L10 - Chapter 29 Sources of Magnetic Fields • Danish...

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Unformatted text preview: Chapter 29 Sources of Magnetic Fields • Danish Physicist: Hans Christian Oersted (1777-1851) discovers a current carrying wire deflects a compass (magnet) in 1819 • English Physicist: Michael Faraday (1791-1867) changing currents can induce a voltage in a nearby circuit • American Physicst: Joseph Henry (1797-1878) shows that the motion of a magnet produces same effect… • Biot and Savart investigate phenomena and discover – – B obeys an inverse square law B is perpendicular to direction of current and displacement from current (wire) – B is proportional to the sine of the angle between current and displacement The B-field near a long straight wire with current i curls around the wire Dipole Field Dipole Magnet or current loop can produce the same magnetic field B-field curls around the current loop! Implies a second right hand rule Moving charge create B-Fields r r ˆ qv × r B = k′ 2 r q r v Moving charges are the source of magnetic fields Write moving charge in terms of current Δ q = iΔ t =i Then, r r ˆ i ds × r dB = k ′ r2 L v Biot - Savart Law r r ˆ ids × r dB = k ′ 2 r μ0 k = 4π ' μ o ≡ 4π ×10 −7 T ⋅m A r r ˆ i ds × r B = k′ 2 r μo is the permeability of free space (vacuum) k ′ is the magnetic constant ∫ r We add up (integrate) d B along r the wire to determine B( p ) Biot-Savart Law applied to important geometries • Straight Wire (perpendicular bisector) • Circular coil Straight Wire r r μ o ids × r ˆ dB = 2 4π r r ˆ ˆ ds × r = dx sinθ n The magnetic field points into the page r μo i ˆ B= n 2π R ˆ n is a unit vector pointing into the page B for a wire with current i right hand rule defines direction of the magnetic field r μ o iΔϕ ˆ B= n 4π R The magnetic field points out of the page at the point c Arc of a circular wire r μ o ids sin 90 ° ˆ nout dB = 2 4π R r r B = dB ∫ 0 ϕ r μ i ˆ B(c) = o nin 8 R r μ o iΔϕ ˆ B= n 4π R ˆ n is a unit vector point " out of the page" Circular coil Examining the symmetry shows will contribute nothing to the final field, so dB⊥ μ o i ds cosα dB = dB|| = 4π (z 2 + R 2 ) cosα = R z2 + R2 r r μ o i ds × r ˆ dB = 4π r 2 μo dB|| = 4π i Rds +R μ o i R ( Rdϕ ) = 3 4π 2 z + R2 2 2 (z ( 3 2 2 ) ) Circular coil μo B|| = 4π iR 2 2π (z μo 2 2 + R2 ) 3 2 0 ∫ dφ B|| ( p ) = iR 2 (z (z 2 + R2 iR 2 ) 3 2 or r μo B( p ) = 2 2 + R2 ) 3 2 ˆ k There is only a zcomponent!!! Second Right Hand Rule Put thumb in direction of the current and curl fingers around the wire. Your fingers give the direction of the Bfield lines Circular Current Loop Force between parallel conductors μ o ia Ba (b ) = 2πd r r Fba = ib L × Ba • First consider the magnetic field generated by wire a current ia in the vicinity of wire b • Then determine the force this field exerts on wire b Fba = ib LB b sin (90 ) Fba μ o ia ib = L 2π d What is the magnitude and direction of Fab? Circular coil as a magnetic dipole r μo B= 2π r μo B= 2π r μo B= 2π NIπ R 2 x2 + R ( 3 2 2 ) ˆ x ˆ NIAx x2 + R ( 3 2 2 ) ) μ r 3 2 2 r μo B= 2 NIR 2 3 2 2 x2 + R ( ) ˆ x r μo μ if x>>R B≈ 3 2π x x2 + R ( r Example Problem 3 • Two electric transmission lines each carry 100 A in opposite directions. If they are separated by 0.50 m, what is the force/ length between them? Is the force one of attraction or repulsion? ...
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