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Reference Static Magnetic Fields (2 Weeks) 5.2 Fundamental postulates of magnetostatics in free space 5.4 Biot-Savart Law and Applications 5.10 Inductances and Inductors Intro.1 V. STATIC MAGNETIC FIELD 5.1 Law of conservation of magnetic flux ∇ .B = 0 Point form ∫ B. d S = 0 Integral form B is called magnetic flux density (unit: tesla T) . The law states that the net magnetic flux leaving a closed surface is equal to zero (compare with the case in electric field). Intro.2 5.2 Biot-Savart Law • Source of magnetic field: current used for calculating the magnetic field intensity dH at a point P due to a current I flowing in an element dl. dH = Id l × r 4π r 3 I dl r P The resultant magnetic field intensity H at P is found by integrating dH for all current elements. (Unit of H is ampere/metre) Intro.3 Relation between B and H: B = µ o µ r H = µH µo is permeability of free space (vacuum), µ r is the relative permeability of the medium material, and µ is the permeability of the medium material. (Unit of µ o and µ are Henry/metre) In free space, µr = 1 B = µoH Intro.4 5.3 Ampere’s Law (Ampere’s circuital law) • an equivalent form of Biot-Savart Law • Integral form: The line integral of H around any closed path C is equal to the current Ienc enclosed by the path (in a right-handed sense). Ienc ∫ H. dl = I C enc dl • Differential form of Ampere’s law: ∇×H = J Intro.5 The above form is very convenient in calculating the magnetic field around cylindrically symmetric current distributions. Example: Show that the field at a point distance r from the centre of a solid cylindrical conductor of radius a carrying a current Io is: rI o aφ 2 2π a Io H = aφ 2π r H = r<a r>a Intro.6 z path of radius r y x Io Solution: (a) For a path of radius r >a, applying Ampere’s law, 2π r H = I o Io H = aφ 2π r Intro.7 (b) For a path of radius r < a, the current enclosed 2 r is equal to I 2 o a (for uniform current density) Applying the Ampere’s law, r 2Io 2π r H = a2 Ior H = aφ 2 2π a Hφ b r Intro.8 Example Example 5-2 – Determine the magnetic flux density inside a closely wound toroidal coil with an air core having N turns of coil and carrying a current I. The toroid has a mean radius of b, and the radius of each turn is a I I Intro.9 Summary Static Electric field Static Magnetic field Source Charges Current Divergence equation (Gauss’s law) ∇ ⋅ D = ρv ∫ D ⋅ ds = Q s Curl equation ∇⋅B = 0 ∫ B ⋅ ds = 0 s ∇×E = 0 ∇×H = J ∫ E ⋅ dl = 0 ∫ H ⋅ dl = I c c Intro.10 5.4 Inductance Lij= flux linking Φ ij due to current in Si current in Si L11 is called the self-inductance; L12 is called the mutualinductance Φ12 = ∫ B1 ⋅ ds 2 s2 S2 S1 I Φ12 L12 = I1 Intro.11 Example: The coaxial cable shown consists of inner and outer conductors of radii b and a separated by a nonmagnetic medium of permeability εo. Find the inductance per unit length of the cable. Solution: Let the inner conductor carries a current I. Apply Ampere’s law to a closed path of radius r where the magnetic field magnitude is B(r). r B(r) a 2π r B (r ) =I µo µo I B (r ) = 2π r Intro.12 The total magnetic flux between the inner and outer conductor (per unit length) is: a φ = ∫ B ( r ) dr b a = ∫ b µ o Idr µoI a ln = 2π r 2π b The self-inductance L per unit length is: µo a L= = ln I 2π b φ Intro.13 Example Example 5-8 I I Intro.14