Phys 012 1st Edition Lecture 13Outline of Last Lecture II. Velocity Selectora. The electric field between a parallel plate capacitor is pointing downward with a force equal to qE.b. The magnetic field coming toward observer with a force equal to qvB (pointing upward).c. The particle will move straight through if the forces are equal.i. qvB = qEii. v = E/B (faster = curves upward, slower = curves downward)III. Problem: A particle accelerates from rest through a potential difference and then enters a magnetic field. Find an expression for the mass of the particle.a. ΔEPE = ΔKEb. qΔV = ½ mv2c. v = [(2q|ΔV|)/m]½d. ΣF = qvB = mace. qB = m/r [(2q|ΔV|)/m]½f. m = (r2B2q)/(2ΔV)IV. Current through a straight conducting wire in a uniform magnetic fielda. Electrons moving downward, feel force to the lefThese 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.b. F = ILBsinθV. Torque on a rectangular electrical circuita. F1 = -F3 = 0b. F2 = -F4 Side view of circuitc. τ = rFsinθd. τnet= τ2 + τ4 = (b/2)(IaB)(sin(90°)) + (b/2)(IaB)(sin(90°)) = abIB = AIBi. A = area of circuit = abe. Generally, τnet = IABsinθf. With multiple loops, τnet = NIABsinθi. N = number of loopsOutline of Current Lecture VI. 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)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] +
View Full Document