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UVM PHYS 012 - Magnetic Fields and Forces
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Physics 012 1st Edition Lecture 11Outline of Last Lecture II. Kirchhoff’s Rulesa. Junction rule: the total current going into a junction equals the total current out of the junctioni. ΣIin = ΣIoutii. Derived from the law of conservation of chargeb. Loop rule: the sum of changes in potential around a closed circuit loop is zeroi. ΣΔVclosed loop = 0c. Problem: Find the values for current through the closed circuit below given: ε1 = 10V, ε2 = 14V, R1 = 6Ω, R2 = 4Ω, R3 = 2Ω.i. First, draw currents through this circuit. Direction does not matter at this point in the problem. When you get values for the current, if a value is negative then current flows in the opposite direction as predicted. The magnitude will still be the same.1. Since ΣIin = ΣIout, for junction A, I1 + I2 = I3.ii. Second, mark each side of every element in the circuit with higher (+) or lower (-) potential. Remember: current flows from high potential to low potential.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.iii. Third, add up the potentials around each part of the circuit (top and bottom). The potentials should add up to zero (loop rule). The magnitude of the change in potential will be negative if you’re going from lower (-) tohigher (+) potential.1. Following top loop clockwise: +I1R1 – ε1 – I2R2 – ε2 = 02. Following bottom loop clockwise: -I3R3 + ε1 – I1R1 = 0iv. Finally, solve for I1, I2, and I3.1. Substituting the values for R1, R2, ε1, and ε2: +6I1 – 10 – 4I2 – 14 = 6I1 - 4I2 – 24 = 02. Substituting the values for R1, R3, and ε1: -2I1 – 2I2 + 10 – 6I1 = -8I1 –2I2 + 10 = (- 1/2)(16I1 + 4I2 - 20) = 03. Adding the two equations together: 22I1 – 44 = 04. Solve for I1: I1 = 44/22 = 2.0 A5. Solve for I2: 6I1 - 4I2 – 24 = 0; 6(2) - 4I2 – 24 = 0; I2 = 12/(-4) = -3.0 A6. Solve for I3: I1 + I2 = I3 ; I3 = (2.0) + (-3.0) = -1.0 AIII. RC Circuits:a. Chargingi. ΔV = Q/Cii. When battery is first connected to the circuit, voltage across the capacitorincreases from zero while voltage across the resistor decreases until the voltage across the resistor is zero and current stops.iii. Time constant (τ) = RC1. When t = τ, I is reduced by a factor of 1/e.2. After a few time constants, I is assumed to be zero, even though it never actually reaches zero.b. Dischargingi. When the battery is removed from the circuit, electric potential energy is stored in the capacitor, which then produces a current in the opposite direction.ii. Electric potential energy is gradually lost through the resistors in the formof thermal energy.c. Problem: Find I at t = 0 (right after closing switch) and as t approaches ∞ (after switch has been closed for a long time) in the circuit below.i. When t = 0, Iε= ε/R1.ii. When t  ∞, Iε = ε/Req.Outline of Current Lecture IV. Magnetic Fieldsa. Magnets have two poles: north and south.b. Like poles repel, opposites attract.c. No magnetic monopoles have been found.d. Magnetic fields exist near moving charge.e. Field lines come out of north and go to south poles.f. Magnetic field at a point is tangent to field line (B).V. Earth’s Magnetic Fielda. Geographic north is magnetic south (but this switches every couple ten thousandyears).b. Compass will point in direction of magnetic field lines (ie. geographic north).c. Magnetic north is a few degrees off of “true north.”VI. Magnetic Forcea. Electric charges moving in a magnetic field can feel a force.i. F = qvBsinθ1. v = velocity2. B = magnetic field strength in teslas (T) = (N*s)/(C*m)3. θ = angle between v and B if they are tail to tailb. Force is always perpendicular to v and B.c. Movement of a particle in magnetic field will be circular.i. ΣF = qvBsinθ = macii. sin (90°) = 1iii. ac= v2/riv. qvB = macv. qvB = mv2/rvi. r = mv/qBd. Right-hand Rule:i. 1. Hold right hand so fingers point in the direction of v (opposite directionif charge is negative)ii. 2. Bend fingers in direction of B.iii. 3. Thumb will be pointing in direction of F.VII. Magnetic Force versus Electric Forcea. FE acts along direction of E fieldb. FB acts perpendicular to B fieldc. FE acts on stationary or moving chargesd. FB acts on moving charges onlye. FE does work in displacing a charged particlef. FB from steady doesn’t do worki. W = Fscosθii. F always perpendicular to viii. cos (90°) = 0iv. W = Fs (0) =


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UVM PHYS 012 - Magnetic Fields and Forces

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