Magnetostatics
Magnetostatics




If charges are moving with constant velocity, a static
magnetic (or magnetostatic) field is produced. Thus,
magnetostatic fields originate from currents (for instance,
direct currents in current-carrying wires).

Most of the equations we have derived for the electric fields
may be readily used to obtain corresponding equations for
magnetic fields if the equivalent analogous quantities are
substituted.




2
Magnetostatics




     Biot – Savart’s Law                    The magnetic field intensity
                                            dH produced at a point P by
                                            the differential current
                                            element Idl is proportional
                                            to the product of Idl and the
                                            sine of the angle α between
                                            the element and the line
                                            joining P to the element and
                                            is inversely proportional to
             Idl sin α                      the square of the distance R
    dH = k
               R2                           between P and the element.
                    1
IN SI units, k = 4π      , so


                                         Idl sin α
                                dH = k
                                          4πR 2

3
Magnetostatics



Using the definition of cross product ( A × B = AB sin θ AB an )
we can represent the previous equation in vector form as
                     I dl × ar I dl × R
               dH =           =              R= R       ar = R
                       4πR 2    4πR 3                            R
The direction of  d H can be determined   by the right-hand rule
or by the right-handed screw rule.

If the position of the field point is specified by r and the
position of the source point by r ′

         I (dl × ar ) I [dl × (r − r ′)]
    dH =            2
                      =             3
         4π r − r ′      4π r − r ′
where er′r is the unit vector directed from the source point to
the field point, and r − r ′ is the distance between these two
points.

4
Magnetostatics




    The field produced by
    a wire can be found
    adding up the fields
    produced by a large
    number of current
    elements placed head
    to tail along the wire
    (the principle of
    superposition).




5
Types of Current Distributions

• Line current density (current) - occurs for
  infinitesimally thin filamentary bodies
  (i.e., wires of negligible diameter).
• Surface current density (current per unit
  width) - occurs when body is perfectly
  conducting.
• Volume current density (current per unit
  cross sectional area) - most general.
               6
Ampere’s Circuital Law in
        Integral Form
• Ampere’s Circuit Law states that the line
  integral of H around a closed path
  proportional to the total current through
  the surface bounding the path




            C
             ∫ H ⋅ d l = I enc

                7
Ampere’s Circuital Law in
      Integral Form (Cont’d)
                   dl       By convention, dS is
                              taken to be in the
              dS           direction defined by the
      S
                           right-hand rule applied
                                     to dl.
                        Since volume current
                         density is the most
I encl = ∫ J ⋅ d s      general, we can write
                           Iencl in this way.
          S
Applying Stokes’s Theorem to
       Ampere’s Law
      ∫ H ⋅ dl = ∫ ∇× H ⋅ d s
      C            S

                 = I encl =    ∫ J ⋅d s
                               S
 ⇒ Because the above must hold for any
   surface S, we must have
                              Differential form
    ∇× H = J
                              of Ampere’s Law
             9
Ampere’s Law in Differential
             Form
• Ampere’s law in differential form implies
  that the B-field is conservative outside of
  regions where current is flowing.




               10
Applications of Ampere’s Law

• Ampere’s law in integral form is an integral
  equation for the unknown magnetic flux
  density resulting from a given current
  distribution.
                                  known
           ∫
           C
             H ⋅ d l = I encl
               unknown


               11
Applications of Ampere’s Law
             (Cont’d)
• In general, solutions to integral equations
  must be obtained using numerical
  techniques.
• However, for certain symmetric current
  distributions closed form solutions to
  Ampere’s law can be obtained.



               12
Applications of Ampere’s Law
             (Cont’d)
• Closed form solution to Ampere’s law
  relies on our ability to construct a suitable
  family of Amperian paths.
• An Amperian path is a closed contour to
  which the magnetic flux density is
  tangential and over which equal to a
  constant value.


                13
Magnetic Flux Density of an Infinite
Line Current Using Ampere’s Law
Consider an infinite line current along the z-
 axis carrying current in the +z-direction:

                      I




               14
Magnetic Flux Density of an Infinite Line
   Current Using Ampere’s Law (Cont’d)

(1) Assume from symmetry and the right-
  hand rule the form of the field

                H = aφ H φ
                    ˆ
(2) Construct a family of Amperian paths
          circles of radius ρ where
                0≤ρ ≤∞
               15
Magnetic Flux Density of an Infinite Line
 Current Using Ampere’s Law (Cont’d)

        y

                               Amperian path

            ρ
                     x
            I

                         I encl = I


                16
Magnetic Flux Density of an Infinite Line
   Current Using Ampere’s Law (Cont’d)

(4) For each Amperian path, evaluate the
  integral

         C
           ∫ H ⋅ d l = Hl length
                              of Amperian
                                  path.
             magnitude of H
              on Amperian
                  path.

                 ∫ H ⋅ d l = Hφ 2π ρ
                 C

                 17
Magnetic Flux Density of an Infinite Line
   Current Using Ampere’s Law (Cont’d)

(5) Solve for B on each Amperian path

                  I encl
               H=
                     l

                       I
            H = aφ
                ˆ
                     2π ρ


              18
Magnetic Flux Density
The magnetic flux density vector is related to the magnetic
                H
field intensity   by the following equation
                   B = µ H,
                                       T (tesla) or Wb/m2
      µ
where    is the permeability of the medium. Except for
ferromagnetic materials ( such as cobalt, nickel, and iron),
                                  µ
most materials have values of very nearly equal to that for
vacuum,          µ o = 4π ⋅ 10 −7
                                   H/m    (henry per meter)
The magnetic flux through a given surface S is given by
                   Ψ = ∫ B ⋅ d s,
                        S            Wb (weber)


19
Law of conservation
  of magnetic flux



           ∫
           s
             B•ds = 0

          Law   of   conservation    of
          magnetic flux or Gauss’s law
          for magnetostatic field
∇•B = 0
∇ × B = µ0 J
    ∇• J = 0
There are no magnetic flow sources, and the
magnetic flux lines always close upon
themselves
Forces due to magnetic field
There are three ways in which force due to magnetic fields can
be experienced

1.Due to moving charge particle in a B field
2.On a current element in an external field
3.Between two current elements




 23
Force on a Moving Charge

    • A moving point charge placed in a
      magnetic field experiences a force
      given by
             Fm = qu × B
                                       The force experienced
                                       by the point charge is
                        v        B     in the direction into the
                  Q
                                       paper.
The force is perpendicular to the direction of charge motion,
and is also perpendicular to B .
Lorentz’s Force equation
  If a point charge is moving in a region where both
  electric and magnetic fields exist, then it experiences a
  total force given by


   Fe = q E                          Fe + Fm = F
   Fm = qu × B

               F = q( E + u × B)
Note: Magnetic force is zero for q moving in the
direction of the magnetic field (sin0=0)
Magnetic Force on a current element
One can tell if a magneto-static field is present because it
exerts forces on moving charges and on currents. If a
   currentIdl                               B
element       is located in a magnetic field , it experiences a
                                __
force
                  Id l ⇔ Q u


        d= l B
        FI ×
          d
                                       Idl
This force, acting on current element , is in a direction
perpendicular to the current element and also perpendicular
  B                       B
to . It is largest when     and the wire are perpendicular.



26
FORCE ON A CURRENT ELEMENT


• The total force exerted on a circuit C
  carrying current I that is immersed in a
  magnetic flux density B is given by


              F = I ∫ dl × B
                    C




               27
Force between two current elements


• Experimental facts:                F21 F12
  – Two parallel wires           ×             ×
    carrying current in         I1             I2
    the same direction
    attract.
  – Two parallel wires    F21                       F12
    carrying current in          ×             •
    the opposite                I1             I2
    directions repel.
               28
Force between two current elements


• Experimental facts:
  – A short current-          F12 = 0
    carrying wire         ×             I2
    oriented             I1
    perpendicular to a
    long current-
    carrying wire
    experiences no
    force.

              29
Force between two current elements


• Experimental facts:
  – The magnitude of the force is inversely
    proportional to the distance squared.
  – The magnitude of the force is proportional to
    the product of the currents carried by the two
    wires.




                30
Force between two current elements


• The force acting on a current element I2 dl2
  by a current element I1 dl1 is given by

                     µ 0 I 2 d l 2 × ( I1d l 1 × aR12 )
                                                 ˆ
       d (d F 12 ) =
                     4π                  2
                                       R12


    Permeability of free space
       µ0 = 4π × 10-7 F/m

                       31
Force between two current elements


• The total force acting on a circuit C2 having
  a current I2 by a circuit C1 having current I1
  is given by
             µ 0 I1 I 2      d l 2 × ( d l 1 × aR12 )
                                               ˆ
      F 12 =
               4π C2 1  ∫ C∫           R122




                  32
Force between two current elements


• The force on C1 due to C2 is equal in
  magnitude but opposite in direction to the
  force on C2 due to C1.

           F 21 = − F 12


               33

Magnetostatics (1)

  • 1.
  • 2.
    Magnetostatics If charges aremoving with constant velocity, a static magnetic (or magnetostatic) field is produced. Thus, magnetostatic fields originate from currents (for instance, direct currents in current-carrying wires). Most of the equations we have derived for the electric fields may be readily used to obtain corresponding equations for magnetic fields if the equivalent analogous quantities are substituted. 2
  • 3.
    Magnetostatics Biot – Savart’s Law The magnetic field intensity dH produced at a point P by the differential current element Idl is proportional to the product of Idl and the sine of the angle α between the element and the line joining P to the element and is inversely proportional to Idl sin α the square of the distance R dH = k R2 between P and the element. 1 IN SI units, k = 4π , so Idl sin α dH = k 4πR 2 3
  • 4.
    Magnetostatics Using the definitionof cross product ( A × B = AB sin θ AB an ) we can represent the previous equation in vector form as I dl × ar I dl × R dH = = R= R ar = R 4πR 2 4πR 3 R The direction of d H can be determined by the right-hand rule or by the right-handed screw rule. If the position of the field point is specified by r and the position of the source point by r ′ I (dl × ar ) I [dl × (r − r ′)] dH = 2 = 3 4π r − r ′ 4π r − r ′ where er′r is the unit vector directed from the source point to the field point, and r − r ′ is the distance between these two points. 4
  • 5.
    Magnetostatics The field produced by a wire can be found adding up the fields produced by a large number of current elements placed head to tail along the wire (the principle of superposition). 5
  • 6.
    Types of CurrentDistributions • Line current density (current) - occurs for infinitesimally thin filamentary bodies (i.e., wires of negligible diameter). • Surface current density (current per unit width) - occurs when body is perfectly conducting. • Volume current density (current per unit cross sectional area) - most general. 6
  • 7.
    Ampere’s Circuital Lawin Integral Form • Ampere’s Circuit Law states that the line integral of H around a closed path proportional to the total current through the surface bounding the path C ∫ H ⋅ d l = I enc 7
  • 8.
    Ampere’s Circuital Lawin Integral Form (Cont’d) dl By convention, dS is taken to be in the dS direction defined by the S right-hand rule applied to dl. Since volume current density is the most I encl = ∫ J ⋅ d s general, we can write Iencl in this way. S
  • 9.
    Applying Stokes’s Theoremto Ampere’s Law ∫ H ⋅ dl = ∫ ∇× H ⋅ d s C S = I encl = ∫ J ⋅d s S ⇒ Because the above must hold for any surface S, we must have Differential form ∇× H = J of Ampere’s Law 9
  • 10.
    Ampere’s Law inDifferential Form • Ampere’s law in differential form implies that the B-field is conservative outside of regions where current is flowing. 10
  • 11.
    Applications of Ampere’sLaw • Ampere’s law in integral form is an integral equation for the unknown magnetic flux density resulting from a given current distribution. known ∫ C H ⋅ d l = I encl unknown 11
  • 12.
    Applications of Ampere’sLaw (Cont’d) • In general, solutions to integral equations must be obtained using numerical techniques. • However, for certain symmetric current distributions closed form solutions to Ampere’s law can be obtained. 12
  • 13.
    Applications of Ampere’sLaw (Cont’d) • Closed form solution to Ampere’s law relies on our ability to construct a suitable family of Amperian paths. • An Amperian path is a closed contour to which the magnetic flux density is tangential and over which equal to a constant value. 13
  • 14.
    Magnetic Flux Densityof an Infinite Line Current Using Ampere’s Law Consider an infinite line current along the z- axis carrying current in the +z-direction: I 14
  • 15.
    Magnetic Flux Densityof an Infinite Line Current Using Ampere’s Law (Cont’d) (1) Assume from symmetry and the right- hand rule the form of the field H = aφ H φ ˆ (2) Construct a family of Amperian paths circles of radius ρ where 0≤ρ ≤∞ 15
  • 16.
    Magnetic Flux Densityof an Infinite Line Current Using Ampere’s Law (Cont’d) y Amperian path ρ x I I encl = I 16
  • 17.
    Magnetic Flux Densityof an Infinite Line Current Using Ampere’s Law (Cont’d) (4) For each Amperian path, evaluate the integral C ∫ H ⋅ d l = Hl length of Amperian path. magnitude of H on Amperian path. ∫ H ⋅ d l = Hφ 2π ρ C 17
  • 18.
    Magnetic Flux Densityof an Infinite Line Current Using Ampere’s Law (Cont’d) (5) Solve for B on each Amperian path I encl H= l I H = aφ ˆ 2π ρ 18
  • 19.
    Magnetic Flux Density Themagnetic flux density vector is related to the magnetic H field intensity by the following equation B = µ H, T (tesla) or Wb/m2 µ where is the permeability of the medium. Except for ferromagnetic materials ( such as cobalt, nickel, and iron), µ most materials have values of very nearly equal to that for vacuum, µ o = 4π ⋅ 10 −7 H/m (henry per meter) The magnetic flux through a given surface S is given by Ψ = ∫ B ⋅ d s, S Wb (weber) 19
  • 20.
    Law of conservation of magnetic flux ∫ s B•ds = 0 Law of conservation of magnetic flux or Gauss’s law for magnetostatic field
  • 21.
    ∇•B = 0 ∇× B = µ0 J ∇• J = 0
  • 22.
    There are nomagnetic flow sources, and the magnetic flux lines always close upon themselves
  • 23.
    Forces due tomagnetic field There are three ways in which force due to magnetic fields can be experienced 1.Due to moving charge particle in a B field 2.On a current element in an external field 3.Between two current elements 23
  • 24.
    Force on aMoving Charge • A moving point charge placed in a magnetic field experiences a force given by Fm = qu × B The force experienced by the point charge is v B in the direction into the Q paper. The force is perpendicular to the direction of charge motion, and is also perpendicular to B .
  • 25.
    Lorentz’s Force equation If a point charge is moving in a region where both electric and magnetic fields exist, then it experiences a total force given by Fe = q E Fe + Fm = F Fm = qu × B F = q( E + u × B) Note: Magnetic force is zero for q moving in the direction of the magnetic field (sin0=0)
  • 26.
    Magnetic Force ona current element One can tell if a magneto-static field is present because it exerts forces on moving charges and on currents. If a currentIdl B element is located in a magnetic field , it experiences a __ force Id l ⇔ Q u d= l B FI × d Idl This force, acting on current element , is in a direction perpendicular to the current element and also perpendicular B B to . It is largest when and the wire are perpendicular. 26
  • 27.
    FORCE ON ACURRENT ELEMENT • The total force exerted on a circuit C carrying current I that is immersed in a magnetic flux density B is given by F = I ∫ dl × B C 27
  • 28.
    Force between twocurrent elements • Experimental facts: F21 F12 – Two parallel wires × × carrying current in I1 I2 the same direction attract. – Two parallel wires F21 F12 carrying current in × • the opposite I1 I2 directions repel. 28
  • 29.
    Force between twocurrent elements • Experimental facts: – A short current- F12 = 0 carrying wire × I2 oriented I1 perpendicular to a long current- carrying wire experiences no force. 29
  • 30.
    Force between twocurrent elements • Experimental facts: – The magnitude of the force is inversely proportional to the distance squared. – The magnitude of the force is proportional to the product of the currents carried by the two wires. 30
  • 31.
    Force between twocurrent elements • The force acting on a current element I2 dl2 by a current element I1 dl1 is given by µ 0 I 2 d l 2 × ( I1d l 1 × aR12 ) ˆ d (d F 12 ) = 4π 2 R12 Permeability of free space µ0 = 4π × 10-7 F/m 31
  • 32.
    Force between twocurrent elements • The total force acting on a circuit C2 having a current I2 by a circuit C1 having current I1 is given by µ 0 I1 I 2 d l 2 × ( d l 1 × aR12 ) ˆ F 12 = 4π C2 1 ∫ C∫ R122 32
  • 33.
    Force between twocurrent elements • The force on C1 due to C2 is equal in magnitude but opposite in direction to the force on C2 due to C1. F 21 = − F 12 33