18. Ampere s law and Gauss s law (for B) Announcements: This Friday, Quiz 1 in-class and during class (training exam)

Transcription

1 18. Ampere s law and Gauss s law (for B) Announcements: This Friday, Quiz 1 in-class and during class (training exam)

2 Where does a B-field come from?

3 Facts: Electrical current produces a magnetic field Current wiki B

4 Demo of current (>100A) through wire or beer can creating a magnetic field

5 Facts: Electrical current segment produces a magnetic field

6 Facts: Electrical current produces a magnetic field A moving charge produces a magnetic field

7 Facts: Electrical current produces a magnetic field A moving charge produces a magnetic field Moving in an electric field produces a magnetic field

8 Facts: Electrical current produces a magnetic field A moving charge produces a magnetic field Moving in an electric field produces a magnetic field Moving speed Electric field B Ԧv E c 2 Magnetic field Speed of light GIPHY

9 Facts: Electrical current produces a magnetic field A moving charge produces a magnetic field Moving in an electric field produces a magnetic field Moving speed Electric field B Ԧv E c 2 Magnetic field Speed of light This effect is not considered in this class! (very small contribution at typical moving speeds) GIPHY

10 Two methods (theories) to find the magnetic field due to a current: 1. Biot-Savart law (magnetic equivalent to Coulomb) 2. Ampere s law (magnetic equivalent to Gauss)

11 Two methods (theories) to find the magnetic field due to a current: 1. Biot-Savart law B = μ oi 4π න d Ԧs r r 2 dԧs is the line element of the integral which is taken along the current I 2. Ampere s law රB dԧs = μ o I encl dԧs is the line element of the integral which is taken along any closed loop around current I encl The constant m o is called the permeability of free space. m o = 4p x 10-7 T. m / A

12 Two methods (theories) to find the magnetic field due to a current (if current line does not move): 1. Biot-Savart law (explicit expression of B) B = μ oi 4π න d Ԧs r r 2 dԧs is the line element of the integral which is taken along the current I 2. Ampere s law (implicit expression of B) රB dԧs = μ o I encl dԧs is the line element of the integral which is taken along any closed loop around current I encl The constant m o is called the permeability of free space. m o = 4p x 10-7 T. m / A

13 Magnetic Field for a Long, Straight Conductor (using Biot-Savart) Find the field contribution from a small element of current and then integrate over the current distribution. The thin, straight wire is carrying a constant current ( sin ) ds ˆr = dx θ Integrating over all the current elements gives μ θ2 o I B = θ dθ 4πa cos θ1 μo I = 4πa kˆ ( sinθ sinθ ) 1 2 B = μ oi 4π න d Ԧs r r 2 (See derivation in example 30.1 in the textbook) Section 30.1

14 Magnetic Field for a Long, Straight Conductor (using Biot-Savart) If the conductor is an infinitely long, straight wire, q 1 = p/2 and q 2 = -p/2 The field becomes B = μ o I 2πa B = μ oi 4π න d Ԧs r r 2 a: distance from wire Section 30.1

15 Magnetic Field for a Long, Straight Conductor (using Ampere) Calculate the magnetic field at a distance r from the center of a wire carrying a steady current I. The current is uniformly distributed through the cross section of the wire. Since the wire has a high degree of symmetry, the problem can be categorized as a Ampère s Law problem. For r R, this should be the same result as obtained from the Biot-Savart Law. B රB dԧs = μ o I encl Section 30.3

16 Magnetic Field for a Long, Straight Conductor (using Ampere) (1) Outside of the wire, r > R μo B d s I = B( 2πr ) = μo I B = 2πr රB dԧs = μ o I encl Section 30.3

17 Magnetic Field for a Long, Straight Conductor (using Ampere) (1) Outside of the wire, r > R μo B d s I = B( 2πr ) = μo I B = 2πr රB dԧs = μ o I encl (2) Inside the wire, we need I encl, the current inside the amperian circle. රB d Ԧs = B 2πr = μ o I encl I encl = r2 R 2 I B = μ oi 2πR 2 r Section 30.3

18 Magnetic Field for a Long, Straight Conductor (using Ampere) The field is proportional to r inside the wire. The field varies as 1/r outside the wire. Both equations are equal at r = R. Section 30.3

19 Magnetic Field of a Wire The circular magnetic field around the wire is shown by the iron filings. Section 30.3

20 Magnetic Field of a Wire A compass can be used to detect the magnetic field. When there is no current in the wire, there is no field due to the current. The compass needles all point toward the Earth s north pole. Due to the Earth s magnetic field Section 30.3

21 From above, with a current pointing upwards, the magnetic field turns A. Clockwise B. Anti-clockwise C. Doesn t turn

22 Magnetic Field of a Wire Here the wire carries a strong current. The compass needles deflect in a direction tangent to the circle. This shows the direction of the magnetic field produced by the wire. If the current is reversed, the direction of the needles also reverse. Section 30.3

23 Right Hand Rules:

24 (1) Magnetic force: q Ԧv B = ԦF Right hand; positive charge v: Towards velocity (Thumbs) Magnetic field (Middle finger) = Force (slap) B Ԧv ԦF

25 (1b) Magnetic force: I Ԧl B = ԦF Right hand; positive charge I: Towards current (Thumbs) Magnetic field (Middle finger) = Force (slap) B Ԧl ԦF

26 (2) Linear current: B = μ 0I 2πr I B

27 (3) circular current: (solenoid) B = μ 0IN L B B I

28 (1) Magnetic force: (2) Linear current: q Ԧv B = ԦF Ԧv I (3) circular current: (solenoid) B B ԦF B I

29 Applying Ampere s law to a solenoid: රB dԧs = μ o I encl

30 Applying Ampere s law to a solenoid: රB dԧs = μ o I encl Chose an integration rectangle c d b a

31 Applying Ampere s law to a solenoid: රB dԧs = μ o I encl Chose an integration rectangle c d b a

32 Applying Ampere s law to a solenoid: රB dԧs = μ o I encl Chose an integration rectangle c d b a b c d a ර B ds = න B ds + න B ds + න B ds + න B ds a b c d

33 Applying Ampere s law to a solenoid: රB dԧs = μ o I encl Chose an integration rectangle c d b a b c d a ර B ds = න B ds + න B ds + න B ds + න B ds a b c BL d 0 න B ds b c

34 Applying Ampere s law to a solenoid: රB dԧs = μ o I encl Chose an integration rectangle c d b a b c d a ර B ds = න B ds + න B ds + න B ds + න B ds = BL a b c BL d 0 න B ds b c

35 Applying Ampere s law to a solenoid: රB dԧs = μ o I encl Chose an integration rectangle c d b a ර B ds = BL = μ 0 I encl B = μ 0 I N L Field inside a solenoid

36 In which direction does the field point to inside the solenoid? A. Left B. Right C. Up D. Down

37 Applying Ampere s law to a solenoid: රB dԧs = μ o I encl ර B ds = BL = μ 0 I encl B = μ 0 I N L Field inside a solenoid

38 In which direction does the field point to outside the solenoid? A. Left B. Right C. Up D. Down

39 Applying Ampere s law to a solenoid: රB dԧs = μ o I encl

40 Electric field Magnetic field gfycat Electric field lines Magnetic field lines: always closed

41 Electric field Electric charge Magnetic field No magnetic charge (monopole) gfycat Electric field lines Magnetic field lines: always closed

42 Electric field Electric charge Magnetic field No magnetic charge (monopole) Gauss s law ර E da = q encl (1) රB da = 0 ε 0 (2) gfycat Electric field lines Magnetic field lines: always closed Gauss s law of magnetism

43 Gauss s law Electric field Magnetic field Electric charge No magnetic charge (monopole) ර E da = q encl (1) රB da = 0 ε 0 (2) Flux= B = B d A Units: Wb = T m 2 gfycat Electric field lines Magnetic field lines: always closed Gauss s law of magnetism

44 Imagine walking and swimming all over the earth and measuring the vertical magnetic field every square meter and adding up the result. What do you get Magnetic field on the surface of earth Swarm data A. A huge positive number B. A huge negative number C. Zero

45

46 Ampere s discovery: Two wires with current exert a force on each other! (Ampere s force law) wiki

47 I 1 I 2 I 1 produces B 1 I 2 produces B 2 B 1 = m 0I 1 2pr B 2 = m 0I 2 2pr

48 I 1 I 2 I 1 produces B 1 I 2 produces B 2 I 1 experiences a force due to B 2 F 1 = I 1 LB 2 F 2 = I 2 LB 1 I 2 experiences a force due to B 1 Total field: B = B 1 +B 2 B 1 = m 0I 1 2pr B 2 = m 0I 2 2pr

49 I 1 I 2 F 1 = I 1 LB 2 F 2 = I 2 LB 1 F 1 = F 2 = m 0 I 1 I 2 L 2pr μ 0 = 4π 10 7 T m/a => typically, fields produced by currents are small B 1 = m 0I 1 2pr B 2 = m 0I 2 2pr

50 I 1 I 2 F 2 = I 2 LB 1 F 1 = I 1 LB 2 F 1 = F 2 = m 0 I 1 I 2 L 2pr μ 0 = 4π 10 7 T m/a => typically, fields produced by currents are small B 1 = m 0I 1 2pr B 2 = m 0I 2 2pr