Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Electrical resistance and conductance wikipedia , lookup
Magnetic monopole wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Magnetic field wikipedia , lookup
Electrostatics wikipedia , lookup
Maxwell's equations wikipedia , lookup
Electromagnetism wikipedia , lookup
Field (physics) wikipedia , lookup
Superconductivity wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
The Biot-Savart Law and Ampere’s Law The Biot-Savart Law r µ0 Ids × rˆ µ0 Ids × r µ0 ( Ids) sin θ dB = = = 2 3 2 4π r 4π r 4π r – µ0 is the “permeability of free space” µ0 = 4π ×10 −7 T ⋅ m/A • Always have to integrate along whole wire (current) to get actual field µ B= 0 4π Ids × rˆ ∫ r2 Field of a Loop of Current • What’s at the center of a loop of current? µ 0 Ids × rˆ B= 4π ∫ r 2 µ0 I B= ds 2 ∫ 4π r µ0 I = 2πr 2 4π r µ0 I = 2r I ds r ds B Line Integral of Field about a Loop • Switch it around • What’s the line integral of the field going around a line current? B ds µ0 ˆ ∫ B ⋅ ds = ∫ 2πr I × r ⋅ ds µ0 (I × rˆ ) ⋅ ds = ∫ 2πr µ0 I = ds ∫ 2πr µ0 I = 2πr 2πr = µ0 I r ds I Ampere’s Law • This is true actually for any loop – Radial segments don’t contribute – Tangential segments contribute by angle • Could add another current – Superposition works for magnetic field too! • General result is Ampere’s Law: B ⋅ d s = µ I 0 enclosed ∫loop – Use it like Gauss’s Law in electrostatics Example: An extended wire • Apply Ampere’s Law to a real (finite diameter) wire – Outside the wire (r > R) ∫ B ⋅ ds = 2πrB = µ I 0 B= µ0 I 2πr – Inside wire (r < R) 2 r ∫ B ⋅ ds = 2πrB = µ0 R I µI r B= 0 2π R 2 Force Between Two Parallel Wires • Can now calculate the force between two long parallel wires – Magnetic force on one wire (1) due to field of the other (2) F1 = lI1 × B 2 F1 = lI1 B2 – Field of other wire (2) µ B 2 = 0 I 2 × aˆ 2πa µI B2 = 0 2 2πa Force Between Two Parallel Wires • Can now calculate the force between two long parallel wires – Putting it together µI F1 = lI1 B2 ; B2 = 0 2 2πa µII F12 = 0 1 2 l 2πa – Attractive for currents in the same direction – This is used to define the amp as a unit (and the coulomb) Example: Field Inside a Solenoid • A solenoid is a wire wrapped in a helix Example: Field Inside a Solenoid • A solenoid is a wire wrapped in a helix • Ideally, the coil spacing is tight… Example: Field Inside a Solenoid • A solenoid is a wire wrapped in a helix • Ideally, the coil spacing is tight… • …and it goes on forever • Field outside is small – Using Ampere’s Law: no net current thru loop Example: Field Inside a Solenoid • Inside: ∫ B ⋅ ds = ∑ ∫ B ⋅ ds π d B ds = 0 B ⋅ s = cos ∫ ∫ ∫ B ⋅ ds = ∫ 0 ds = 0 ∫ B ⋅ ds = Bl ∫ B ⋅ ds = Bl = µ NI i i 2 2, 4 3 1 0 N B = µ 0 I = µ 0 nI l N turns inside loop n turns per unit length Magnetic Flux • Just as with the electric field, we define the flux of the magnetic field thru a surface S Φ B = ∫ B ⋅ dA S – dA is a vector perpendicular to the surface – Units of flux: Weber (Wb) 1 Wb ≡ 1 T ⋅ m 2 Gauss’s Law in Magnetism • Since all magnetic field lines form loops (no magnetic charges), net flux thru any closed surface is zero! ∫ B ⋅ dA = 0 S Review of Magnetostatics • If fields are constant (in time), we have – Gauss’s Law ∫ E ⋅ dA = S – Gauss’s Law Magnetism Qenc ε0 ∫ B ⋅ dA = 0 ∫ E ⋅ ds = 0 ∫ B ⋅ ds = µ I S – Conservative Electric Force loop – Ampere’s Law loop 0 enclosed Displacement Current • Consider the magnetic field around a capacitor being charged ∫ B ⋅ ds = µ I ∫ B ⋅ ds = 0 S1 S2 0 – But S1 and S2 have the same boundary! Displacement Current • Consider the magnetic field around a capacitor being charged ∫ B ⋅ ds = µ I ∫ B ⋅ ds = 0 S1 S2 0 – But S1 and S2 have the same boundary! – Have to add something to make these consistent Displacement Current • If there is a current flowing, the field (and electric flux) in the capacitor is changing! Φ E = ∫ E ⋅ dA S2 q =∫ dAC + 0 Cε A 0 C = q ε0 dΦ E 1 dq I = = ε 0 dt ε 0 dt Define the Displacement Current I displacement dΦ E ≡ ε0 dt Displacement Current • If there is a current flowing, the field (and electric flux) in the capacitor is changing! • Use displacement current to make the two integrals consistent ∫ B ⋅ ds = µ I S1 0 dΦ E ∫SB2 ⋅ ds = µ0ε 0 dt = µ0 I displacement Define the Displacement Current I displacement dΦ E ≡ ε0 dt General Form of Ampere’s Law • In general Ampere’s Law (Ampere-Maxwell’s Law) should be ( ) µ B s ⋅ d = I + I 0 enclosed disp ∫loop dΦ E B s µ µ ε ⋅ d = I + 0 enclosed 0 0 ∫loop dt