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Magnetostatics (1)


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Magnetostatics (1)

  1. 1. Magnetostatics
  2. 2. MagnetostaticsIf charges are moving with constant velocity, a staticmagnetic (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 fieldsmay be readily used to obtain corresponding equations formagnetic fields if the equivalent analogous quantities aresubstituted.2
  3. 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. 1IN SI units, k = 4π , so Idl sin α dH = k 4πR 23
  4. 4. MagnetostaticsUsing 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 RThe direction of d H can be determined by the right-hand ruleor by the right-handed screw rule.If the position of the field point is specified by r and theposition 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 tothe field point, and r − r ′ is the distance between these twopoints.4
  5. 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. 6. 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
  7. 7. 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
  8. 8. 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 mostI encl = ∫ J ⋅ d s general, we can write Iencl in this way. S
  9. 9. 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
  10. 10. 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
  11. 11. 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
  12. 12. 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
  13. 13. 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
  14. 14. Magnetic Flux Density of an InfiniteLine Current Using Ampere’s LawConsider an infinite line current along the z- axis carrying current in the +z-direction: I 14
  15. 15. 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
  16. 16. Magnetic Flux Density of an Infinite Line Current Using Ampere’s Law (Cont’d) y Amperian path ρ x I I encl = I 16
  17. 17. 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
  18. 18. 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
  19. 19. Magnetic Flux DensityThe magnetic flux density vector is related to the magnetic Hfield intensity by the following equation B = µ H, T (tesla) or Wb/m2 µwhere is the permeability of the medium. Except forferromagnetic materials ( such as cobalt, nickel, and iron), µmost materials have values of very nearly equal to that forvacuum, 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. 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. 21. ∇•B = 0∇ × B = µ0 J ∇• J = 0
  22. 22. There are no magnetic flow sources, and themagnetic flux lines always close uponthemselves
  23. 23. Forces due to magnetic fieldThere are three ways in which force due to magnetic fields canbe experienced1.Due to moving charge particle in a B field2.On a current element in an external field3.Between two current elements 23
  24. 24. 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 .
  25. 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 thedirection of the magnetic field (sin0=0)
  26. 26. Magnetic Force on a current elementOne can tell if a magneto-static field is present because itexerts forces on moving charges and on currents. If a currentIdl Belement is located in a magnetic field , it experiences a __force Id l ⇔ Q u d= l B FI × d IdlThis force, acting on current element , is in a directionperpendicular to the current element and also perpendicular B Bto . It is largest when and the wire are perpendicular.26
  27. 27. 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
  28. 28. 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
  29. 29. 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
  30. 30. 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
  31. 31. 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
  32. 32. 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
  33. 33. 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