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Sources of the Magnetic Field

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Sources of the Magnetic Field. Review of Magnetic Fields. FB = q v x B FB = |q| vB sin q ... Moving charged particle in a magnetic field does uniform circular motion ... – PowerPoint PPT presentation

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Title: Sources of the Magnetic Field


1
Chapter 30
  • Sources of the Magnetic Field

2
Review of Magnetic Fields
  • FB q v x B ? FB q vB sin q
  • FB is the magnetic force, q is the charge, v is
    the velocity of the moving charge, B is the
    magnetic field, direction by RHR
  • Force on a wire F I L x B ? dF I ds x B?
  • Torque on a current loop t IA x B ?t m x B.
    Analogous to t p x E for electric dipole
  • Moving charged particle in a magnetic field does
    uniform circular motion
    ?
  • Angular speed and period ?

3
Biot-Savart Law Set-Up
  • The magnetic field is dB at some point P
  • The length element is ds
  • The wire is carrying a steady current of I

4
Biot-Savart Law Equation
  • The observations are summarized in the
    mathematical equation called the Biot-Savart law
  • The magnetic field described by the law is the
    field due to the current-carrying conductor
  • The vector dB is perpendicular to both ds and to
    the unit vector r-hat directed from ds toward P
  • The magnitude of dB is inversely proportional to
    r2, where r is the distance from ds to P
  • The magnitude of dB is proportional to the
    current and to the magnitude ds of the length
    element ds
  • The magnitude of dB is proportional to sin ?,
    where ? is the angle between the vectors ds and

5
Permeability of Free Space
  • The constant mo is called the permeability of
    free space
  • mo 4p x 10-7 T. m / A

6
Total Magnetic Field
  • dB is the field created by the current in the
    length segment ds
  • To find the total field, sum up the contributions
    from all the current elements I ds
  • The integral is over the entire current
    distribution

7
B Compared to E
  • Distance Both are inverse square laws
  • Direction
  • The electric field created by a point charge is
    radial in direction
  • The magnetic field created by a current element
    is perpendicular to both the length element ds
    and the unit vector
  • Source
  • An electric field is established by an isolated
    electric charge
  • The current element that produces a magnetic
    field must be part of an extended current
    distribution
  • Ends
  • Magnetic field lines have no beginning and no end
  • They form continuous circles
  • Electric field lines begin on positive charges
    and end on negative charges

8
B for a Long, Straight Conductor
  • The thin, straight wire is carrying a constant
    current
  • Integrating over all the current elements gives

9
B for a Long, Straight Conductor, Special Case
  • If the conductor is an infinitely long, straight
    wire, q1 0 and q2 p
  • The field becomes

10
B for a Long, Straight Conductor, Direction
  • The magnetic field lines are circles concentric
    with the wire
  • The field lines lie in planes perpendicular to to
    wire
  • The magnitude of B is constant on any circle of
    radius a
  • The right-hand rule for determining the direction
    of B is shown
  • DEMO

11
B for a Curved Wire Segment
  • Find the field at point O due to the wire segment
  • I and R are constants
  • q will be in radians

12
B for a Circular Loop of Wire
  • Consider the previous result, with q 2p
  • This is the field at the center of the loop

13
B for a Circular Current Loop
  • The loop has a radius of R and carries a steady
    current of I. See Ex. 30.3
  • Find B at point P

14
Comparison of Loops
  • Consider the field at the center of the current
    loop
  • At this special point, x 0
  • Then, (note mistake x?R)
  • This is exactly the same result as from the
    circular wire

15
Magnetic Field Lines for a Loop
  • Figure (a) shows the magnetic field lines
    surrounding a current loop
  • Figure (b) shows the field lines in the iron
    filings
  • Figure (c) compares the field lines to that of a
    bar magnet (Conclusion? magnetic materials have
    currents)

16
Magnetic Force Between Two Parallel Conductors
  • Two parallel wires each carry a steady current
  • The field B2 due to the current in wire 2 exerts
    a force on wire 1 of F1 I1l B2

17
Magnetic Force Between Two Parallel Conductors,
cont.
  • Substituting the equation for B2 gives
  • Parallel conductors carrying currents in the same
    direction attract each other
  • Parallel conductors carrying current in opposite
    directions repel each other
  • Definition of Ampere and Coulomb
  • DEMO

18
Amperes Law
  • The product of B . ds can be evaluated for small
    length elements ds on the circular path defined
    by the compass needles for the long straight wire
  • Amperes law states that the line integral of B .
    ds around any closed path equals moI where I is
    the total steady current passing through any
    surface bounded by the closed path.
  • Similar to Gauss Law

19
Amperes Law, cont.
  • Amperes law describes the creation of magnetic
    fields by all continuous current configurations
  • Most useful for this course if the current
    configuration has a high degree of symmetry
  • Put the thumb of your right hand in the direction
    of the current through the amperian loop and your
    fingers curl in the direction you should
    integrate around the loop

20
Field Due to a Long Straight Wire From Amperes
Law
  • Want to 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

21
Field Due to a Long Straight Wire Results From
Amperes Law
  • Outside of the wire, r gt R
  • Inside the wire, we need I, the current inside
    the amperian circle

22
Field Due to a Long Straight Wire Results
Summary
  • 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

23
Magnetic Field of a Toroid
  • Find the field at a point at distance r from the
    center of the toroid
  • The toroid has N turns of wire
  • DEMO

24
Magnetic Field of an Infinite Sheet
  • Assume a thin, infinitely large sheet
  • Carries a current of linear current density Js
  • The current is in the y direction
  • Js represents the current per unit length along
    the z direction

25
Magnetic Field of an Infinite Sheet, cont.
  • Use a rectangular amperian surface
  • The w sides of the rectangle do not contribute to
    the field
  • The two l sides (parallel to the surface)
    contribute to the field.
  • Similar to infinitely charged plates.

26
Magnetic Field of a Solenoid
  • A solenoid is a long wire wound in the form of a
    helix
  • A reasonably uniform magnetic field can be
    produced in the space surrounded by the turns of
    the wire
  • The interior of the solenoid

27
Magnetic Field of a Solenoid, Description
  • The field lines in the interior are
  • approximately parallel to each other
  • uniformly distributed
  • close together
  • This indicates the field is strong and almost
    uniform

28
Magnetic Field of a Tightly Wound Solenoid
  • The field distribution is similar to that of a
    bar magnet
  • As the length of the solenoid increases
  • the interior field becomes more uniform
  • the exterior field becomes weaker
  • DEMO

29
Ideal Solenoid Characteristics
  • An ideal solenoid is approached when
  • the turns are closely spaced
  • the length is much greater than the radius of the
    turns

30
Amperes Law Applied to a Solenoid
  • Amperes law can be used to find the interior
    magnetic field of the solenoid
  • Consider a rectangle with side l parallel to the
    interior field and side w perpendicular to the
    field
  • The side of length l inside the solenoid
    contributes to the field
  • This is path 1 in the diagram

31
Amperes Law Applied to a Solenoid, cont.
  • Applying Amperes Law gives
  • The total current through the rectangular path
    equals the current through each turn multiplied
    by the number of turns

32
Magnetic Field of a Solenoid, final
  • Solving Amperes law for the magnetic field is
  • n N / l is the number of turns per unit length
  • This is valid only at points near the center of a
    very long solenoid

33
Magnetic Flux
  • The magnetic flux associated with a magnetic
    field is defined in a way similar to electric
    flux
  • Consider an area element dA on an arbitrarily
    shaped surface

34
Magnetic Flux, cont.
  • The magnetic field in this element is B
  • dA is a vector that is perpendicular to the
    surface
  • dA has a magnitude equal to the area dA
  • The magnetic flux FB is
  • The unit of magnetic flux is T.m2 Wb
  • Wb is a weber

35
Magnetic Flux Through a Plane, 1
  • A special case is when a plane of area A makes an
    angle q with dA
  • The magnetic flux is FB BA cos q
  • In this case, the field is parallel to the plane
    and F 0

36
Magnetic Flux Through A Plane, 2
  • The magnetic flux is FB BA cos q
  • In this case, the field is perpendicular to the
    plane and
  • F BA
  • This will be the maximum value of the flux

37
Gauss Law in Magnetism
  • Magnetic fields do not begin or end at any point
  • The number of lines entering a surface equals the
    number of lines leaving the surface
  • Gauss law in magnetism says

38
Displacement Current
  • Amperes law in the original form is valid only
    if any electric fields present are constant in
    time
  • Maxwell modified the law to include time-changing
    electric fields
  • Maxwell added an additional term which includes a
    factor called the displacement current, Id

39
Displacement Current, cont.
  • The displacement current is not the current in
    the conductor
  • Conduction current will be used to refer to
    current carried by a wire or other conductor
  • The displacement current is defined as
  • FE òE . dA is the electric flux and eo is the
    permittivity of free space

40
Amperes Law General Form
  • Also known as the Ampere-Maxwell law

41
Amperes Law General Form, Example
  • The electric flux through S2 is EA
  • A is the area of the capacitor plates
  • E is the electric field between the plates
  • If q is the charge on the plate at any time, FE
    EA q/eo

42
Amperes Law General Form, Example, cont.
  • Therefore, the displacement current is
  • The displacement current is the same as the
    conduction current through S1
  • The displacement current on S2 is the source of
    the magnetic field on the surface boundary

43
Ampere-Maxwell Law, final
  • Magnetic fields are produced both by conduction
    currents and by time-varying electric fields
  • This theoretical work by Maxwell contributed to
    major advances in the understanding of
    electromagnetism
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