Physics 012 1st Edition Lecture 14 Outline of Last Lecture II. Ampere’s Lawa. Analogous to Gauss’s Law, but for magnetic fields instead of electric fieldsb. ΣBΔlcosθ = μ0Ienc.i. Δl = length around the surface that you are measuringii. μ0 = permeability of free space = 4π x 10-7 (T*m)/Ac. Right hand rule for magnetic field about a current-carrying wirei. Point thumb in direction of currentii. Fingers will curl pointing toward direction of magnetic field d. B is weaker with greater r (magnetic field strength decreases with greater distance from source)i. Around a circle about a long, straight current-carrying wire:1. ΣBΔlcosθ = μ0Ienc.2. BΣΔl = μ0Ienc. (B always perpendicular to r so cosθ = 1; B is constant so ΣB = B)3. B(2πR) = μ0Ienc. (total length around circle [Δl] = 2πR)4. B = (μ0Ienc.)/2πRe. For a circular, solenoid coil, B = (μ0Ienc.N)/L or μ0nIenc. (where n = N/L)These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.i. N = number of wraps around the coilii. L = length of coiliii. B will point straight through coil at center of concentric circlesf. At the center of a circular current, B = (μ0I)/2r.g. Two parallel current-carrying wires with current travelling in the same direction will attract each other with equal force (even if the two currents are not equal in magnitude).i. F21 = I2LBsinθ = I2L[(μ0I1)/2r]ii. F12 = I1LBsinθ = I1L[(μ0I2)/2r]iii. F21 = F12 (action-reaction pair)h. Problem: Find an expression for Bnet at the center of a circular loop with radius a in a wire carrying current I.i. Bnet = Iwire + Iloop = [(μ0I)/2πa] + [(μ0I)/2a]Outline of Current Lecture III. Induced Currenta. When a conductive object (metal bar, wire, etc.) moves through a magnetic field B, a current is induced in the direction of the magnetic force FB.b. Metal bar through a magnetic fieldc. Bar magnet through a coiled conductive wired. Closed loop inside a circuitIV. Magnetic Flux (ΦB)a. ΦB = ΣBΔAcosθ in T/m2 (weber, Wb)b. Change in magnetic flux causes an induced currenti. Can be due to change in magnetic field strength (B), area (A), or angle (θ)V. Faraday’s Lawa. εind = -N(ΔΦB/Δt)b. Problem: Find an expression for Fapp so bar moves at a constant velocity:i. Fapp = FBii. Iind = |εind|/Riii. Φ = ΣBΔAcosθ = BΣΔA = BLy (y is changing)iv. εind = -N(ΔΦB/Δt) = -(ΔBLy)/Δt = -BL (Δy/Δt) = -BLvv. Iind = (-BLv)/Rvi. FB = [(BLv)/R] LBsinθ =
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