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Ampere’s Law Andre Ampere AP Physics C Mrs. Coyle Remember: Biot-Savart Law: Field produced by current carrying wires ds rˆ dB 0 I 2 4 r – Distance a from long straight wire – Centre of a wire loop radius R – Centre of a Solenoid with N turns 0 I B 2a B B 0 I 2R 0 NI 2R Remember: There are two ways to find the electric field around a charged object. • Coulomb’s Law (Superposition) • kdq E 2 r Gauss’s Law • This is used for high symmetry cases. surface EdA cos qin 0 There are two ways to calculate magnetic field. • Biot-Savart Law • Ampere’s Law 0 ds rˆ dB I 2 4 r B ds μ o I – Used for high symmetry cases. Ampere’s Law • For small length elements ds on a closed path (not necessarily circular) B ds μ o I • I is enclosed current passing through any surface bounded by the closed path. • Note: dot product • Use where there is high symmetry B ds I Sign Convention for the Current in Ampere’s Law I I negative I positive I B B ds r ds r The current I passing through a loop is positive if the direction of B from the right hand rule is the same as the direction of the integration (ds). Field Outside a Long Straight Wire at a distance r from the center, r > R • The current is uniformly distributed through the cross section of the wire B ds B( 2πr ) μ o μo I B 2πr I Field Inside a Long Straight Wire at a distance r from the center, r< R • Inside the wire, the current considered is inside the amperian circle 2 π r A enclosed by r r2 B ds = μ 0 Iencl = μ 0 I A enclosed by R = μ 0 I πR 2 = μ 0 I R 2 r2 B 2 r μ 0 I 2 B μ I 0 R r Note the linear 2 R 2 relationship of B with r Field Due to a Long Straight Wire • 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 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 B ds B( 2πr ) μ N I o μo N I B 2πr Magnetic Field of an Thin Infinite Sheet • 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 • Js =I/l is the linear current density along the z direction • The current is in the y direction B ds μ o Js B μo 2 I μo Js Magnetic Field of a Solenoid • The field lines in the interior are – approximately parallel to each other – uniformly distributed – close together • The field is strong and almost uniform in the interior 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 Ideal Solenoid – The turns are closely spaced – The length is much greater than the radius of the turns Magnetic Field Inside a Long Solenoid B ds = B ds B ds B ds B ds 1 B ds = 2 B 3 0 4 0 -The total current through the rectangular path equals the current through each turn multiplied by the number of turns B = μ0 N I B μo N I μo n I 0 = μ 0 Ienclosed Note • The magnetic field inside a long solenoid does not depend on the position inside the solenoid (if end effects are neglected). Magnetic Field – At a distance a from long straight wire B 0 I 2a – At the centre of a wire loop radius R B – At the centre of a solenoid with N turns -In the interior of a toroid 0 I 2R B 0 NI B 2 R 0 NI 2R