ( ) 10 7 T = T (b) µ = π(4m) 2 (5A) π(2m) 2 (3A) = (251 38)Am 2 µ = 213Am 2 r 2 r /10/2015 [tsl381 32/33]

43 Unit Exam : Problem #2 (Spring 09) Two semi-infinite straight wires are connected to a curved wire in the form of a full circle, quarter circle, or half circle of radius R = 1m in four different configurations. A current = 1A flows in the directions shown. Find magnitude a, b, c, d and direction ( / ) of the magnetic field thus generated at the points a, b, c, d. a c b d 27/10/2015 [tsl396 33/33]

44 Unit Exam : Problem #2 (Spring 09) Two semi-infinite straight wires are connected to a curved wire in the form of a full circle, quarter circle, or half circle of radius R = 1m in four different configurations. A current = 1A flows in the directions shown. Find magnitude a, b, c, d and direction ( / ) of the magnetic field thus generated at the points a, b, c, d. a c Solution: a = µ 0 4πR + µ 0 2R + µ 0 4πR = 100nT + 628nT + 100nT = 828nT b d 27/10/2015 [tsl396 33/33]

45 Unit Exam : Problem #2 (Spring 09) Two semi-infinite straight wires are connected to a curved wire in the form of a full circle, quarter circle, or half circle of radius R = 1m in four different configurations. A current = 1A flows in the directions shown. Find magnitude a, b, c, d and direction ( / ) of the magnetic field thus generated at the points a, b, c, d. a c Solution: b a = µ 0 4πR + µ 0 2R + µ 0 4πR = 100nT + 628nT + 100nT = 828nT b = µ 0 4πR + µ 0 4R µ 0 4πR = 100nT + 314nT 100nT = 314nT d 27/10/2015 [tsl396 33/33]

46 Unit Exam : Problem #2 (Spring 09) Two semi-infinite straight wires are connected to a curved wire in the form of a full circle, quarter circle, or half circle of radius R = 1m in four different configurations. A current = 1A flows in the directions shown. Find magnitude a, b, c, d and direction ( / ) of the magnetic field thus generated at the points a, b, c, d. a c Solution: b a = µ 0 4πR + µ 0 2R + µ 0 4πR = 100nT + 628nT + 100nT = 828nT b = µ 0 4πR + µ 0 4R µ 0 4πR = 100nT + 314nT 100nT = 314nT c = µ 0 4πR + µ 0 8R + 0 = 100nT + 157nT = 257nT d 27/10/2015 [tsl396 33/33]

47 Unit Exam : Problem #2 (Spring 09) Two semi-infinite straight wires are connected to a curved wire in the form of a full circle, quarter circle, or half circle of radius R = 1m in four different configurations. A current = 1A flows in the directions shown. Find magnitude a, b, c, d and direction ( / ) of the magnetic field thus generated at the points a, b, c, d. a c Solution: a = µ 0 4πR + µ 0 2R + µ 0 4πR = 100nT + 628nT + 100nT = 828nT b = µ 0 4πR + µ 0 4R µ 0 4πR = 100nT + 314nT 100nT = 314nT c = µ 0 4πR + µ 0 8R + 0 = 100nT + 157nT = 257nT d = µ 0 4πR µ 0 2R + µ 0 4πR = 100nT 628nT + 100nT = 428nT b d 27/10/2015 [tsl396 33/33]