Download presentation

Presentation is loading. Please wait.

Published byCory Garrett Modified over 8 years ago

2
Chapter 20 The Production and Properties of Magnetic Fields

3
Main points Biot–Savart law: Ampère’s law: Gauss’ law for magnetism:

4
What is the source of magnetic fields, if there is no magnetic charge? Answer: electric charge in motion! bar magnet current in wire moving point charge Therefore, understanding source of field generated by bar magnet lies in understanding currents at atomic level within bulk matter.

5
20-1 The Biot-Savart Law The electric field: The magnetic field ： Permeability of Free Space ? the magnetic field of a straight section of wire is given by the Biot–Savart law: It is a defined value

6
The magnitude of the field ： The direction of the field is given by the cross product. P Example ： P

7
Using the Biot-Savart Law a I Find the magnetic field at a point P located a distance a from a straight wire carrying a current I. P Field around a current-carrying wire: we integrated dl over the whole wire ?

8
I a P l O Magnitude: Direction: determined by a right-hand rule

9
Field around a current- carrying wire The magnetic field lines are circles around the wire a right-hand rule

10
(1) For an infinite long wire (2) A very long wire carrying a current I is bent into the shape shown in figure P a I 1 2 Discussion I P The direction of the field is given by the right-hand rule

11
I 1 2 3 R O Example Consider the wire shown in the figure below ， find the magnetic field at a point O. Solution The magnetic field at point O is the sum of three fields They are directed out of page

12
ACTA current I flows in the +y direction in an infinite wire; a current I also flows in the loop as shown in the diagram. – What is F x, the net force on the loop in the x-direction? (a) F x < 0 (b) F x = 0 (c) F x > 0 y x You may have remembered that the net force on a current loop in a uniform B-field is zero, but the B- field produced by an infinite wire is not uniform!

13
Example Two parallel, infinitely long wires, separated by a distance a, carry parallel currents I 1 and I 2, find the force exerted by one long, straight current on another. Solution : The magnetic field of wire 1 at the position of wire 2 is The force on per unit length of wire 2 is Similarly, the force on per unit length of wire 1 is attract

14
(1)The force between two currents is in accordance with Newton’s third law Discussion Currents in the same direction attract Currents in the opposite directions repel

15
(2) Defining the ampere One ampere is the current which, following through two infinitely long, thin, straight wires placed one meter apart in a vacuum, produces a force of newtons on each meter length of wire This definition gives

16
(3) The force between two current element Similarly, The force between two current element doesn’t abide by Newton’s third law.

17
(4) Two infinite line charges of linear charge density 1 and 2 move along the length direction with velocity of v 1 and v 2. Find the forces between the two wires. The current of the two wires is The electrical force on per unit length of the wire is Commonly, the magnetic force is much less than the electrical force. The magnetic force on per unit length of the wire is

18
Example The figure below shows an infinite straight wire and a current loop. Find the torque on the loop. x Solution 1 Use

19
x Solution 2 Use

20
I b P x y O Field around a long current-carrying flat : Break the flat into strips of width dx

21
(1) (2) (3) (1) (2) magnetic shield i i Discussion Infinite Straight Wire Infinite Current Sheet

22
P x R x O Find the magnetic field at a point P located a distance x from the center along the axis of a circular current loop of radius R and current I. P x I The direction of the field is given by the right-hand rule. Field due to a current loop :

23
(1) At the center of the circular loop If the circular loop has N turns Discussion (2) The magnetic field at the center of the arc segment I

24
I (3) A Define Magnetic dipole moment Compare with electric dipole

25
The magnetic field of a magnetic dipole The electric field of an electric dipole

26
The magnetic field lines of a circular loop of a current and a bar magnet

27
Solution Example: A nonconducting disk of radius R has charge q and rotates with angular velocity about its axis, find the magnetic field at a point on the axis of the rotating disk. x O R q P r At the center of the disk

28
Example A hydrogen atom may be described as consisting of an electron that moves in a circular orbit around a proton. The force that gives rise to the motion is the Coulomb attraction between the proton and the electron, which have charges respectively, where.The motion is further constrained by the requirement that the angular momentum has the value where n is an integer and, Planck’s constant. Calculate the magnitude and direction of the magnetic field at the location of the proton. What is the magnetic moment of the current loop?

29
Solution The Coulomb attraction provides the centripetal force The constraint on the angular momentum The magnetic field at the location of the proton Perpendical to the orbit The magnetic moment Perpendical to the orbit

30
I P R A solenoid of radius R has n turns in per unit length, with each turn carrying a current I. l Magnetic field in a solenoid : A solenoid is a length of wire coiled into a cylinder

31
P R l (1) For a long solenoid Discussion (2) If the origin is at one end of the solenoid

32
20-2 Gauss’ Law for Magnetism Electrostatic field ： Magnetic field ： Magnetic flux In the Magnetic field, the number of Magnetic field lines crossing a curved surface S is defined as magnetic flux.

33
For a curved surface If the magnetic field lines enter the surface For a closed surface If the magnetic field lines leave the surface SI: weber (Wb)

34
No magnetic monopoles (single magnetic charge) have ever been observed Magnetic field lines must be continuous The magnetic “charge” inside any closed surface must always be zero. Gauss’s Law For Magnetism

35
The magnetic field lines must be continuous

36
20-3 Ampere’s Law Ampere’s Law Electrostatic field ： conservative field Magnetic field: Take a line integral over the magnetic field along a circular path C Path C circles a current-carrying wire Current I is through the surface whose edge is defined by the path

37
Path C’ consists of C 1 and C 2 Take a line integral over the magnetic field along a closed path C’(C 1 +C 2 ) In this case, there is no current through the surface whose edge is defined by the path

38
In the static magnetic field, the path integral of the total magnetic field around a closed path is equal to times the algebraic sum of the currents threading the path. Ampere’s law Integral around a closed pathCurrent “enclosed” by that path

39
(2) The enclosed current can be positive or negative contribution. Discussion Curl your right hand around the closed path, with the fingers pointing in the direction of integration. A current through the path in the general direction of your outstretched thumb is assigned a plus sign, and a current generally in the opposite direction is assigned a minus sign (1)Ampere’s Law applies to an arbitrary closed path The direction of the closed path is arbitrary

40
(3) Ampere’s law is valid only if the currents are steady. A finite current segment, if For example The integral of the magnetic field around the closed path L is

41
Using Ampere’s law to find the magnetic field Ampere's Law can simplify the calculation if there is symmetry of the current! (and the magnetic field ) Choose a closed path with direction Evaluate line integral in Ampere’s Law (The integral is simple) Calculate the current enclosed by the path Apply Ampere’s Law

42
Example: A long, hollow, cylindrical surface of radius R carries a current I that is uniformly distributed over the surface. Find the magnetic field both outside and inside the cylindrical surface. P L Solution Because of the high degree of symmetry, we know that is tangent to the circle of radius r about the axis and constant in magnitude everywhere on the circle. P

43
ACT For a long cylinder that current is uniformly distributed over the cross-sectional area, find the magnetic field.

44
Example Find the magnetic field of an infinite sheet of current, the current per unit length (along the direction which is perpendicular to the current) is i. Solution

45
If the thickness of the sheet is d j is the current density outside the sheet inside the sheet x

46
Choose the rectangle abcd as the integral path. Example Find the the magnetic field of a long ideal solenoid carrying a current i. Solution The field inside is parallel to the cylinder axis The field outside is insignificant

47
The Magnetic field lines for a real solenoid of finite length. The field is strong and uniform at interior points such as P 1 but relatively weak at external points such as P 2.

48
Solution If the cross-sectional width is much less than the radius of the toroidal solenoid, outside the torus Example Find the magnetic field of a toroidal solenoid The Magnetic field within the torus must be parallel to the walls

49
If the cross-sectional surface is rectangle with width h. Find the magnetic flux in it.

50
Example Figure below shows a cross section of a long cylindrical conductor containing a long cylindrical hole.The axes of the cylinder and the hole are parallel and are a distance d apart. The current is uniformly distributed in the conductor. The current density is Find the magnetic field in the hole. O1O1 O2O2 Solution Regard the cylindrical hole as resulting from the superposition of a complete cylinder (no hole) carrying a current in one direction and a small cylinder carrying a current in the opposite direction, both cylinders having the same current density.

51
O P The magnetic field in a cylinder O1O1 O2O2 Use superposition

52
Ampere’s law Gauss’s Law For Magnetism Gauss equation Stokes equation

53
20-4 The Magnetic Field of Moving Point Charges P + q A —— The magnetic field of a moving charges

54
The Electric and Magnetic Field of Moving Point Charges

55
O a b Example A closed circuit consists of two semicircles of radii a and b that are connected by straight segments as shown in the figure, the linear charge density is (>0) ， and rotates with angular velocity about O. Find the magnetic field at point O. 1 2 3 4 Segment 1 ： Solution Segment 2 ：

56
O a b 1 2 3 4 Segment 3 ： Segment 4 ：

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google